10.2 Polynomial and Radical Expressions

Polynomials

A term is a constant or the product of a constant and one or more variables.  A monomial is an algebraic expression with one term.  When it is of the form $ax^n$, where $a$ is a constant, $n$ is a whole number, and $x$ is a variable, it is called a monomial in one variable.

A monomial, or two or more monomials combined by addition or subtraction, is a polynomial.  Some polynomials have special names, based on the number of terms.  A monomial is a polynomial with exactly one term.  A binomial has exactly two terms, and a trinomial has exactly three terms.  There are no special names for polynomials with more than three terms.

Definition: POLYNOMIALS

• polynomial—A monomial, or two or more algebraic terms combined by addition or subtraction
• monomial—A polynomial with exactly one term
• binomial—A polynomial with exactly two terms
• trinomial—A polynomial with exactly three terms

Every monomial, binomial, and trinomial is also a polynomial.  They are just special members of the “family” of polynomials and so they have special names. 

The degree of a polynomial and the degree of each of its terms are determined by the exponents of the variable.  A monomial that is just a constant is a special case.  The degree of a constant is 0.

Definition: DEGREE OF A POLYNOMIAL

• The degree of a term is the sum of the exponents of its variables.
• The degree of a constant is 0.
• The degree of a polynomial is the highest degree of all its terms.

Working with polynomials is easier when you list the terms in descending order of degrees.  When a polynomial is written this way, it is said to be in standard form.  Get in the habit of writing the term with the highest degree first.

Example $\PageIndex{1}$

Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial.  Then find the degree of each polynomial.

1. $7y^2−5y+3$
2. $−2a^4b^2$
3. $3x^5−4x^3−6x^2+x−8$
4. $2y−8xy^3$
5. $15$

Solution

	Polynomial	Number of terms	Type	Degree of terms	Degree of polynomial
a.	$7y^2−5y+3$	3	Trinomial	2, 1, 0	2
b.	$−2a^4b^2$	1	Monomial	4 + 2	6
c.	$3x^5−4x^3−6x^2+x−8$	5	Polynomial	5, 3, 2, 1, 0	5
d.	$2y−8xy^3$	2	Binomial	1, 4	4
e.	$15$	1	Monomial	0	0

Adding and Subtracting Polynomials

We can think of adding and subtracting polynomials as just combining and simplifying two or more polynomials by combining like terms. Remember, like terms must have the same variables with the same exponent.  Look for the like terms—those with the same variables and the same exponent. The Commutative Property of Addition allows us to rearrange the terms to put like terms together.

Example $\PageIndex{2}$

Find the sum: $(7y^2−2y+9)\;+\;(4y^2−8y−7)$.

Solution

$\begin{align*} &\text{Identify like terms.} & & (\underline{\underline{7y^2}}−\underline{2y}+9)+(\underline{\underline{4y^2}}−\underline{8y}−7) \\[6pt] &\text{Rewrite without the parentheses,} \\ &\text{rearranging to get the like terms together.} & & \underline{\underline{7y^2+4y^2}}−\underline{2y−8y}+9−7\\[6pt] &\text{Combine like terms.} & & 11y^2−10y+2 \end{align*}$

$\PageIndex{1}$

Find the sum: $(7x^2−4x+5)\;+\;(x^2−7x+3)$

Answer

$8x^2−11x+8$

Be careful with the signs as you distribute while subtracting the polynomials in the next example.

Example $\PageIndex{3}$

Find the difference: $(9w^2−7w+5)\;−\;(2w^2−4)$

Solution

$\begin{align*} & & & (9w^2−7w+5)\;−\;(2w^2−4) \\[6pt] &\text{Distribute and identify like terms.} & & \underline{\underline{9w^2}}−\underline{7w}+5-\underline{\underline{2w^2}}+4 \\[6pt] &\text{Rearrange the terms.} & & \underline{\underline{9w^2-2w^2}}−\underline{7w}+5+4\\[6pt] &\text{Combine like terms.} & & 7w^2−7w+9 \end{align*}$

$\PageIndex{2}$

Find the difference: $(8x^2+3x−19)\;−\;(7x^2−14)$

Answer

$x^2+3x−5$

When we add and subtract more than two polynomials, the process is the same.  When more than one variable is present, make sure you are combining like terms!

Example $\PageIndex{4}$

Simplify: $(a^3−a^2b)\;−\;(ab^2+b^3)\;+\;(a^2b+ab^2)$

Solution

$\begin{align*} & & & (a^3−a^2b)\;−\;(ab^2+b^3)\;+\;(a^2b+ab^2) \\[6pt] &\text{Distribute} & & a^3−a^2b − ab^2 - b^3 + a^2b+ab^2\\[6pt] &\text{Rearrange the terms to put like terms together.} & & a^3−a^2b + a^2b− ab^2 + ab^2 - b^3 \\[6pt] &\text{Combine like terms.} & & a^3−b^3 \end{align*}$

$\PageIndex{3}$

Simplify: $(x^3−x^2y)\;−\;(xy^2 - y^3)\;+\;(x^2y+xy^2)$

Answer

$x^3+y^3$

Multiplying Polynomials

Multiplying a monomial by a monomial

We can use the properties of exponents to multiply monomials.

Example $\PageIndex{5}$

Multiply:

1. $(3x^2)(−4x^3)$
2. $\left(\frac{5}{6}x^3y\right)(12xy^2).$

Solution

$\begin{array} {ll} {\bf{a.}} &{(3x^2)(−4x^3)} \\ {\text{Use the Commutative Property of Multiplication to rearrange the factors.}} &{3·(−4)·x^2·x^3} \\ {\text{Multiply.}} &{−12x^5} \\ \end{array}$

 $\begin{array} {ll} {\bf{b.}} &{\left(\frac{5}{6}x^3y\right)(12xy^2)} \\ {\text{Use the Commutative Property of Multiplication to rearrange the factors.}} &{\frac{5}{6}·12·x^3·x·y·y^2} \\ {\text{Multiply.}} &{10x^4y^3} \\ \end{array}$

$\PageIndex{4}$

Multiply:

1. $(5y^7)(−7y^4)$
2. $(25a^4b^3)(15ab^3)$

Answer a

$−35y^{11}$

Answer b

$6a^5b^6$

Multiplying a polynomial by a monomial

Multiplying a polynomial by a monomial is really just applying the Distributive Property of Multiplication.

Example $\PageIndex{6}$

Multiply:

1. $−2y(4y^2+3y−5)$
2. $3x^3y(x^2−8xy+y^2)$.

Solution a

Distribute.
Multiply.

Solution b

$\begin{array} {ll} {} &{3x^3y(x^2−8xy+y^2)} \\ {\text{Distribute.}} &{3x^3y⋅x^2+(3x^3y)⋅(−8xy)+(3x^3y)⋅y^2} \\ {\text{Multiply.}} &{3x^5y−24x^4y^2+3x^3y^3} \\ \end{array}$

$\PageIndex{5}$

Multiply:

1. $-3y(5y^2+8y^{−7})$
2. $4x^2y^2(3x^2−5xy+3y^2)$

Answer a

$−15y^3−24y^2+21y$

Answer b

$12x^4y^2−20x^3y^3+12x^2y^4$

Multiplying a binomial times a binomial

To multiply a binomial by a binomial, we use the Distributive Property of Multiplication twice, and then combine like terms.

Example $\PageIndex{7}$

Multiply:

1. $(y+5)(y+8)$
2. $(4y+3)(2y−5)$.

Solution

ⓐ

Distribute $(y+8)$.
Distribute again.
Combine like terms.

ⓑ

Distribute.
Distribute again.
Combine like terms.

$\PageIndex{6}$

Multiply:

1. $(x+8)(x+9)$
2. $(3c+4)(5c−2)$

Answer a

$x^2+17x+72$

Answer b

$15c^2+14c−8$

Patterns

FOIL

If you multiply binomials often enough you may notice a pattern.  Notice that the first term in the result is the product of the first terms in each binomial.  The second and third terms are the product of multiplying the two outer terms and then the two inner terms.  The last term results from multiplying the two last terms in each binomial.

We abbreviate “First, Outer, Inner, Last” as FOIL.  The word FOIL is easy to remember and ensures we find all four products.  If you are comfortable with the FOIL method, you save yourself one step, as you can see in the following comparison.

Multiply $(x+3)(x+7)$.

NOTE: The FOIL method only applies to multiplying binomials, not other polynomials!

The square of a binomial

When a binomial is squared (multiplied by itself), the result always follows a very particular pattern; namely, the square of the first term added to twice the product of the two terms plus the square of the last term.  We refer to the resulting trinomial as a perfect square trinomial.  If you know the pattern, you can save yourself one step.

$\begin{aligned} (a+b)^2 &= (a+b)(a+b) = a^2+ab +ab +b^2= a^2+2ab+b^2 \qquad \text{ and}\\[2pt] (a-b)^2 &= (a-b)(a-b) = a^2-ab -ab +(-b)^2= a^2-2ab+b^2  \end{aligned}$

For example,

$\begin{aligned} (2x+7)^2 &= (2x)^2+2(2x)(7)+7^2 =  4x^2+28x+49 \qquad \text{ and}\\[4pt] (5y-4)^2 &= (5y)^2 - 2(5y)(4) +4^2 =  25y^2-40y+16. \end{aligned}$

Multiplying a binomial by its conjugate

Given a binomial $a+b$, we refer to the related binomial $a-b$ as its conjugate.  Likewise, the conjugate of $a-b$ is $a+b$.  When a binomial is multiplied by its conjugate, we get another pattern, a binomial which is referred to as the difference of squares.  Notice that the middle term always adds to zero and drops out; if you know this pattern, you can save yourself one step.

$$(a+b)(a-b) =  a^2+ab -ab -b^2= a^2-b^2 \nonumber$$

For example,

$$(2x+7)(2x-7) =  4x^2-49. \nonumber$$

Factoring Polynomials

Recall the definition of the greatest common factor (GCF) of two or more expressions, and the "official" way, shown below, to find the GCF.   (If you have had a lot of practice with finding a GCF, you may save the "official" rubric for the most difficult cases.)

GREATEST COMMON FACTOR

The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of each of the expressions.

FIND THE GREATEST COMMON FACTOR (GCF) OF TWO EXPRESSIONS.

1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
2. List all factors—matching common factors in a column. In each column, circle the common factors.
3. Bring down the common factors that all expressions share.
4. Multiply the factors.

The next example will show us the steps to find the greatest common factor of three expressions.

Example $\PageIndex{8}$

Find the greatest common factor of $21x^3,\space 9x^2,\space 15x$.

Solution

Factor each coefficient into primes and write the variables with exponents in expanded form. Circle the common factors in each column. Bring down the common factors.
Multiply the factors.
	The GCF of $21x^3$, $9x^2$ and $15x$ is $3x$.

$\PageIndex{7}$

Find the greatest common factor: $25m^4,\space 35m^3,\space 20m^2.$

Answer

$5m^2$

Factor the Greatest Common Factor

It is sometimes useful to represent a number as a product of factors; for example, 12 as $2·6$ or $3·4$.  In algebra, it can also be useful to represent a polynomial in factored form.  We will start with a polynomial written in standard form, such as $3x^2+15x$, and rewrite it in factored form as a product, $3x(x+5)$.  To do this we apply the Distributive Property “in reverse.”

Recall the Distributive Property, which we write here once and then again “in reverse," just switching the sides of the equality.

DISTRIBUTIVE PROPERTY

If a, b, and c are real numbers, then

$$a(b+c)=ab+ac \quad \text{and} \quad ab+ac=a(b+c).\nonumber$$

The form on the left is used to multiply.  The form on the right is used to factor.

So how do you use the Distributive Property to factor a polynomial? You just find the GCF of all the terms and rewrite the polynomial as a product!

FACTOR THE GREATEST COMMON FACTOR FROM A POLYNOMIAL.

1. Find the GCF of all the terms of the polynomial.
2. Rewrite each term as a product using the GCF.
3. Use the “reverse” Distributive Property to factor the expression.
4. Check by multiplying the factors.

FACTOR AS A NOUN AND A VERB

We use “factor” as both a noun and a verb:

$$\begin{array} {ll} \text{Noun:} &\hspace{50mm} 7 \text{ is a factor of }14 \\ \text{Verb:} &\hspace{50mm} \text{factor }3 \text{ from }3a+3\end{array}\nonumber$$

Example $\PageIndex{9}$

Factor: $5x^3−25x^2$.

Solution

Find the GCF of $5x^3$ and $25x^2$.
Rewrite each term.
Factor the GCF.
Check: $$5x^2(x−5) \nonumber$$ $$5x^2·x−5x^2·5 \nonumber$$ $$5x^3−25x^2 \checkmark\nonumber$$

$\PageIndex{8}$

Factor: $2x^3+12x^2$.

Answer

$2x^2(x+6)$

When the leading coefficient is negative, factor the negative out as part of the GCF.

Example $\PageIndex{10}$

Factor: $−4a^3+36a^2−8a$.

Solution

The leading coefficient is negative, so the GCF will be negative.

Rewrite each term using the GCF, $−4a$.
Factor the GCF.
Check: $$−4a(a^2−9a+2)\nonumber$$ $$−4a·a^2−(−4a)·9a+(−4a)·2\nonumber$$ $$−4a^3+36a^2−8a\,\checkmark\nonumber$$

$\PageIndex{9}$

Factor: $−4b^3+16b^2−8b$.

Answer

$−4b(b^2−4b+2)$

$\PageIndex{10}$

Factor: $−7a^3+28a^2−14a$.

Answer

$−7a(a^2−4a+2)$

So far our greatest common factors have been monomials.  In the next example, the greatest common factor is a binomial.

Example $\PageIndex{11}$

Factor: $3y(y+7)−4(y+7)$.

Solution

The GCF is the binomial $y+7$.

Factor the GCF, $(y+7)$.
Check on your own by multiplying.

$\PageIndex{11}$

Factor: $4m(m+3)−7(m+3)$.

Answer

$(m+3)(4m−7)$

$\PageIndex{12}$

Factor: $8n(n−4)+5(n−4)$.

Answer

$(n−4)(8n+5)$

Factor by Grouping

Factoring by grouping is a technique that enables us to factor polynomials with four terms into a product of binomials.  This involves an intermediate step where a common binomial factor will be factored out.  For example, 

$$\begin{align*} x^{3}−12x^{2}+2x−8  &= \underbrace{  3x^3−12x^2 }                                     +   \underbrace{2x−8 } && \qquad \text{Begin by grouping the first two terms and the last two terms} \\ &= 3x^2 {\color{Cerulean}{(x-4)}} +  2   {\color{Cerulean}{(x−4)}  } && \qquad \text{Factor out the GCF of each grouping }\\ &   &&  \qquad \text{A common binomial factor appears} \\ &= {\color{Cerulean}{(x-4)}} (3x^2+2) &&  \qquad \text{Factor out the common binomial } \\ \nonumber \end{align*}$$

We can check by multiplying.

$\begin{aligned} ( x - 4 ) \left( 3 x ^ { 2 } + 2 \right) & = 3 x ^ { 3 } + 2 x - 12 x ^ { 2 } - 8  \end{aligned} \:\:\color{Cerulean}{✓}$ 

When factoring any polynomial expression, our first step should always be to check for a GCF.  Look for the GCF of the coefficients, and then look for the GCF of the variables.  Then if there are four terms in the polynomial, try factoring by grouping pairs.

Can every four term polynomial be factored by grouping?

No.  Some four term polynomials can be factored by starting with a different pairing of the four terms; however some four term polynomials simply cannot be factored with grouping by pairs.  Sometimes four term polynomials can be factored by grouping three terms together first.  Some can be factored by determining the possible rational zeros the polynomial might have (see Chapter 3.?).  And then, there are some four term polynomials that simply are not factorable.  These polynomials are said to be prime.

Given a four term polynomial, factor it

1. Determine if there are any factors common to all four terms.  If so, factor the GCF from each of the four terms.
2. Factor the GCF from the first two terms.  Factor out the GCF from the last two terms.  Sometimes the four terms can be rearranged in order for this to be possible to do.
3. If both expressions from step 2 produced the same binomial factor, then factor out this common binomial GCF.
4. Check by multiplying the factors.

Example $\PageIndex{12}$: Factor by Grouping

Factor

1. $24a^4 − 18a^3 − 20a + 15$
2. $xy -2x^2y+x^3-2y^2$

Solution

$$\begin{align*} 1.\quad 24a^4 − 18a^3 − 20a + 15  &= \underbrace{  24a^4 − 18a^3 }_{ {\color{Cerulean}{group}} }                                       \text{ - } \underbrace{20a + 15}_{ {\color{Cerulean}{group}} } && \text{ group} \\ &= 6a^3{\color{Red}{(4a-3)}}+5{\color{Red}{(-4a+3)}} & &\text{Binomial factors are different} \\ &= 6a^3{\color{Cerulean}{(4a-3)}}-5{\color{Cerulean}{(4a-3)}} && \text{The binomial factors are the same} \\ &= {\color{Cerulean}{(4a-3)}} (6a^3-5) && \text{Factor out the common binomial } \\ \nonumber \end{align*}$$

$$\begin{align*} 2.\quad xy -2x^2y+x^3-2y^2  &= \underbrace{  xy -2x^2y }_{ {\color{Cerulean}{group}} }                                       \text{ + } \underbrace{x^3-2y^2 }_{ {\color{Cerulean}{group}} } && \text{ group} \\ &= xy{\color{Red}{(1-2x)}}+1{\color{Red}{(x^3-2y^2)}} & &\text{Binomial factors are different} \\ &= x^3 -2x^2y+xy-2y^2  && \text{change the order of the terms}\\ &= \underbrace{  x^3 -2x^2y }_{ {\color{Cerulean}{group}} }                                       \text{  } \underbrace{+xy-2y^2 }_{ {\color{Cerulean}{group}} } && \text{ group} \\ &= x^2{\color{Cerulean}{(x-2y)}}+y{\color{Cerulean}{(x-2y)}} && \text{The binomial factors are the same} \\ &= {\color{Cerulean}{(x-2y)}} (x^2+y) && \text{Factor out the common binomial } \\ \nonumber \end{align*}$$

Analysis. After factoring, we can check our work by multiplying.  Use the distributive property to confirm that 
$$24a^4 − 18a^3 − 20a + 15=(4a-3)(6a^3-5) \qquad \text{ and } xy -2x^2y+x^3-2y^2=(x-2y)(x^2+y) \nonumber$$

$\PageIndex{13}$

Factor $x^3-x^2y-xy+y^2$    

Answer

$(x-y)(x^2-y)$     

Factoring a Trinomial

There are several techniques we can try when we want to factor a polynomial with three terms.  Choosing the right technique may depend on the degree of the polynomial, the value of the leading coefficient, and personal preference. 

Factor a Trinomial with Leading Coefficient 1  

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored.  The polynomial $x^2+5x+6$ has a GCF of $1$, but it can be written as the product of the factors $(x+2)$ and $(x+3)$.

To factor a trinomial of the form $x^2+bx+c$, we would like to find two numbers whose product is $c$ and whose sum is $b$.  The trinomial $x^2+10x+16$, for example, can be factored using the numbers $2$ and $8$ because the product of those numbers is $16$ and their sum is $10$.  The trinomial can be rewritten as the product of $(x+2)$ and $(x+8)$.  Here, we are using a Reverse FOIL technique.

FACTORING A SECOND DEGREE TRINOMIAL WITH LEADING COEFFICIENT $1$

A trinomial of the form $x^2+bx+c$ can be written in factored form as $(x+p)(x+q)$ where $pq=c$ and $p+q=b$.

 Can every trinomial be factored as a product of binomials?

No.  Some polynomials cannot be factored.  These polynomials are said to be prime.

  Given a trinomial in the form $x^2+bx+c$, factor it

1. List factor pairs of $c$.
2. Find $p$ and $q$, a pair of factors of $c$ with a sum of $b$.
3. Write the factored expression $(x+p)(x+q)$.
4. Check by multiplying.

Example $\PageIndex{13}$: Factoring a Second Degree Trinomial with Leading Coefficient 1

Factor $x^2+2x−15$.

Solution

We have a trinomial with leading coefficient $a=1$, $b=2$, and $c=−15$.  We need to find two numbers with a product of $−15$ and a sum of $2$.  In Table $\PageIndex{1}$, we list factor pairs until we find a pair with the desired sum.

Table $\PageIndex{1}$

Factors of −15	Sum of Factors
1,−15	−14
−1,15	14
3,−5	−2
−3,5	2

Now that we have identified $p$ and $q$ as $−3$ and $5$, write the factored form as $(x−3)(x+5)$.

Analysis. We can check our work by multiplying.  Use FOIL to confirm that $(x−3)(x+5)=x^2+2x−15$.

 Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.  For example, $(x-3)(x+5) = (x+5)(x-3)$.

$\PageIndex{14}$

Factor $x^2−7x+6$.

Answer

$(x−6)(x−1)$

Factor a Second Degree Trinomial with Leading Coefficient Not 1

Trinomials with leading coefficients other than $1$ are somewhat more complicated to factor.

Remember to always check for a GCF first!  Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1, the degree becomes 2, and you can factor it by the methods we’ve used so far.  Let’s do an example to see how this works.

Example $\PageIndex{14}$

Factor completely: $4x^3+16x^2−20x$.

Solution

$\begin{array} {lll} \text{Is there a greatest common factor?} &\qquad &4x^3+16x^2−20x \\ \quad \text{Yes, }GCF=4x.\text{ Factor it.} & &4x(x^2+4x−5) \\ & & \\ & & \\ \text{Binomial, trinomial, or more than three terms?} & & \\ \quad \text{It is a trinomial. So “reverse FOIL.”} & &4x(x\qquad)(x\qquad) \\ & & \\ & & \\ \text{Use a table like the one shown to find two numbers that} & &4x(x−1)(x+5) \\ \text{multiply to }−5\text{ and add to }4. & & \\ & & \\ & & \end{array}$

Factors of $−5$	Sum of factors
$−1,5$ $1,−5$	$−1+5=4 \,\checkmark$ $1+(−5)=−4$

$\begin{array} {l} \text{Check:}\\ \hspace{27mm}4x(x−1)(x+5) \\ \hspace{25mm} 4x(x^2+5x−x−5) \\ \hspace{30mm} 4x(x^2+4x−5) \\ \hspace{25mm} 4x^3+16x2−20x\,\checkmark \end{array}$

$\PageIndex{15}$

Factor completely: $5x^3+15x^2−20x$.

Answer

$5x(x−1)(x+4)$

$\PageIndex{16}$

Factor completely: $6y^3+18y^2−60y$.

Answer

$6y(y−2)(y+5)$

 

What happens when the leading coefficient is not 1 and there is no GCF?  Two different methods for factoring are Reverse FOIL and AC Grouping.  The choice of which method to use is largely a matter of personal preference.  In some cases, one approach may be more efficient than the other, but this is not really predictable beforehand.  The Reverse FOIL Method is illustrated below; the AC grouping method will follow.

Reverse FOIL

Let’s factor the trinomial $3x^2+5x+2$.

From our earlier work, we expect this will factor into two binomials.

$$3x^2+5x+2\nonumber$$ $$(\qquad)(\qquad)\nonumber$$

We know the first terms of the binomial factors will multiply to give us $3x^2$.  The only reasonable factors of $3x^2$ are $1x,\, 3x$.  We can place them in the binomials. Note: Can you see why the factor pair $3, x^2$ could not possibly work, even though their product is $3x^2$?

Check: Does $1x·3x=3x^2$?

We know the last terms of the binomials will multiply to 2.  Since this trinomial has all positive terms, we only need to consider positive factors.  The only factors of 2 are 1, 2.  But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.

Which factors are correct?  To decide that, we multiply the inner and outer terms.

Since the middle term of the trinomial is $5x$, the factors in the first case will work.  Let’s use FOIL to check.

$$(x+1)(3x+2)\nonumber$$ $$3x^2+2x+3x+2\nonumber$$ $$3x^2+5x+2\,\checkmark\nonumber$$

Given a trinomial in the form $ax^2+bx+c$, factor by Reverse FOIL.

Determine if there are any factors common to all three terms.  If so, factor the GCF from each of the three terms.  This GCF should be negative if the leading coefficient is negative.  Start with $ax^2 + bx + c$ where any GCF has been factored out.

1. Find all the positive factor pairs, $S$ and $T$, of the leading coefficient $a$.
   Use these pairs to begin constructing possible templates for the final factored result: $(Sx \qquad)(Tx \qquad)$.
2. Find all the factor pairs, $s$ and $t$, of the coefficient $c$ of the last term.  Revise the template to look like: $(Sx \quad s)(Tx \quad t)$.
   If the middle term $b$ is positive and $c$ is positive, the signs of $s$ and $t$ must both be positive.
   If the middle term $b$ is negative and $c$ is positive, the signs of $s$ and $t$ must both be negative.
   If $c$ is negative, the signs of $s$ and $t$ must be opposites; one must be positive and the other negative.
3. Determine the signs for $s$ and $t$ so that the sum of the inner and outer products is equal to the coefficient of the middle term: $sTx + Stx = bx$.
4. Check by multiplying.

Example $\PageIndex{15}$

Factor completely using Reverse FOIL: $6b^2−13b+5$.

Solution

Find the factors of the first term.
Find the factors of the last term.  Consider the signs. Since the last term, 5, is positive, its factors must both be positive or both be negative.  The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

$6b^2−13b+5$
Possible factors	Product
$(b−1)(6b−5)$	$6b^2−11b+5$
$(b−5)(6b−1)$	$6b^2−31b+5$
$(2b−1)(3b−5)$	$6b^2−13b+5 \,\checkmark$
$(2b−5)(3b−1)$	$6b^2−17b+5$

$\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial.} &(2b−1)(3b−5) \\ \text{Check by multiplying:} & \\ \hspace{50mm} (2b−1)(3b−5) & \\ \hspace{47mm} 6b^2−10b−3b+5 & \\ \hspace{50mm} 6b^2−13b+5\,\checkmark & \end{array}$

$\PageIndex{17}$

Factor completely using Reverse FOIL: $8x^2−13x+3$.

Answer

$(2x−3)(4x−1)$

$\PageIndex{18}$

Factor completely using Reverse FOIL: $10y^2−37y+7$.

Answer

$(2y−7)(5y−1)$

When we factor an expression, we always look for a greatest common factor first.  If the expression does not have a greatest common factor, there cannot be one in its factors either.  This may help us eliminate some of the possible factor combinations.

Example $\PageIndex{16}$

Factor completely using Reverse FOIL: $18x^2−37xy+15y^2$.

Solution

The trinomial is already in descending order, and has no common factors.
Find the factors of the first term.
Find the factors of the last term.  Consider the signs.  Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Consider all the combinations of factors.

$\begin{array} {ll} \text{The correct factors are those whose product is} & \\ \text{the original trinomial.} &(2x−3y)(9x−5y) \\ \text{Check by multiplying:} & \\ & \\ & \\ & \\ \hspace{50mm} (2x−3y)(9x−5y) & \\ \hspace{45mm}18x^2−10xy−27xy+15y^2 & \\ \hspace{47mm}18x^2−37xy+15y^2\,\checkmark & \end{array}$

$\PageIndex{19}$

Factor completely using Reverse FOIL $18x^2−3xy−10y^2$.

Answer

$(3x+2y)(6x−5y)$

Don’t forget to look for a GCF first, and remember: if the leading coefficient is negative, so is the GCF.

Example $\PageIndex{17}$

Factor completely using Reverse FOIL: $−10y^4−55y^3−60y^2$.

Answer

Notice the greatest common factor, so factor it first.
Factor the trinomial.

Consider all the combinations.

$\begin{array} {ll} \text{The correct factors are those whose product} & \\ \text{is the original trinomial. Remember to include} & \\ \text{the factor }−5^y2. &−5y^2(y+4)(2y+3) \\ \text{Check by multiplying:} & \\ \hspace{50mm} −5y^2(y+4)(2y+3) & \\ \hspace{45mm} −5y^2(2y^2+8y+3y+12) & \\ \hspace{47mm}−10y^4−55y^3−60y^2\,\checkmark & \end{array}$

$\PageIndex{20}$

Factor completely using Reverse FOIL: $15n^3−85n^2+100n$.

Answer

$5n(n−4)(3n−5)$

$\PageIndex{21}$

Factor completely using Reverse FOIL: $56q^3+320q^2−96q$.

Answer

$8q(q+6)(7q−2)$

Factor using the AC Grouping Method

The other method for factoring trinomials with leading coefficients other than $1$ is the AC Grouping Method.  In this method, the middle $x$ term is rewritten as two terms, and then the resulting four term expression is factored by grouping pairs.  For example, the trinomial $2x^2+5x+3$ can be rewritten as $2x^2+2x+3x+3$.  Now we can begin to factor by grouping, obtaining $2x(x+1)+3(x+1)$.  The GCF of $(x+1)$ is pulled out to finally obtain the factored expression $(x+1)(2x+3)$.

Given a trinomial in the form $ax^2+bx+c$, factor by grouping.

1. List factor pairs of $ac$.
2. Find $p$ and $q$, a pair of factors of $ac$ with a sum of $b$.
3. Rewrite the original expression as $ax^2+px+qx+c$.
4. Factor the resulting four term expression using factoring by grouping.

Example $\PageIndex{18}$: Factoring a Trinomial by Grouping

Factor $5x^2+7x−6$ by grouping.

Solution

We have a trinomial with $a=5$, $b=7$, and $c=−6$.  First, determine $ac=−30$.  We need to find two numbers with a product of $−30$ and a sum of $7$.  In the table below, we list factor pairs until we find a pair with the desired sum.

Table $\PageIndex{2}$

Step 1. List Factors of −30	Step 2. Find factor pair whose sum is 7
1,−30	1+ -30 = −29 $\quad \color{Red}{✗}$
−1,30	1- + 30 = 29 $\quad \color{Red}{✗}$
2,−15	2 + -15 = −13 $\quad \color{Red}{✗}$
−2,15	-2 + 15 = 13 $\quad \color{Red}{✗}$
3,−10	3 + -10 = −7 $\quad \color{Red}{✗}$
−3,10	-3 + 10 = 7 $\quad \color{Cerulean}{✓}$

Step 3. Rewrite $5x^2+{\color{Cerulean}{7x}}−6$ as $5x^2  {\color{Cerulean}{−3x+10x}}−6$

Step 4. Factor by grouping. 

$x(5x−3)+2(5x−3)$ Factor out the GCF of each part

$(5x−3)(x+2)$ Factor out the GCF of the expression.

Analysis. We can check our work by multiplying.  Use FOIL to confirm that $(5x−3)(x+2)=5x^2+7x−6$.

Try It $\PageIndex{22}$

Factor:

a.  $2x^2+9x+9$ Answer a $(2x+3)(x+3)$

b.  $6x^2+x−1$ Answer b $(3x-1)(2x+1)$

Factor a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

$\begin{aligned} (a+b)^2 &= (a+b)(a+b) = a^2+ab +ab +b^2= a^2+2ab+b^2 \qquad \text{ and}\\ (a-b)^2 &= (a-b)(a-b) = a^2-ab -ab +b^2= a^2-2ab+b^2  \\ \end{aligned}$

We can use these equations to factor any perfect square trinomial.

Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

$$a^2+2ab+b^2=(a+b)^2 \nonumber$$

$$a^2-2ab+b^2=(a-b)^2 \nonumber$$

Given a perfect square trinomial, factor it into the square of a binomial

1. Confirm that the first term $a^2$ and last term $b^2$ are perfect squares.
2. Confirm that the middle term is twice the product of $ab$.
3. Write the factored form as $(a+b)^2$ if the middle term is positive and $(a-b)^2$ if the middle term is negative.

Example $\PageIndex{19}$: Factoring a Perfect Square Trinomial

Factor $25x^2+20x+4$.

Solution

Notice that $25x^2$ and $4$ are perfect squares because $25x^2={(5x)}^2$ and $4=2^2$.  Check to see if the middle term is twice the product of $5x$ and $2$.  The middle term is, indeed, twice the product: $2(5x)(2)=20x$.  Therefore, the trinomial is a perfect square trinomial and can be written as $(5x+2)^2$.

$\PageIndex{23}$

Factor $49x^2−14x+1$.

Answer

$(7x−1)^2$

Quadratic form

Sometimes a trinomial does not appear to be in the $ax^2+bx+c$ form, but it follows a similar pattern, which is known as quadratic form.  We can use a technique called factoring by substitution that will make it fit the $ax^2+bx+c$ form.  It is standard to use u for the substitution.

In $ax^2+bx+c$, the middle term has a variable, x, and its square, $x^2$, is the variable part of the first term.  Look for this quadratic form pattern as you try to find a substitution.

Example $\PageIndex{20}$

Factor by substitution: $x^4−4x^2−5$.

Solution

The variable part of the middle term is $x^2$ and its square, $x^4$, is the variable part of the first term.  (We know $(x^2)^2=x^4)$.  If we let $u=x^2$, we can put our trinomial in the $ax^2+bx+c$ form we need to factor it.

Rewrite the trinomial to prepare for the substitution.
Let $u=x^2$ and substitute.
Factor the trinomial.
Replace u with $x^2$.
Check: $\begin{array} {l} \hspace{37mm} (x^2+1)(x^2−5) \\ \hspace{35mm}x^4−5x^2+x^2−5 \\ \hspace{40mm}x^4−4x^2−5\,\checkmark\end{array}$

$\PageIndex{24}$

Factor by substitution: $h^4+4h^2−12$.

Answer

$(h^2−2)(h^2+6)$

$\PageIndex{25}$

Factor by substitution: $y^4−y^2−20$.

Answer

$(y^2+4)(y^2−5)$

Factor a Difference of Squares

A difference of squares is a binomial consisting of a perfect square subtracted from a perfect square.  Recall that when two binomials containing the same terms but opposite signs (conjugates) are multiplied together, the middle terms cancel each other out, and all that is left is a difference of squares.  As usual, the order of the two factors does not matter, since multiplication is commutative.

$\begin{aligned} (a-b)(a+b) &= a^2+ab -ab +b^2= a^2-b^2 \qquad \text{ and}\\ (a+b)(a-b) &= a^2-ab +ab +b^2= a^2-b^2  \\ \end{aligned}$

Difference of Squares

A difference of squares can be rewritten as the product of two binomial conjugates.

$$a^2−b^2=(a−b)(a+b) \nonumber$$

Given a difference of squares, factor it into binomials

1. Confirm that there are only two terms, both are perfect squares, and the second term is being subtracted from the first.
2. Write the factored form as $(a−b)(a+b)$.

Example $\PageIndex{21}$: Factoring a Difference of Squares

Factor $9x^2−25$.

Solution

Notice that $9x^2$ and $25$ are perfect squares because $9x^2={(3x)}^2$ and $25=5^2$, so if the formula is used, $a =3x$ and $b = 5$.  The polynomial represents a difference of squares and can be rewritten as $(3x−5)(3x+5)$.

$\PageIndex{26}$

Factor $81y^2−100$.

Answer

$(9y−10)(9y+10)$

 Is there a formula to factor the sum of squares?

No. A sum of squares is prime and cannot be factored if only real numbers are used. 

When factoring, always check the resulting factors to see if any can be factored further.

Example $\PageIndex{22}$

Factor $x^4 - 81y^4$ completely .

Solution

First, identify what is being squared.  To do this, recall the power rule for exponents, $(x^{m})^{n}=x^{mn}$.  Thus, $x ^ { 4 } - 81 y ^ { 4 } = \left( \color{Cerulean}{x ^ { 2} } \right) ^ { 2 } - \left(\color{Cerulean}{ 9 y ^ { 2} } \right) ^ { 2 }$.  Therefore, substitute $a=x^{2}$ and $b=9y^{2}$ into the formula for difference of squares.

$x ^ { 4 } - 81 y ^ { 4 } = \left( x ^ { 2 } - 9 y ^ { 2 } \right) \left( x ^ { 2 } + 9 y ^ { 2 } \right)$

At this point, notice that the factor $(x^{2}−9y^{2})$ is itself a difference of squares and thus can be further factored using $a=x^{2}$ and $b=3y$.  The factor $(x^{2}+9y^{2})$ is prime and cannot be factored using real numbers.

$\begin{aligned} x ^ { 4 } - 81 y ^ { 4 } & =\left( x ^ { 2 } - 9 y ^ { 2 } \right) \left( x ^ { 2 } + 9 y ^ { 2 } \right) \\ & = ( x - 3 y )( x + 3 y ) \left( x ^ { 2 } + 9 y ^ { 2 } \right) \end{aligned}$

Factor the Sum and Difference of Cubes

Now, we will look at two new special products: the sum and difference of cubes.  Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.  Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs.

Sum and Difference of Cubes

The sum of two cubes can be factored as

$$a^3+b^3=(a+b)(a^2−ab+b^2) \nonumber$$

The difference of two cubes can be factored as

$$a^3−b^3=(a−b)(a^2+ab+b^2) \nonumber$$

The acronym SOAP can be used to remember the signs when factoring the sum or difference of cubes.  The first letter of each word relates to the signs: Same Opposite AlwaysPositive.  Consider the following example.

$$x^3−2^3=(x−2)(x^2+2x+4) \nonumber$$

The sign of the first 2 is the Same as the sign between $x^3−2^3$.  The sign of the $2x$ term is Opposite the sign between $x^3−2^3$.  And the sign of the last term, $4$, is Always Positive.

Given a sum of cubes or difference of cubes, factor it

1. Confirm both terms are cubes, $a^3+b^3$ or $a^3−b^3$.  Identify $a$ and $b$.
2. For a sum of cubes, write the factored form as $(a+b)(a^2−ab+b^2)$.
   For a difference of cubes, write the factored form as $(a−b)(a^2+ab+b^2)$.

Example $\PageIndex{23}$: Factoring a Sum of Cubes

Factor $x^3+512$.

Solution

Notice that $x^3$ and $512$ are cubes because $8^3=512$, so $a = x$ and $b = 8$.  Rewrite the sum of cubes as $(x+8)(x^2−8x+64)$.

Analysis.  After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further.  However, the second degree trinomial portion cannot be factored, so we do not need to check.

$\PageIndex{27}$

Factor the sum of cubes: $216a^3+b^3$.

Answer

$(6a+b)(36a^2−6ab+b^2)$

Example $\PageIndex{24}$: Factoring a Difference of Cubes

Factor $8x^3−125$.

Solution

Notice that $8x^3$ and $125$ are cubes because $8x^3=(2x)^3$ and $125=5^3$.  Write the difference of cubes as $(2x−5)(4x^2+10x+25)$.

Analysis. Just as with the sum of cubes, we will not be able to further factor the second degree trinomial portion.

$\PageIndex{28}$

Factor the difference of cubes: $1000x^3−1$

Answer

$(10x−1)(100x^2+10x+1)$

If a binomial is both a difference of squares and a difference of cubes, then first factor it as a difference of squares.  This will result in a more complete factorization.

Example $\PageIndex{25}$

Factor $64 x ^6 - y ^6$ completely.

Solution

This binomial is both a difference of squares and a difference of cubes.

$\begin{array} { l } { 64 x ^ { 6 } - y ^ { 6 } = \left( \color{Cerulean}{4 x ^ { 2} } \right) ^ { 3 } - \left( \color{Cerulean}{y ^ { 2} } \right) ^ { 3 } \quad\color{Cerulean} \text{ Difference of cubes } } \\ { 64 x ^ { 6 } - y ^ { 6 } = \left( \color{Cerulean}{8 x ^ { 3} } \right) ^ { 2 } - \left( \color{Cerulean}{y ^ { 3} } \right) ^ { 2 } \quad \color{Cerulean} \text{ Difference of squares } } \end{array}$

When confronted with a binomial that is a difference of both squares and cubes, as this is, factor using difference of squares first.

$$\begin{align*} 64 x ^ 6 - y ^ 6  &=  ( 8 x ^ 3 ) ^ 2 - (y ^ 3 ) ^ 2  &&  \qquad \text{Factor first as a difference of squares} \\ &= ( 8 x ^ 3 - y ^ 3 )( 8 x ^ 3 + y ^ 3 ) \\ &= \underbrace{   ( 8 x ^ 3 - y ^ 3 ) }_{ { ( 2x-y)(4x^2+2xy+y^2 )   }  }   \cdot \underbrace{   ( 8 x ^ 3 + y ^ 3 ) }_{ { ( 2x+y)(4x^2-2xy+y^2 )   }  } && \qquad \text{Then factor a sum and difference of cubes} \\ &= ( 2x-y)(4x^2+2xy+y^2 )( 2x+y)(4x^2-2xy+y^2 ) &&  \\ \nonumber \end{align*}$$

All four factors are prime and the expression is completely factored.

$\PageIndex{29}$

Factor: $a ^ { 6 } b ^ { 6 } - 1$

Answer

$( a b + 1 ) \left( a ^ { 2 } b ^ { 2 } - a b + 1 \right) ( a b - 1 ) \left( a ^ { 2 } b ^ { 2 } + a b + 1 \right)$

Challenge

Refer to Example 25.  You may wonder why we do not get the same result if we factor using a difference of squares first as opposed to using a difference of cubes first.  You may be familiar with the unique factorization property of polynomials, even though this is not part of our precalculus curriculum.  However, if you factor the polynomial in Example 25 using a difference of cubes first, the result is $64 x ^6 - y ^6 = (2x-y)(2x+y)(16x^4+4x^2+y^4)$. (Try it!)  Comparing the two different results, what can we conclude?

Answer

$16x^4+4x^2+y^4 = (4x^2+2xy+y^2 )(4x^2-2xy+y^2 )\quad$ The fourth degree trinomial on the left can be factored as a product of the two second trinomials on the right!

Factoring Summary 

The following outlines a useful strategy for factoring polynomials.

USE A GENERAL STRATEGY FOR FACTORING POLYNOMIALS.

1. Is there a greatest common factor?
   Factor it out.
2. Is the polynomial a binomial, trinomial, or are there more than three terms?
   If it is a binomial:
   • Is it a sum?
     Of squares? Sums of squares do not factor.
     Of cubes? Use the sum of cubes pattern.
   • Is it a difference?
     Of squares? Factor as the product of conjugates.
     Of cubes? Use the difference of cubes pattern.
   If it is a trinomial:
   • Is it of the form $x^2+bx+c$? Reverse FOIL.
   • Is it of the form $ax^2+bx+c$?
     If a and c are squares, check if it fits the trinomial square pattern.
     Use the Reverse FOIL or “$ac$” method.
   If it has four terms:
   • Use the grouping method.
3. Check.
   Is it factored completely?
   Do the factors multiply back to the original polynomial?

Remember, a polynomial is completely factored if, other than monomials, its factors are prime!

Example $\PageIndex{26}$

Factor completely: $7x^3−21x^2−70x$.

Solution:

$\begin{array}  {ll}  \text{1. Is there a GCF? Yes, }7x. \text{ Factor out the GCF.} & 7x(x^2−3x−10) \\ \text{2. In the parentheses, is a trinomial } & \\ \hspace{5mm}\text{with leading coefficient 1, so “Undo” FOIL.} &7x(x+2)(x−5) \\ \text{3. Is the expression factored completely? Yes. } & \text{Neither binomial can be factored.}  \end{array}$
$\hspace{5mm}\text{Check.  } \quad 7x(x+2)(x−5)  \text{  =  } 7x(x^2−5x+2x−10) \text{  =  } 7x(x^2−3x−10)  \text{  =  } 7x^3−21x^2−70x\,\checkmark$

Be careful when you are asked to factor a binomial, as there are several options! 

Example $\PageIndex{27}$

Factor completely: $24y^2−150$

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, }6. \text{ Factor out the GCF.} &6(4y^2−25) \\ \text{2. Two terms are in the parentheses.} & \\ \hspace{5mm}\text{The binomial is a difference of squares.} &6((2y)^2−(5)^2) \\ \hspace{5mm}\text{Write as a product of conjugates.} &6(2y−5)(2y+5)  \\ \text{3. Is the expression factored completely? Yes.} & \text{Neither binomial can be factored.}  \end{array}$
$\quad \text{Check. } \quad 6(2y−5)(2y+5)  \text{  =  } 6(4y^2−25)  \text{  =  } 24y^2−150\,\checkmark$ 

The next example can be factored using several methods. Recognizing the Perfect Square Trinomial pattern will make your work easier.

Example $\PageIndex{28}$

Factor completely: $4a^2−12ab+9b^2$.

Solution:

$\begin{array} {ll}  \text{1. Is there a GCF? No.} & \\ \text{2. A trinomial with }a\neq 1.\text{ But the first term is a perfect square.} \\  \hspace{5mm}\text{Is the last term a perfect square? Yes.} &(2a)^2−12ab+(3b)^2 \\  \hspace{5mm}\text{Does it fit the pattern, }a^2−2ab+b^2?\text{  Yes. } &(2a)^2 −12ab+ (3b)^2 \\ &\hspace{7mm} {\,}^{\searrow}{\,}_{−2(2a)(3b)}{\,}^{\swarrow}\\  \hspace{5mm}\text{Write it as a binomial square.} &(2a−3b)^2 \\ \text{3. Is the expression factored completely? Yes.} & \text{The binomial cannot be factored.} \end{array}$ $\hspace{5mm} \text{Check.  } \quad (2a−3b)^2 \text{  =  } (2a)^2−2·2a·3b+(3b)^2 \text{  =  } 4a^2−12ab+9b^2\,\checkmark$

Remember, sums of squares do not factor, but sums of cubes do!

Example $\PageIndex{29}$

Factor completely $12x^3y^2+75xy^2$.

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, }3xy^2. \text{ Factor out the GCF.} & 3xy^2(4x^2+25) \\ \text{2. In the parentheses is a binomial - a sum of squares} &  \text{Sums of squares cannot be factored.}  \\ \text{3. Is the expression factored completely? Yes. Check.  } & 3xy^2(4x^2+25) \text{  =  } 12x^3y^2+75xy^2 \,\checkmark  \end{array}$

When using the sum or difference of cubes pattern, be careful with the signs.

Example $\PageIndex{30}$

Factor completely: $24x^3+81y^3$.

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, }3. \text{ Factor out the GCF.} &3(8x^3+27y^3) \\ \text{2. Two terms are in the parentheses.} & \\ \quad\text{The binomial is a sum of cubes.} &3((2x)^3+(3y)^3) \\ \quad \text{Factor it using the sum of cubes formula.} & 3(2x+3y)((2x)^2 - 2x \cdot  3y +(3y)^2) \\ & 3(2x+3y)(4x^2-6xy+9y^2) \\ \text{3. Is the expression factored completely? Yes.} & \text{Neither binomial nor trinomial can be factored.}  \end{array}$
$\quad \text{Check. } \quad 3(2x+3y)(4x^2-6xy+9y^2)$
$\hspace{20mm} \ \text{  =  } (6x+9y)(4x^2-6xy+9y^2) \text{  =  } 24x^3 -36x^2y +54xy^2 +36x^2y - 54xy^2 +81y^3  \text{  =  } 24x^3+81y^3 \,\checkmark$ 

Example $\PageIndex{31}$

Factor completely: $3x^5y−48xy$.

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, }3xy. \text{ Factor out the GCF.} &3xy(x^4−16) \\ \text{2. The binomial in parentheses is a difference of squares.} &3xy\left((x^2)^2−(4)^2\right) \\ \quad \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\ \text{3. Is the expression factored completely? NO!} \\ \text{2. The first binomial is a difference of squares.} &3xy\left((x)^2−(2)^2\right)(x^2+4) \\ \quad \text{Factor it as a product of conjugates.} &3xy(x−2)(x+2)(x^2+4) \\ \text{3. Is the expression factored completely? Yes.} \end{array}$

$\quad  \text{Check.  } \quad 3xy(x−2)(x+2)(x^2+4)  \text{  =  } 3xy(x^2−4)(x^2+4)  \text{  =  } 3xy(x^4−16)  \text{  =  } 3x^5y−48xy\,\checkmark$

Example $\PageIndex{32}$

Factor completely: $4x^2+8bx−4ax−8ab$.

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, } 4 \text{. Factor out the GCF.  } &4(x^2+2bx−ax−2ab) \\ \text{2. There are four terms. Factor each pair.} &4 \big( x {\color{Cerulean}{(x+2b)}} −a {\color{Cerulean}{(x+2b)}} \big) \\ \quad \text{Factor out the common binomial GCF} & 4 {\color{Cerulean}{(x+2b)}} (x−a) \\ \text{3. Is the expression factored completely? Yes.} \end{array}$
$\quad \text{Check.  } \quad 4(x+2b)(x−a)  \text{  =  } (4x + 8b)(x- a)  \text{  =  } 4x^2−4ax+8bx−8ab \,\checkmark$

Example $\PageIndex{33}$

Factor completely: $40x^2y+44xy−24y$.

Solution:

$\begin{array} {ll} \text{1. Is there a GCF? Yes, } 4y \text{. Factor out the GCF.  } &4y(10x^2+11x−6) \\ \text{2. Factor the trinomial with }a\neq 1.   & 4y(5x−2)(2x+3) \\ \quad \text{Use the Reverse FOIL or AC Grouping Method}  & \\ \text{3. Is the expression factored completely? Yes.} \end{array}$
$\quad \text{Check.  } \quad 4y(5x−2)(2x+3)  \text{  =  } 4y(10x^2+11x−6)  \text{  =  } 40x^2y+44xy−24y \,\checkmark$

 

Radical Expressions

Definition: Square and Square Root of a Number

Square

If $n^{2}=m$, then $m$ is the square of $n$.

Square Root

If $n^{2}=m$, then $n$ is a square root of $m$.

Notice that both $13^2=169$ and $(−13)^{2} = 169$.  Therefore, both $13$ and $−13$ are square roots of $169$.

Every positive number has two square roots—one positive and one negative.  When we use a radical sign, and write $\sqrt{m}$, it denotes the positive square root of $m$. The positive square root is also called the principal square root.  If we want to find the negative square root of a number, we place a negative in front of the radical sign.  For example, $-\sqrt{169}=-13$.

Because $0^2=0, \sqrt{0}=0$.  Notice that zero has only one square root.

Definition: Square Root Notation

$\sqrt{m}$ is read "the square root of $m$."

If $n^2=m$, then $n=\sqrt{m}$, for $n\geq 0$.

Example $\PageIndex{34}$

Simplify:

1. $\sqrt{144}$
2. $-\sqrt{289}$

Solution:

a. Since $12^{2}=144$, $\sqrt{144} = 12.$

b. Since $17^{2}=289$ and the negative is in front of the radical sign, $-\sqrt{289} = -17.$

$\PageIndex{30}$

Simplify:

1. $-\sqrt{64}$
2. $\sqrt{225}$

Answer

1. $-8$
2. $15$

Can we simplify $\sqrt{-49}$?  Is there a number whose square is $-49$?

$($___$)^{2}=-49$

Since any positive number squared is positive and any negative number squared is positive, there is no real number equal to $\sqrt{-49}$. 

So far we have only talked about squares and square roots.  Let’s now extend our work to include higher powers and higher roots.

It will be helpful to have a table of the powers of the integers from $−5$ to $5$.

 

 

Notice the signs in the table.  All powers of positive numbers are positive, of course.  But when we have a negative number, the even powers are positive and the odd powers are negative.  We’ll copy the row with the powers of $−2$ to help you see this.

 

Definition $\PageIndex{3}$: n^th Root of a Number

For any whole number $n$ greater than 1, if $b^{n}=a$, then $b$ is an $n^{th}$ root of $a$.

The principal $n$th root of $a$ is written $\sqrt[n]{a}$ and has the same sign as $a$.
When $a$ is negative and $n$ is even, the principal $n$th root is undefined as a real number.

The $n$ is called the index of the radical.

 Just as we use the word ‘cubed’ for $b^{3}$, we use the term ‘cube root’ for $\sqrt[3]{a}$.

We know that the square root of a negative number is not a real number.  The same is true for any even root.  Even roots of negative numbers are not real numbers.  Odd roots of negative numbers are real numbers.

When $n$ is an even number and

• $a \geq 0$, then $\sqrt[n]{a}$ is a real number.
• $a<0$, then $\sqrt[n]{a}$ is not a real number.

When $n$ is an odd number, $\sqrt[n]{a}$ is a real number for all the values of $a$.

Example $\PageIndex{35}$

Simplify:

1. $\sqrt[3]{-125}$
2. $\sqrt[4]{-16}$
3. $\sqrt[5]{-243}$

Solution:

a. Since $(-5)^{3}=-125$, $\sqrt[3]{-125} = -5$.

b. Think, $(?)^{4}=-16$. No real number raised to the fourth power is negative.  Thus, $\sqrt[4]{16}$ is not a real number.

c. Since $(-3)^{5}=-243$, $\sqrt[5]{-243} = -3$.

$\PageIndex{31}$

Simplify:

1. $\sqrt[3]{-27}$
2. $\sqrt[4]{-256}$
3. $\sqrt[5]{-32}$

Answer

1. $-3$
2. not real
3. $-2$

Estimate and Approximate Roots

When we see a number with a radical sign, we often don’t think about its numerical value.  While we probably know that the $\sqrt{4}=2$, what is the value of $\sqrt{21}$ or $\sqrt[3]{50}$?  In some situations a quick estimate is meaningful and in others it is important to have a decimal approximation.

To get a numerical estimate of a square root, we look for perfect square numbers closest to the radicand (the number inside the radical sign).  To find an estimate of $\sqrt{11}$, we see $11$ is between perfect square numbers $9$ and $16$, closer to $9$.  Its square root will be between $3$ and $4$, but closer to $3$.  Similarly, to estimate $\sqrt[3]{91}$, we see $91$ is between perfect cube numbers $64$ and $125$.  The cube root will be between $4$ and $5$.

 

  $\PageIndex{32}$

Estimate each root between two consecutive whole numbers:

1. $\sqrt{38}$
2. $\sqrt[3]{200}$

Answer

1. $6<\sqrt{38}<7$
2. $5<\sqrt[3]{200}<6$

There are mathematical methods to approximate square roots, but nowadays most people use a calculator to find square roots.  To find a square root you will use the $\sqrt{x}$ key on your scientific calculator.  To find a cube root, or any root with higher index, you will use the $\sqrt[y]{x}$ key.

When you use these keys, you get an approximate value if the radicand is not a perfect square.  The approximation is accurate to the number of digits shown on your calculator’s display.  The symbol for an approximation is $≈$ and it is read ‘approximately’.

Suppose your calculator has a $10$ digit display.  You would see that

$\sqrt{5} \approx 2.236067978$, rounded to two decimal places is $\sqrt{5} \approx 2.24$

$\sqrt[4]{93} \approx 3.105422799$, rounded to two decimal places is $\sqrt[4]{93} \approx 3.11$

How do we know these values are approximations and not the exact values?  Look at what happens when we square the first and raise the next to the fourth power:

$\begin{aligned}(2.236067978)^{2} &=5.000000002 &(3.105422799)^{4}&=92.999999991 \\(2.24)^{2} &=5.0176 & (3.11)^{4}&=93.54951841 \end{aligned}$

The squares are close to $5$, but are not exactly equal to $5$.  The fourth powers are close to $93$, but not equal to $93$.  If we had a calculator that displays more digits, could we find the exact value of $\sqrt{5}$ or $\sqrt[4]{93}$?  No, because these are irrational numbers, which have no exact decimal representation, no matter how many digits are displayed.

Note: Most scientific calculators can be programmed to display results rounded to however many decimal places you want (below the maximum).

  $\PageIndex{33}$

Round to two decimal places:

1. $\sqrt{11}$
2. $\sqrt[3]{71}$
3. $\sqrt[4]{127}$

Answer

1. $\approx 3.32$
2. $\approx 4.14$
3. $\approx 3.36$

Using the Product Rule to Simplify Square Roots

Note: Although we are concerned with simplifying square roots, the following rules and procedures can be modified to apply to any $n$th root. 

"Simplify" a square root means rewrite it so that we can compare and/or combine expressions with square roots.  There are several properties of square roots that allow us to simplify complicated radical expressions.  The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions.  For instance, we can rewrite $\sqrt{15}$ as $\sqrt{3}\times\sqrt{5}$.  We can also use the product rule to express the product of multiple radical expressions as a single radical expression.

The Product Rule For Simplifying Square Roots

If $a$ and $b$ are nonnegative, the square root of the product $ab$ is equal to the product of the square roots of $a$ and $b$

$$\sqrt{ab}=\sqrt{a}\times\sqrt{b}$$

Simplify the square root of a product

1. Rewrite the radicand as a product of a perfect square and another factor.
2. Write the radical expression as a product of radical expressions.
3. Simplify.

Because negative numbers do not have real square roots, assume for the rest of this section that all variables represent nonnegative numbers.

Example $\PageIndex{36}$: Using the Product Rule to Simplify Square Roots

Simplify the radical expression.

1. $\sqrt{300}$
2. $\sqrt{162a^5b^4}$

Solution

a. $\quad \sqrt{300}=\sqrt{100\times3} \quad$ Rewrite 300 as a product of a perfect square and another factor.

$\qquad = \sqrt{100}\times\sqrt{3} \quad$ Write radical expression as a product of radical expressions.

$\qquad = 10 \times \sqrt{3} \quad$ Simplify by finding the square root of 100.

b. $\quad \sqrt{162a^5b^4}=\sqrt{81a^4b^4\times2a} \quad$ Rewrite radicand as a product of a perfect square and another factor.

$\qquad =\sqrt{81a^4b^4}\times\sqrt{2a} \quad$ Write radical expression as product of radical expressions.

$\qquad = 9a^2b^2 \times \sqrt{2a} \quad$ Simplify.

$\PageIndex{34}$

Simplify $\sqrt{50x^2y^3z}$.

Answer

$5xy \times \sqrt{2yz}$

Sometimes we are given a product of radical expressions, and want to simplify if possible.

Simplify the product of multiple radical expressions

1. Express the product of multiple radical expressions as a single radical expression.
2. Simplify.

Example $\PageIndex{5}$: Using the Product Rule to Simplify the Product of Multiple Square Roots

Multiply.  Simplify the radical expression.

a. $\sqrt{12}\times\sqrt{3}$

b. $\sqrt { 6 x ^ { 3 } y ^ { 3 } } \times \sqrt{2 x^3}$.

Solution

a. $\quad \text{Express the product as a single radical expression:  } \sqrt{12\times3} =  \sqrt{36} = 6$

b. $\quad$Begin by writing as a single radical expression:  $\sqrt { 12 x ^ { 6 } y ^ { 3 } }$.

Determine the square factors of $12, x^{ 6}$ , and $y^{ 3}$.

$\left. \begin{array} { l } { 12 = \color{Cerulean}{2 ^ { 2} }\color{black}{ \cdot} 3 } \\ { x ^ { 6 } = \color{Cerulean}{\left( x ^ { 3 } \right) ^ { 2 } }} \\ { y ^ { 3 } = \color{Cerulean}{y ^ { 2} }\color{black}{ \cdot} y } \end{array} \right\} \quad\color{Cerulean}\text{Square factors}$

Make these substitutions, and then apply the product rule for radicals and simplify.

$\begin{aligned} \sqrt { 12 x ^ { 6 } y ^ { 3 } } & = \sqrt { \color{Cerulean}{2 ^ { 2} }\color{black}{ \cdot} 3 \cdot \color{Cerulean}{\left( x ^ { 3 } \right) ^ { 2 }}\color{black}{ \cdot}\color{Cerulean}{ y ^ { 2} }\color{black}{ \cdot} y } \\ & = \sqrt { 2 ^ { 2 } } \cdot \sqrt { \left( x ^ { 3 } \right) ^ { 2 } } \cdot \sqrt { y ^ { 2 } } \cdot \sqrt { 3 y } \quad\quad\quad\quad\:\color{Cerulean} \text{ Apply the  product rule for radicals. } \\ & = 2 \cdot x ^ { 3 } \cdot  y  \cdot \sqrt { 3 y } \\ & = 2 x ^ { 3 }  y  \sqrt { 3 y } \quad\color{Cerulean} \text{ Simplify. } \end{aligned}$

$\PageIndex{5}$

Simplify $\sqrt{50x}\times\sqrt{2x}$.

Answer

$10x$

Adding and Subtracting Square Roots

We can combine terms of radical expressions only when they have the same radicand and when they have the same radical type such as square roots.  For example, the sum of $\sqrt{2}$ and $3\sqrt{2}$ is $4\sqrt{2}$.

The radical expression $\sqrt{2}+\sqrt{18}$ seemingly cannot be combined since the radicands.  However, this is where simplifying radical expressions is valuable. The radical expression $\sqrt{18}$ can be written with a $2$ in the radicand, as $3\sqrt{2}$, so $\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}$.

Simplify a radical expression requiring addition or subtraction of square roots

1. Simplify each radical expression.
2. Add or subtract expressions with equal radicands.

Often, we will have to simplify before we can identify the like radicals within the terms. If the radicand and the index are not exactly the same, then the radicals are not similar and we cannot combine them.

Example​ $\PageIndex{8}$: Add or Subtract Square Roots

Perform the indicated operation and simplify.

1. $4 \sqrt { 10 } - 5 \sqrt { 10 }$
2. $5\sqrt{12}+2\sqrt{3}$.
3. $20\sqrt{72a^3b^4c}-14\sqrt{8a^3b^4c}$

Solution

a. $\begin{aligned}  \text{ }& \text{ } \\ & \text{ } &4 \sqrt { 10 } - 5 \sqrt { 10 } & = ( 4 - 5 ) \sqrt { 10 } \\ & & &= - 1 \sqrt { 10 } \\ & & &= - \sqrt { 10 } \end{aligned}$ 

b. $\quad$We can rewrite $5\sqrt{12}$ as $5\sqrt{4\times3}$. According to the product rule, this becomes $5\sqrt{4}\sqrt{3}$. The square root of $\sqrt{4}$ is $2$ , so the expression becomes $5\times2\sqrt{3}$, which is $10\sqrt{3}$. Now the terms have the same radicand so we can add.

$$10\sqrt{3}+2\sqrt{3}=12\sqrt{3} \nonumber$$

c. $\quad$Rewrite each term so they have equal radicands.

$$\begin{align*} 20\sqrt{72a^3b^4c} &= 20\sqrt{9}\sqrt{4}\sqrt{2}\sqrt{a}\sqrt{a^2}\sqrt{(b^2)^2}\sqrt{c}\\ &= 20(3)(2)ab^2\sqrt{2ac}\\ &= 120ab^2\sqrt{2ac} \end{align*}$$

$$\begin{align*} 14\sqrt{8a^3b^4c} &= 14\sqrt{2}\sqrt{4}\sqrt{a}\sqrt{a^2}\sqrt{(b^2)^2}\sqrt{c}\\ &= 14(2)ab^2\sqrt{2ac}\\ &= 28|a|b^2\sqrt{2ac} \end{align*}$$

Now the terms have the same radicand so we can subtract.

$$120ab^2\sqrt{2ac}-28ab^2\sqrt{2ac}=92ab^2\sqrt{2ac}$$

$\PageIndex{8}$

1. Add $\sqrt{5}+6\sqrt{20}$
2. Subtract $3\sqrt{80x}-4\sqrt{45x}$

Answers

a. $13\sqrt{5}$ $\qquad$ b. $0$

Example $\PageIndex{9}$: Adding and Subtracting Square Roots

Simplify.

1. $10 \sqrt { 5 } + 6 \sqrt { 2 } - 9 \sqrt { 5 } - 7 \sqrt { 2 }$
2. $\sqrt { 32 } - \sqrt { 18 } + \sqrt { 50 }$
3. $2 a \sqrt { 125 a ^ { 2 } b } - a ^ { 2 } \sqrt { 80 b } + 4 \sqrt { 20 a ^ { 4 } b }$.

Solution

a. 

$\begin{aligned}   & \text{ } && = \color{Cerulean}{10 \sqrt { 5 } - 9 \sqrt { 5 }}\color{black}{ +}\color{OliveGreen}{ 6 \sqrt { 2 } - 7 \sqrt { 2 }} \\ &&& = \sqrt { 5 } - \sqrt { 2 } \end{aligned}$

We cannot simplify any further because $\sqrt{5}$ and $\sqrt{2}$ are not like radicals; the radicands are not the same. 
$\color{YellowOrange}{\text{Caution:}}$ It is important to point out that $\sqrt { 5 } - \sqrt { 2 } \neq \sqrt { 5 - 2 }$. We can verify this by calculating the value of each side with a calculator.

$\sqrt { 5 } - \sqrt { 2 } \approx 0.82 \qquad \text{ is not the same as } \qquad  \sqrt { 5 - 2 } = \sqrt { 3 } \approx 1.73$

b.

$\begin{aligned} & \text{ } && = \sqrt { 16  \cdot 2 }- \sqrt { 9 \cdot 2 } + \sqrt { 25 \cdot 2 } \\ &&& = 4 \sqrt { 2 } - 3 \sqrt { 2 } + 5 \sqrt { 2 } \\ &&& = 6 \sqrt { 2 } \end{aligned}$

At first glance, the radicals do not appear to be similar. However, after simplifying completely, we see that we can combine them.

c. $\;$ Step 1: Simplify the radical expression. 
$\quad$ Step 2: Combine all like radicals. Remember to add only the coefficients; the variable parts remain the same.

$\begin{array} { l } { 2 a \sqrt { 125 a ^ { 2 } b } - a ^ { 2 } \sqrt { 80 b } + 4 \sqrt { 20 a ^ { 4 } b } } \\ { = 2 a \sqrt { 25 \cdot 5 \cdot a ^ { 2 } \cdot b } - a ^ { 2 } \sqrt { 16 \cdot 5 \cdot b } + 4 \sqrt { 4 \cdot 5 \cdot \left( a ^ { 2 } \right) ^ { 2 } b } }\quad\color{Cerulean}{Factor.} \\ { = 2 a \cdot 5 \cdot a \sqrt { 5 b } - a ^ { 2 } \cdot 4 \sqrt { 5 b } + 4 \cdot 2 \cdot a ^ { 2 } \sqrt { 5 b } } \quad\quad\quad\quad\quad\:\color{Cerulean}{ Simplify.} \\ { = 10 a ^ { 2 } \sqrt { 5 b } - 4 a ^ { 2 } \sqrt { 5 b } + 8 a ^ { 2 } \sqrt { 5 b } } \quad\quad\quad\quad\quad\quad\quad\quad\quad \color{Cerulean}{Combine\:like\:terms.} \\ { = 14 a ^ { 2 } \sqrt { 5 b } } \end{array}$

 Try It $\PageIndex{9}$

Simplify: $\sqrt { 20 } + \sqrt { 27 } - 3 \sqrt { 5 } - 2 \sqrt { 12 }$.

Answer

$- \sqrt { 5 } - \sqrt { 3 }$

Caution: Simplifying Radicals

Take careful note of the differences between products and sums within a radical. Assume both $x$ and $y$ are nonnegative.

$\begin{array} { l } Products \quad \quad\quad\quad Sums\\\hline { \sqrt { x ^ { 2 } y ^ { 2 } } = x y \quad\sqrt { x ^ { 2 } + y ^ { 2 } } \neq x + y } \end{array}$

The property says that we can simplify radicals when the operation in the radicand is multiplication. There is no corresponding property for addition.

Rationalizing Denominators

An expression involving a square root radical in the denominator is considered to be unsimplified. To simplify, we can remove radicals from the denominators of fractions using a process called rationalizing the denominator.

We know that multiplying by $1$ does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of $1$ that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is $b\sqrt{c}$, multiply by $\dfrac{\sqrt{c}}{\sqrt{c}}$.

Given an expression with a single square root radical term in the denominator, rationalize the denominator

1. Multiply the numerator and denominator by the radical in the denominator.
2. Simplify.

 Sometimes, we will need to continue simplifying by reducing the fraction after rationalizing the denominator.

Example $\PageIndex{11}$: Rationalize a Denominator Containing a Single Term

Write in simplest form (rationalize the denominator).

a.  $\dfrac{2\sqrt{3}}{3\sqrt{10}}$

b.  $\dfrac { \sqrt { 2 } } { \sqrt { 5 x } }$

c.  $\dfrac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } }$

Solution

a. The radical in the denominator is $\sqrt{10}$. So multiply the fraction by $\dfrac{\sqrt{10}}{\sqrt{10}}$. Then simplify.

$\dfrac{2\sqrt{3}}{3\sqrt{10}}\times   {\color{Cerulean}{  \dfrac{\sqrt{10}}{\sqrt{10}} } } = \dfrac{2\sqrt{30}}{30} =  \dfrac{\sqrt{30}}{15}$

b. The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by $1$ in the form $\frac { \sqrt { 5 x } } { \sqrt { 5 x } }$.

$\begin{aligned} \frac { \sqrt { 2 } } { \sqrt { 5 x } } & = \frac { \sqrt { 2 } } { \sqrt { 5 x } } \cdot \color{Cerulean}{\frac { \sqrt { 5 x } } { \sqrt { 5 x } } }\\ & = \frac { \sqrt { 10 x } } { \sqrt { 25 x ^ { 2 } } } \quad\quad\: \color{Cerulean} { Simplify. } \\ & = \frac { \sqrt { 10 x } } { 5 x } \end{aligned}$

c. In this example, we will multiply by $1$ in the form $\frac { \sqrt { 6 a b } } { \sqrt { 6 a b } }$.

$\begin{aligned} \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } & = \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } \cdot \color{Cerulean}{\frac { \sqrt { 6 a b } } { \sqrt { 6 a b } }} \\ & = \frac { 3 a \sqrt { 12 a b } } { \sqrt { 36 a ^ { 2 } b ^ { 2 } } } \quad\quad\color{Cerulean}{Simplify.}\\ & = \frac { 3 a \sqrt { 4 \cdot 3 a b} } { 6 ab } \\ & = \frac { 6 a \sqrt { 3 a b } } { b }\quad\quad\:\:\color{Cerulean}{Cancel.} \\ & = \frac { \sqrt { 3 a b } } { b } \end{aligned}$

Notice that $b$ does not cancel in this example. Do not cancel factors inside a radical with those that are outside.

 Try It $\PageIndex{11}$

Write in simplest form (rationalize the denominator).

a.  $\dfrac{12\sqrt{3}}{\sqrt{2}}$

b.  $\sqrt { \dfrac { 9 x } { 2 y } }$

Answer

a. $6\sqrt{6}$ $\qquad$ b. $\frac { 3 \sqrt { 2xy } } { 2 y }$