0.5: Review  Factoring
 Page ID
 44372
Factor the Greatest Common Factor of a Polynomial
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, \(4\) is the GCF of \(16\) and \(20\) because it is the largest number that divides evenly into both \(16\) and \(20\) The GCF of polynomials works the same way: \(4x\) is the GCF of \(16x\) and \(20x^2\) because it is the largest polynomial that divides evenly into both \(16x\) and \(20x^2\) .
When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.
Definition: Greatest Common Factor
The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.
How to: Given a polynomial expression, factor out the greatest common factor
 Find the GCF of the expression
 Identify the GCF of the coefficients.
 Identify the GCF of the variables and of the variable expressions.
 Combine to find the GCF of the expression.
 Determine what the GCF needs to be multiplied by to obtain each term in the expression.
 Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.
Example \(\PageIndex{1}\): Factor out the GCF of constants and variables
Factor \(6x^3y^3 +45x^2y^2+21xy\).
Solution
Step 1. First, find the GCF of the expression. The GCF of \(6\), \(45\), and \(21\) is \(\color{ForestGreen}{3}\). The GCF of \(x^3\) , \(x^2\), and \(x\) is \({\color{ForestGreen}{x}}\) . (Note that the GCF of a set of expressions in the form \(x^n\) will always be the exponent of lowest degree.) And the GCF of \(y^3\), \(y^2\), and \(y\) is \({\color{ForestGreen}{y}}\). Combine these to find the GCF of the polynomial, \({\color{ForestGreen}{3xy}}\).
Step 2. Next, determine \( {\color{Cerulean}{\text{what the GCF needs to be multiplied by}}} \) to obtain each term of the polynomial. We find that
 \({\color{ForestGreen}{3xy}}( {\color{Cerulean}{2x^2y^2}} ) =6x^3y^3\),
 \({\color{ForestGreen}{3xy}} ({\color{Cerulean}{15xy}}) =45x^2y^2\), and
 \({\color{ForestGreen}{3xy}} ({\color{Cerulean}{7}}) =21xy\).
Notice that for each term, \({ \color{Cerulean}{\text{WhattheGCFneedstobemultipliedby }}} \) = \( \dfrac {\text{A Term} }{ \color{ForestGreen}{\text{The GCF}}} \)
So, \( \dfrac{6x^3y^3}{\color{Cerulean}{\color{ForestGreen}{3xy}}} =\color{Cerulean}{2x^2y^2}\), \( \dfrac{45x^2y^2}{\color{Cerulean}{\color{ForestGreen}{3xy}}} =\color{Cerulean}{15xy}\), and \( \dfrac{21xy}{\color{ForestGreen}{3xy}}=\color{Cerulean}{7} \)
Step 3. Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
\[({\color{Green}{3xy}})({\color{Cerulean}{2x^2y^2}} +{\color{Cerulean}{15xy}}+{\color{Cerulean}{7}} ) \nonumber\]
Analysis. After factoring, multiply to confirm result: \(({\color{Green}{3xy}})({\color{Cerulean}{2x^2y^2+15xy+7}})=6x^3y^3+45x^2y^2+21xy \)
Try It \(\PageIndex{1}\): Factor out the GCF of a common variable expression
Factor \(x(b^2−a)+6(b^2−a)\) by pulling out the GCF.
 Answer
 \((b^2−a)(x+6)\)
Factor Four Terms by Grouping Pairs
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,
\[ \begin{align*}
x^{3}−12x^{2}+2x−8 &= \underbrace{ 3x^{3}−12x^{2} }_{ {\Large 3x^2 {\color{Cerulean}{(x4)}} } }
\text{ + } \underbrace{2x−8 }_{ {\Large +2 \color{Cerulean}{(x−4)} } } && \qquad \text{Begin by grouping the first two terms and the last two terms} \\
& && \qquad \text{ factor out the GCF of each grouping }\\
&= 3x^2{\color{Cerulean}{(x4)}}+2{\color{Cerulean}{(x4)}} && \qquad \text{A common binomial factor appears} \\
&= {\color{Cerulean}{(x4)}} (3x^2+2) && \qquad \text{Factor out the common binomial } \\
\nonumber \end{align*}\]
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 a 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. And then there are some four term polynomials that simply are not factorable. These polynomials are said to be prime.
How to: Given a four term polynomial, factor it
 Determine if there are any factors common to all four terms. If so, factor the GCF from each of the four terms.
 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.
 If both expressions from step 2 produce the same binomial factor, then factor out this common binomial GCF.
Example \(\PageIndex{2x}\): Factor by Grouping Pairs
Factor
 \(24a^4 − 18a^3 − 20a + 15\)
 \(xy 2x^2y+x^32y^2 \)
Solution
\[ \begin{align*}
1.\quad 24a^4 − 18a^3 − 20a + 15 &= \underbrace{ 24a^4 − 18a^3 }_{ {\color{Cerulean}{group}} }
\text{  } \underbrace{20a + 15}_{ {\color{Cerulean}{group}} } && \text{ group} \\
&= 6a^3{\color{Red}{(4a3)}}+5{\color{Red}{(4a+3)}} & &\text{Binomial factors are different} \\
&= 6a^3{\color{Cerulean}{(4a3)}}5{\color{Cerulean}{(4a3)}} && \text{The binomial factors are the same} \\
&= {\color{Cerulean}{(4a3)}} (6a^35) && \text{Factor out the common binomial } \\
\nonumber \end{align*}\]
\[ \begin{align*}
2.\quad xy 2x^2y+x^32y^2 &= \underbrace{ xy 2x^2y }_{ {\color{Cerulean}{group}} }
\text{ + } \underbrace{x^32y^2 }_{ {\color{Cerulean}{group}} } && \text{ group} \\
&= xy{\color{Red}{(12x)}}+1{\color{Red}{(x^32y^2)}} & &\text{Binomial factors are different} \\
x^3 2x^2y+xy2y^2
&= \underbrace{ x^3 2x^2y }_{ {\color{Cerulean}{group}}}
\text{ } \underbrace{+xy2y^2 }_{ {\color{Cerulean}{group}} } && \text{Change the order of the terms and group}\\
&= x^2{\color{Cerulean}{(x2y)}} +y{\color{Cerulean}{(x2y)}} && \text{The binomial factors are the same} \\
&= {\color{Cerulean}{(x2y)}} (x^2+y) && \text{Factor out the common binomial } \\
\nonumber \end{align*}\]
Analysis. After factoring, we can check our work by multiplying. Use the distributive property to confirm that \(24a^4 − 18a^3 − 20a + 15=(4a3)(6a^35) \) and \(xy 2x^2y+x^32y^2=(x2y)(x^2+y) \)
Try It \(\PageIndex{2x}\)
Factor \(x^3x^2yxy+y^2\)
 Answer
 \( (xy)(x^2y) \)
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 \((x+3)\).
Trinomials of the form \(x^2+bx+c\) can be factored by finding two numbers with a product of \(c\) and a sum of \(b\). 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 \((x+8)\).
FACTORING A 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.
How to: Given a trinomial in the form \(x^2+bx+c\), factor it
 List factor pairs of \(c\).
 Find \(p\) and \(q\), a pair of factors of \(c\) with a sum of \(b\).
 Write the factored expression \((x+p)(x+q)\).
Example \(\PageIndex{3}\): Factoring a Trinomial with Leading Coefficient 1
Factor \(x^2+2x−15\).
Solution
We have a trinomial with leading coefficient \(1\), \(b=2\), and \(c=−15\). We need to find two numbers with a product of \(−15\) and a sum of \(2\). In the table below, we list factor pairs until we find a pair with the desired sum.
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, \( (x3)(x+5) = (x+5)(x3) \).
Try It \(\PageIndex{3}\)
Factor \(x^2−7x+6\).
 Answer

\((x−6)(x−1)\)
Factor a Trinomial using the Reverse FOIL Method
Trinomials with leading coefficients other than \(1\) are slightly more complicated to factor. Two different methods 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 the efficacy of one method over the other is not really predictable beforehand. The Reverse Foil Method will be illustrated below; the AC grouping method will follow.
Factor by using Reverse FOIL
The Reverse FOIL method is derived from assuming that the trinomial \(ax^2+bx+c\) can be factored into the form \( (Sx+s)(Tx+t) \). When multiplied out, the factored form becomes \( STx^2+Stx+sTx+st, \) and for this expression to match the original trinomial means that we are seeking four numbers \( S, s, T,\) and \( t \) such that \(ST = a \), \( st = c \) and \(St + sT = b \).
How to: 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 high order term is negative. Start with \( ax^2 + bx + c\) where the GCF has been factored out.
 List all combinations of positive factor pairs, \(S\) and \(T\), for which \(ST = a\).
Use these pairs of value to begin constructing possible templates for the final factored result: \( (Sx ...)(Tx ...) \).
List all permutations of positive factor pairs, \(s\) and \( t\), for which \(st = c\).  Find \(s\) and \(t\) to revise the template to look like: \( (Sx ... s)(Tx ... t) \).
If \(c\) is positive, look for a sum of products \( St+Ts = b \).
If \(c\) is negative, look for a difference of products \( St  Ts = b \).  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: \(sTx + Stx = bx \).
Example \(\PageIndex{4.1x}\): Factoring a Trinomial by Reverse FOIL
Factor \(3x^2−7x−20\) by Reverse Foil.
Solution
Step 1  Step 2  

Factors of 3  Factors of 20  20 is negative so find a difference of 7  Proposed Factored Form 
1,3  1,20 \(\:\:\color{Red}{✗}\) 2,10 \(\:\:\color{Red}{✗}\) 4,5 \(\:\:\color{Red}{✗}\) 5,4\(\:\:\color{Cerulean}{✓}\) 10,2\( \qquad \) 20,1 \( \quad \;\;\; \) 
\( {\color{Cerulean}{1}} \cdot {\color{Green}{1}} = 1 \qquad {\color{Cerulean}{3}} \cdot {\color{Green}{20}} = 60 \qquad 601 \ne 7 \) 
\( ({\color{Cerulean}{1}}x \text{.....} )({\color{Cerulean}{3}}x \text{.....} ) \) \( \) \( \) \( ({\color{Cerulean}{1}}x \text{...} {\color{Green}{4}} )({\color{Cerulean}{3}}x \text{...} {\color{Green}{5}} ) \) 
Step 3. Determine the signs: Inner product  outer product = middle term: \({\color{Red}{}}12x {\color{Red}{+}}5x =7x \). Thus our final factored result is: \( ({\color{Cerulean}{1}}x {\color{Red}{}} {\color{Green}{4}} )({\color{Cerulean}{3}}x {\color{Red}{+}} {\color{Green}{5}} ) \)
Analysis. We can check our work by multiplying. Use FOIL to confirm that \((3x+5)(x−4)=3x^2−7x−20\).
Example \(\PageIndex{4.2x}\): Factoring a Trinomial by Reverse FOIL
Factor \(15x^2+17x+4\) by Reverse Foil.
Solution
Step 1  Step 2  

Factors of 15  Factors of 4  17 is positive so find a sum of 17  Proposed Factored Form 
1,15  \( \color{Green}{1, 4} \:\:\color{Red}{✗} \) \( \color{Green}{2, 2} \:\:\color{Red}{✗}\) \( \color{Green}{4, 1} \:\:\color{Red}{✗}\) 
\( {\color{Cerulean}{1}} \cdot {\color{Green}{1}} = 1 \qquad {\color{Cerulean}{15}} \cdot {\color{Green}{4}} = 60 \qquad 1+60 \ne 17 \) 
\( ({\color{Cerulean}{1}}x \text{.....} )({\color{Cerulean}{15}}x \text{.....} ) \) \( \\ \) \( \text{Reject! Try other template.}\) 
3,5  \( \color{Green}{1, 4} \:\:\color{Red}{✗} \) \( \color{Green}{2, 2} \:\:\color{Red}{✗}\) \( \color{Green}{4, 1} \:\:\color{Cerulean}{✓}\) 
\( {\color{Cerulean}{3}} \cdot {\color{Green}{1}} = 3 \qquad {\color{Cerulean}{5}} \cdot {\color{Green}{4}} = 20 \qquad 3+20 \ne 17 \) 
\( ({\color{Cerulean}{3}}x \text{.....} )({\color{Cerulean}{5}}x \text{.....} ) \) \( \\ \) \( ({\color{Cerulean}{3}}x \text{...} {\color{Green}{1}} )({\color{Cerulean}{5}}x \text{...} {\color{Green}{4}} ) \) 
Step 3. Determine the signs: Inner product  outer product = middle term: \({\color{Red}{+}}5x {\color{Red}{+}}12x =17x \). Thus our final factored result is: \( ({\color{Cerulean}{3}}x {\color{Red}{+}} {\color{Green}{1}} )({\color{Cerulean}{5}}x {\color{Red}{+}} {\color{Green}{4}} ) \)\( ({\color{Cerulean}{3}}x {\color{Red}{+}} {\color{Green}{1}} )({\color{Cerulean}{5}}x {\color{Red}{+}} {\color{Green}{4}} ) \)
Analysis. We can check our work by multiplying. Use FOIL to confirm that \((3x+1)(5x+4)=15x^2+17x+4\).
Try It \(\PageIndex{4x}\)
Factor:a. \(2x^2+9x+9\)

b. \(6x^2+x−1\)

Factor a Trinomial using the AC Grouping Method
The other method trinomials with leading coefficients other than \(1\) are factored is by 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 \(2x^2+2x+3x+3\). This is then factored by grouping to obtain \(2x(x+1)+3(x+1)\). The GCF of \((x+1)\) is pulled out to finally obtain the factored expression \( (x+1)(2x+3) \).
Factor by Grouping
To factor a trinomial in the form \(ax^2+bx+c\) by grouping, we find two numbers with a product of \(ac\) and a sum of \(b\). We use these numbers to divide the \(x\) term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.
This method works because if we write the factored result as \((a_1 x + c_1)(a_2 x +c_2)\) and multiplied it out we would get \( a_1 a_2 x^2 + (a_1 c_2 + a_2 c_1)x + c_1 c_2\), where \( a_1 a_2 = a,\) \(c_1 c_2 = c.\) The product of the leading coefficients in each factor is \( a_1 a_2 \cdot c_1 c_2 = ac\). The product of the two terms that comprise the \(x\) coefficient sum is \(a_1 c_2 \cdot a_2 c_1= ac\).
How to: Given a trinomial in the form \(ax^2+bx+c\), factor by grouping.
 List factor pairs of \(ac\).
 Find \(p\) and \(q\), a pair of factors of \(ac\) with a sum of \(b\).
 Rewrite the original expression as \(ax^2+px+qx+c\).
 Factor the resulting four term expression using factoring by grouping.
Example \(\PageIndex{5}\): 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.
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 binomial 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 \(\PageIndex{5}\)
Factor:
a. \(2x^2+9x+9\)

b. \(6x^2+x−1\)

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.
\( \begin{aligned} (a+b)^2 &= (a+b)(a+b) = a^2+ab +ab +b^2= a^2+2ab+b^2 \qquad \text{ and}\\
(ab)^2 &= (ab)(ab) = a^2ab ab +b^2= a^22ab+b^2 \\
\end{aligned} \)
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:
\(a^2+2ab+b^2=(a+b)^2 \)
\( a^22ab+b^2=(ab)^2 \)
How to: Given a perfect square trinomial, factor it into the square of a binomial
 Confirm that the first term \(a^2\) and last term \(b^2\) are perfect squares.
 Confirm that the middle term is twice the product of \(ab\).
 Write the factored form as \({(a+b)}^2\) if the middle term is positive and \({(ab)}^2\) if the middle term is negative.
Example \(\PageIndex{6}\): 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 \(4=2^2\). Then 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 \({(5x+2)}^2\).
Try It \(\PageIndex{6}\)
Factor \(49x^2−14x+1\).
 Answer

\({(7x−1)}^2\)
Factor a Difference of Squares
A difference of squares is a perfect square subtracted from a perfect square. Recall that when two factors containing the same terms but opposite signs is multiplied out, the middle terms cancel each other out, and all that is left is a difference of squares. Furthermore, the order of the two factors does not matter.
\( \begin{aligned} (ab)(a+b) &= a^2+ab ab +b^2= a^2b^2 \qquad \text{ and}\\
(a+b)(ab) &= a^2ab +ab +b^2= a^2b^2 \\
\end{aligned} \)
Differences of Squares
A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
\[a^2−b^2=(a−b)(a+b)\]
How to: Given a difference of squares, factor it into binomials
 Confirm that there are only two terms, both are perfect squares, and they are being subtracted.
 Write the factored form as \((a−b)(a+b)\).
Example \(\PageIndex{7.1}\): 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 \((3x−5)(3x+5)\).
Try It \(\PageIndex{7}\)
Factor \(81y^2−100\).
 Answer

\((9y−10)(9y+10)\)
Is there a formula to factor the sum of squares?
No. A sum of squares cannot be factored if only real numbers are used.
Example \(\PageIndex{7.2x}\)
Factor: \(x ^ { 2 }  ( 2 x  1 ) ^ { 2 }\).
Solution
First, identify this expression as a difference of squares. Use \(a=x\) and \(b=\color{Cerulean}{(2 x  1)}\) in the formula for a difference of squares and then simplify.
\(\begin{aligned} x ^ { 2 }  ( 2 x  1 ) ^ { 2 } & = [ x  {\color{Cerulean}{(2 x  1)}} ] [ x + {\color{Cerulean}{(2 x  1)}} ] \\ & = ( x  2 x + 1 )( x + 2 x  1 ) \\ & = (  x + 1 )( 3 x  1 ) \\ &= (x1)(3x1) \end{aligned}\)
When factoring, always check the resulting factors to see if they can be factored further.
Example \(\PageIndex{7.3x}\)
Factor \(x ^ { 4 }  81 y ^ { 4 }\) 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 two 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 Four Terms by a 3 and 1 Grouping
Sometimes a four term polynomial is the result of a difference between a binomial square and another squared expression. Expressions that can be factored in this manner are characterized by having three terms that are perfect squares.
How to: Given a four term polynomial containing 3 perfect squares, factor.
Determine if there are any factors common to all three terms. If so, factor the GCF from each of the three terms.
 Confirm that three of the four terms are perfect squares. Two are positive, one is negative.
 Confirm that the two positive perfect squares and the nonperfect square term, can be factored as a binomial square.
 Factor the difference of the binomial square result and the remaining negative square, as a difference of squares.
Example \(\PageIndex{8x}\): Factoring a Trinomial by 3 and 1 Grouping
Factor \(9x^2+30x+25  81\).
Solution. Identify three perfect squares, \(9x^2\), \(25\), and \(81\).
\( \begin{align*} 9x^2+30x+25  81 &= (9x^2+30x+25)  81&& \text{Factor the 2 positive squares and the nonsquare}\\
&= (3x+5)(3x+5)  81&&\text{As a binomial square}\\
&= (3x+5)^2  9^2&&\text{This produces a difference of squares }\\
&= (3x+5 9)(3x+5 +9)&&\text{Use }a=(3x+5) \text{ and }b=9 \text{ in the difference of squares formula }\\
&= (3x4)(3x+14) && \text{And simplify}
\end{align*} \)
Try It \(\PageIndex{8x}\)
Factor \(16x^2  24x + 9  144x^2\).
 Answer
 \( (8x+3)(16x3) \)
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
\[a^3+b^3=(a+b)(a^2−ab+b^2) \]
The difference of two cubes can be factored as
\[a^3−b^3=(a−b)(a^2+ab+b^2) \]
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. For example, 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.
How to: Given a sum of cubes or difference of cubes, factor it
 Confirm both terms are cubes, \(a^3+b^3\) or \(a^3−b^3\). Identify \(a\) and \(b\).
 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 \((a−b)(a^2+ab+b^2)\).
If the terms could be factored both as a difference of squares or a difference of cubes, factor as a difference of squares first.
Example \(\PageIndex{9}\): 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 trinomial portion cannot be factored, so we do not need to check.
Try It \(\PageIndex{9}\)
Factor the sum of cubes: \(216a^3+b^3\).
 Answer

\((6a+b)(36a^2−6ab+b^2)\)
Example \(\PageIndex{10}\): 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 trinomial portion.
Try It \(\PageIndex{10}\)
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 cubes, then first factor it as difference of squares. This will result in a more complete factorization.
Example \(\PageIndex{11x}\)
Factor \(64 x ^ { 6 }  y ^ { 6 }\) completely.
Solution
This binomial is both a difference of squares and difference of cubes.
\(\begin{array} { l } { 64 x ^ { 6 }  y ^ { 6 } = \left( \color{Cerulean}{4 x ^ { 2} } \right) ^ { 3 }  \left( \color{Cerulean}{y ^ { 2} } \right) ^ { 3 } \quad\color{Cerulean} { Difference\: of\: cubes } } \\ { 64 x ^ { 6 }  y ^ { 6 } = \left( \color{Cerulean}{8 x ^ { 3} } \right) ^ { 2 }  \left( \color{Cerulean}{y ^ { 3} } \right) ^ { 2 } \quad \color{Cerulean} { Difference\: of\: squares } } \end{array}\)
When confronted with a binomial that is a difference of both squares and cubes, as this is, factor using difference of squares first.
\[ \begin{align*}
64 x ^ 6  y ^ 6 &= ( 8 x ^ 3 ) ^ 2  (y ^ 3 ) ^ 2 && \qquad \text{Factor first as a difference of squares} \\
&= ( 8 x ^ 3  y ^ 3 )( 8 x ^ 3 + y ^ 3 ) \\
&= \underbrace{ ( 8 x ^ 3  y ^ 3 ) }_{ { ( 2xy)(4x^2+2xy+y^2 ) } } \cdot \underbrace{ ( 8 x ^ 3 + y ^ 3 ) }_{ { ( 2x+y)(4x^22xy+y^2 ) } }
&& \qquad \text{Then factor a sum and difference of cubes} \\
&= ( 2xy)(4x^2+2xy+y^2 )( 2x+y)(4x^22xy+y^2 ) && \\
\nonumber \end{align*}\]
The trinomial factors are prime and the expression is completely factored.
As an exercise, factor the previous example as a difference of cubes first. The result is \( (2xy)(2x+y)(16x^4+4x^2+y^4) \). Notice in this case, the result is not as complete as the solution obtained above when the difference of squares factoring was done first.
Try It \(\PageIndex{11x}\)
Factor: \(a ^ { 6 } b ^ { 6 }  1\)
 Answer

\(( a b + 1 ) \left( a ^ { 2 } b ^ { 2 }  a b + 1 \right) ( a b  1 ) \left( a ^ { 2 } b ^ { 2 } + a b + 1 \right)\)
Factoring Expressions with Fractional or Negative Exponents
Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, \(2x^{\tfrac{1}{4}}+5x^{\tfrac{3}{4}}\) can be factored by pulling out \(x^{\tfrac{1}{4}}\) and rewritten as \(x^{\tfrac{1}{4}}(2+5x^{\tfrac{1}{2}})\).
Example \(\PageIndex{12}\): Factoring an Expression with Fractional or Negative Exponents
Factor \(3x{(x+2)}^{\tfrac{1}{3}}+4{(x+2)}^{\tfrac{2}{3}}\).
Solution
Factor out the term with the lowest value of the exponent. In this case, that would be \({(x+2)}^{\tfrac{1}{3}}\).
\[\begin{align*} &(x+2)^{\tfrac{1}{3}}(3x+4(x+2))\qquad \text{Factor out the GCF }\\ &(x+2)^{\tfrac{1}{3}}(3x+4x+8)\qquad \text{Simplify } \\ &(x+2)^{\tfrac{1}{3}}(7x+8) \end{align*}\]
Try It \(\PageIndex{12}\)
Factor \(2{(5a−1)}^{\tfrac{3}{4}}+7a{(5a−1)}^{−\tfrac{1}{4}}\).
 Answer

\({(5a−1)}^{−\tfrac{1}{4}}(17a−2)\)
Factoring Summary
The following outlines a useful strategy for factoring polynomials.
USE A GENERAL STRATEGY FOR FACTORING POLYNOMIALS.
 Is there a greatest common factor?
Factor it out.  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.
 Is it of the form \(x^2+bx+c\)? Undo 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.
 Use the grouping method. (pairs, or (if three terms are perfect squares), 3and1)
 Is it a sum?
 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 \(\PageIndex{13.1x}\)
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 \(\PageIndex{13.2x}\)
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 trinomial squares pattern will make your work easier.
Example \(\PageIndex{13.3x}\)
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 \(\PageIndex{13.4x}\)
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, being careful with the signs.
Example \(\PageIndex{13.5x}\)
Factor completely: \(24x^3+81y^3\).
Solution:
\(\begin{array} {ll} \text{1. Is there a GCF? Yes, }3. \text{ Factor out the GCF.} &3(8x^3+27y^3) \\ \text{2. Two terms are in the parentheses.} & \\ \quad\text{The binomial is a sum of cubes.} &3((2x)^3+(3y)^3) \\ \quad \text{Factor it using the sum of cubes formula.} & 3(2x+3y)((2x)^2  2x \cdot 3y +(3y)^2) \\ & 3(2x+3y)(4x^26xy+9y^2) \\ \text{3. Is the expression factored completely? Yes.} & \text{Neither binomial nor trinomial can be factored.} \end{array}\)
\( \quad \text{Check. } \quad 3(2x+3y)(4x^26xy+9y^2) \)
\( \hspace{20mm} \ \text{ = } (6x+9y)(4x^26xy+9y^2) \text{ = } 24x^3 36x^2y +54xy^2 +36x^2y  54xy^2 +81y^3 \text{ = } 24x^3+81y^3 \checkmark \)
Example \(\PageIndex{13.6x}\)
Factor completely: \(3x^5y−48xy\).
Solution:
\(\begin{array} {ll}
\text{1. Is there a GCF? Yes, }3xy. \text{ Factor out the GCF.} &3xy(x^4−16) \\
\text{2. The binomial in parentheses is a difference of squares.} &3xy\left((x^2)^2−(4)^2\right) \\
\quad \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\
\text{3. Is the expression factored completely? NO!} \\
\end{array}\)
\(\begin{array} {ll}
\text{2. The first binomial is a difference of squares.} &3xy\left((x)^2−(2)^2\right)(x^2+4) \\
\quad \text{Factor it as a product of conjugates.} &3xy(x−2)(x+2)(x^2+4) \\
\text{3. Is the expression factored completely? Yes.}
\end{array}\)
\( \quad \text{Check. } \quad 3xy(x−2)(x+2)(x^2+4) \text{ = } 3xy(x^2−4)(x^2+4) \text{ = } 3xy(x^4−16) \text{ = } 3x^5y−48xy\checkmark \)
Example \(\PageIndex{13.7x}\)
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 \(\PageIndex{13.8x}\): Remember to first look for a GCF
Factor completely: \(40x^2y+44xy−24y\).
Solution:
\(\begin{array} {ll} \text{1. Is there a GCF? Yes, } 4y \text{. Factor out the GCF. } &4y(10x^2+11x−6) \\
\text{2. Factor the trinomial with }a\neq 1. & 4y(5x−2)(2x+3) \\
\quad \text{Use the Reverse FOIL or AC Grouping Method} & \\
\text{3. Is the expression factored completely? Yes.} \end{array}\)
\( \quad \text{Check. } \quad 4y(5x−2)(2x+3) \text{ = } 4y(10x^2+11x−6) \text{ = } 40x^2y+44xy−24y \checkmark \)
Example \(\PageIndex{13.9x}\): 4 terms as a difference of a binomial square and another square
Factor completely: \(9x^2−12xy+4y^2−49\).
Solution:
\(\begin{array} {ll} \text{1. Is there a GCF? No.} & \\
\text{2. Four terms, use grouping. Grouping pairs doesn't work} \\
\quad \text{Three terms are squares. One square is negative. } & {\color{Cerulean}{9x^2}}−12xy+{\color{Cerulean}{4y^2}}−{\color{Red}{49}} \\
\quad \text{Group the other 3 terms.} & {\color{Cerulean}{(3x)^2 −12xy+ (2y)^2}} −49 \\
\quad \text{The trinomial fits the pattern, }a^2−2ab+b^2&\hspace{7mm} {\,}^{\searrow}{\,}_{−2(3x)(2y))}{\,}^{\swarrow} \\
\quad \text{Write the trinomial as a binomial square.} & {\color{Cerulean}{ (3x−2y)^2 }}−49 \\
\text{2. Two terms  a difference of squares.} & {\color{Red}{ (3x−2y) }}^2−7^2\\
\quad \text{Write it as a product of conjugates.} &({\color{Red}{ (3x−2y) }}−7)({\color{Red}{ (3x−2y) }}+7) \\
\text{3. Is the expression factored completely? Yes.} &(3x−2y−7)(3x−2y+7) \end{array}\)
\( \quad \text{Check. } \quad (3x−2y−7)(3x−2y+7) \\
\hspace{25mm} \text{ = } 9x^2−6xy−21x−6xy+4y^2+14y+21x−14y−49 \text{ = } 9x^2−12xy+4y^2−49\checkmark \)