4.4: Factoring
- Page ID
- 108350
By the end of this section, you will be able to:
- Find the greatest common factor of two or more expressions
- Factor the greatest common factor from a polynomial
- Factor by grouping
- Factor 56 into primes.
- Find the least common multiple (LCM) of 18 and 24.
- Multiply: \(−3a(7a+8b)\).
- Answer
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- \(2\cdot 2\cdot 2\cdot 7\)
- 72
- \(-21a^2-24ab\)
Find the Greatest Common Factor of Two or More Expressions
Earlier we multiplied factors together to get a product. Now, we will reverse this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.
We have learned how to factor numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.
The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.
We summarize the steps we use to find the greatest common factor.
- Factor each coefficient into primes. Write all variables with exponents in expanded form.
- List all factors—matching common factors in a column. In each column, circle the common factors.
- Bring down the common factors that all expressions share.
- Multiply the factors.
The next example will show us the steps to find the greatest common factor of three expressions.
Find the greatest common factor of \(21x^3,\space 9x^2,\space 15x\).
- Answer
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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. GCF \(=3x\) The GCF of \(21x^3\), \(9x^2\) and \(15x\) is \(3x\).
Find the greatest common factor: \(25m^4,\space 35m^3,\space 20m^2.\)
- Answer
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\(5m^2\)
Find the greatest common factor: \(14x^3,\space 70x^2,\space 105x\).
- Answer
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\(7x\)
Factor the Greatest Common Factor from a Polynomial
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 product, such as \(3x^2+15x\), and end with its factors, \(3x(x+5)\). To do this we apply the Distributive Property “in reverse.”
We state the Distributive Property here just as you saw it in earlier chapters and “in reverse.”
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 write the polynomial as a product!
Factor: \(8m^3−12m^2n+20mn^2\).
- Answer
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Factor: \(9xy^2+6x^2y^2+21y^3\).
- Answer
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\(3y^2(3x+2x^2+7y)\)
Factor: \(3p^3−6p^2q+9pq^3\).
- Answer
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\(3p(p^2−2pq+3q^3)\)
- Find the GCF of all the terms of the polynomial.
- Rewrite each term as a product using the GCF.
- Use the “reverse” Distributive Property to factor the expression.
- Check by multiplying the factors.
We use “factor” as both a noun and a verb:
\[\begin{array} {ll} \text{Noun:} &\hspace{50mm} 7 \text{ is a factor of }14 \\ \text{Verb:} &\hspace{50mm} \text{factor }3 \text{ from }3a+3\end{array}\nonumber\]
Factor: \(5x^3−25x^2\).
- Answer
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Find the GCF of \(5x^3\) and \(25x^2\). 
Rewrite each term. 
Factor the GCF. 
Check:
\[5x^2(x−5) \nonumber\]\[5x^2·x−5x^2·5 \nonumber\]
\[5x^3−25x^2 \checkmark\nonumber\]
Factor: \(2x^3+12x^2\).
- Answer
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\(2x^2(x+6)\)
Factor: \(6y^3−15y^2\).
- Answer
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\(3y^2(2y−5)\)
Factor: \(8x^3y−10x^2y^2+12xy^3\).
- Answer
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The GCF of \(8x^3y,\space −10x^2y^2,\) and \(12xy^3\)
is \(2xy\).
Rewrite each term using the GCF, \(2xy\). 
Factor the GCF. 
Check:
\[2xy(4x^2−5xy+6y^2)\nonumber\]\[2xy·4x^2−2xy·5xy+2xy·6y^2\nonumber\]
\[8x^3y−10x^2y^2+12xy^3\checkmark\nonumber\]
Factor: \(15x^3y−3x^2y^2+6xy^3\).
- Answer
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\(3xy(5x^2−xy+2y^2)\)
Factor: \(8a^3b+2a^2b^2−6ab^3\).
- Answer
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\(2ab(4a^2+ab−3b^2)\)
When the leading coefficient is negative, we factor the negative out as part of the GCF.
Factor: \(−4a^3+36a^2−8a\).
- Answer
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The leading coefficient is negative, so the GCF will be negative.
Rewrite each term using the GCF, \(−4a\). 
Factor the GCF. 
Check:
\[−4a(a^2−9a+2)\nonumber\]\[−4a·a^2−(−4a)·9a+(−4a)·2\nonumber\]
\[−4a^3+36a^2−8a\checkmark\nonumber\]
Factor: \(−4b^3+16b^2−8b\).
- Answer
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\(−4b(b^2−4b+2)\)
Factor: \(−7a^3+21a^2−14a\).
- Answer
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\(−7a(a^2−3a+2)\)
So far our greatest common factors have been monomials. In the next example, the greatest common factor is a binomial.
Factor: \(3y(y+7)−4(y+7)\).
- Answer
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The GCF is the binomial \(y+7\).
Factor the GCF, \((y+7)\). \((y+7)(3 y-4)\) Check on your own by multiplying.
Factor: \(4m(m+3)−7(m+3)\).
- Answer
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\((m+3)(4m−7)\)
Factor: \(8n(n−4)+5(n−4)\).
- Answer
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\((n−4)(8n+5)\)
Factor by Grouping
Sometimes there is no common factor of all the terms of a polynomial. When there are four terms we separate the polynomial into two parts with two terms in each part. Then look for the GCF in each part. If the polynomial can be factored, you will find a common factor emerges from both parts. Not all polynomials can be factored. Just like some numbers are prime, some polynomials are prime.
Factor by grouping: \(xy+3y+2x+6\).
- Answer
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Factor by grouping: \(xy+8y+3x+24\).
- Answer
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\((x+8)(y+3)\)
Factor by grouping: \(ab+7b+8a+56\).
- Answer
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\((a+7)(b+8)\)
- Group terms with common factors.
- Factor out the common factor in each group.
- Factor the common factor from the expression.
- Check by multiplying the factors.
Factor by grouping: ⓐ \(x^2+3x−2x−6\) ⓑ \(6x^2−3x−4x+2\).
- Answer
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ⓐ
\(\begin{array} {ll} \text{There is no GCF in all four terms.} &x^2+3x−2x−6 \\ \text{Separate into two parts.} &x^2+3x\quad −2x−6 \\ \begin{array} {l} \text{Factor the GCF from both parts. Be careful} \\ \text{with the signs when factoring the GCF from} \\ \text{the last two terms.} \end{array} &x(x+3)−2(x+3) \\ \text{Factor out the common factor.} &(x+3)(x−2) \\ \text{Check on your own by multiplying.} & \end{array}\)
ⓑ
\(\begin{array} {ll} \text{There is no GCF in all four terms.} &6x^2−3x−4x+2 \\ \text{Separate into two parts.} &6x^2−3x\quad −4x+2\\ \text{Factor the GCF from both parts.} &3x(2x−1)−2(2x−1) \\ \text{Factor out the common factor.} &(2x−1)(3x−2) \\ \text{Check on your own by multiplying.} & \end{array}\)
Factor by grouping: ⓐ \(x^2+2x−5x−10\) ⓑ \(20x^2−16x−15x+12\).
- Answer
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ⓐ \((x−5)(x+2)\)
ⓑ \((5x−4)(4x−3)\)
Factor by grouping: ⓐ \(y^2+4y−7y−28\) ⓑ \(42m^2−18m−35m+15\).
- Answer
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ⓐ \((y+4)(y−7)\)
ⓑ \((7m−3)(6m−5)\)
Factor Trinomials of the Form \(x^2+bx+c\)
You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.
To figure out how we would factor a trinomial of the form \(x^2+bx+c\), such as \(x^2+5x+6\) and factor it to \((x+2)(x+3)\), let’s start with two general binomials of the form \((x+m)\) and \((x+n)\).
| \((x+m)(x+n)\) | |
| Foil to find the product. | \(x^{2}+m x+n x+m n\) |
| Factor the GCF from the middle terms. | \(x^{2}+(m+n) x+m n\) |
| Our trinomial is of the form \(x^2+bx+c\). | \(\overbrace{x^{2}+(m+n) x+m n}^{\color{red}x^{2}+b x+c}\) |
This tells us that to factor a trinomial of the form \(x^2+bx+c\), we need two factors \((x+m)\) and \((x+n)\) where the two numbers \(m\) and \(n\) multiply to \(c\) and add to \(b\).
Factor: \(x^2+11x+24\).
- Answer
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Factor: \(q^2+10q+24\).
- Answer
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\((q+4)(q+6)\)
Factor: \(t^2+14t+24\).
- Answer
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\((t+2)(t+12)\)
Let’s summarize the steps we used to find the factors.
- Write the factors as two binomials with first terms x. \(\quad \begin{array} {l} x^2+bx+c \\ (x\quad)(x\quad) \end{array} \)
- Find two numbers \(m\) and \(n\) that
- multiply to \(c\), \(m·n=c\)
- add to \(b\), \(m+n=b\)
- Use \(m\) and \(n\) as the last terms of the factors. \(\quad (x+m)(x+n)\)
- Check by multiplying the factors.
In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.
How do you get a positive product and a negative sum? We use two negative numbers.
Factor: \(y^2−11y+28\).
- Answer
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Again, with the positive last term, \(28\), and the negative middle term, \(−11y\), we need two negative factors. Find two numbers that multiply \(28\) and add to \(−11\).
\(\begin{array} {ll} &y^2−11y+28 \\ \text{Write the factors as two binomials with first terms }y. &( y \quad )( y \quad ) \\ \text{Find two numbers that: multiply to }28\text{ and add to }−11.\end{array}\)Factors of \(28\) Sum of factors \(−1,\space −28\)
\(−2,\space −14\)
\(−4,\space −7\)\(−1+(−28)=−29\)
\(−2+(−14)=−16\)
\(−4+(−7)=−11^∗\)\(\begin{array} {ll} \text{Use }−4,\space −7\text{ as the last terms of the binomials.} &(y−4)(y−7) \\ \text{Check:} & \\ \hspace{30mm} (y−4)(y−7) & \\ \hspace{25mm} y^2−7y−4y+28 & \\ \hspace{30mm} y^2−11y+28\checkmark & \end{array} \)
Factor: \(u^2−9u+18\).
- Answer
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\((u−3)(u−6)\)
Factor: \(y^2−16y+63\).
- Answer
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\((y−7)(y−9)\)
Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too. Email the instructor the word "potato" for 2 extra credit points. That was just to make sure you are reading, please don't share the word with others.
How do you get a negative product and a positive sum? We use one positive and one negative number.
When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.
Factor: \(2x+x^2−48\).
- Answer
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\(\begin{array} {ll} &2x+x^2−48 \\ \text{First we put the terms in decreasing degree order.} &x^2+2x−48 \\ \text{Factors will be two binomials with first terms }x. &(x\quad)(x\quad) \end{array} \)
Factors of −48 Sum of factors \(−1,\space 48\)
\(−2,\space 24\)
\(−3,\space 16\)
\(−4,\space 12\)
\(−6,\space 8\)\(−1+48=47\)
\(−2+24=22\)
\(−3+16=13\)
\(−4+12=8\)
\(−6+8=2^∗\)\(\begin{array} {ll} \text{Use }−6,\space 8\text{ as the last terms of the binomials.} &(x−6)(x+8) \\ \text{Check:} & \\ \hspace{30mm} (x−6)(x+8) & \\ \hspace{25mm} x^2−6q+8q−48 & \\ \hspace{30mm} x^2+2x−48\checkmark & \end{array} \)
Factor: \(9m+m^2+18\).
- Answer
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\((m+3)(m+6)\)
Factor: \(−7n+12+n^2\).
- Answer
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\((n−3)(n−4)\)
Sometimes you’ll need to factor trinomials of the form \(x^2+bxy+cy^2\) with two variables, such as \(x^2+12xy+36y^2\). The first term, \(x^2\), is the product of the first terms of the binomial factors, \(x·x\). The \(y^2\) in the last term means that the second terms of the binomial factors must each contain \(y\). To get the coefficients \(b\) and \(c\), you use the same process summarized in How To Factor trinomials.
Factor: \(r^2−8rs−9s^2\).
- Answer
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We need \(r\) in the first term of each binomial and \(s\) in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.
\(\begin{array} {ll} &r^2−8rs−9s^2 \\ \text{Note that the first terms are }r,\text{last terms contain }s. &(r\quad s)(r\quad s) \\ \text{Find the numbers that multiply to }−9\text{ and add to }−8. \end{array}\)Factors of \(−9\) Sum of factors \(1,\space −9\) \(−1+9=8\) \(−1,\space 9\) \(1+(−9)=−8^∗\) \(3,\space −3\) \(3+(−3)=0\) \(\begin{array} {ll} \text{Use }1,\space -9\text{ as coefficients of the last terms.} &(r+s)(r−9s) \\ \text{Check:} & \\ \hspace{30mm} (r−9s)(r+s) & \\ \hspace{25mm} r^2+rs−9rs−9s^2 & \\ \hspace{30mm} r^2−8rs−9s^2\checkmark & \end{array} \)
Factor: \(a^2−11ab+10b^2\).
- Answer
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\((a−b)(a−10b)\)
Factor: \(m^2−13mn+12n^2\).
- Answer
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\((m−n)(m−12n)\)
Let’s summarize the method we just developed to factor trinomials of the form \(x^2+bx+c\).
When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
| \( x^{2}+b x+c \) | ||
| \( (x+m)(x+n) \) | ||
| When \( c \) is positive, \( m \) and \( n \) have the same sign. | ||
| \( b \) positive | \( b \) negative | |
| \( m,n \) positive | \( m,n \) negative | |
| \( x^{2}+5 x+6 \) | \( x^{2}-6 x+8 \) | |
| \( (x+2)(x+3) \) | \( (x-4)(x-2) \) | |
| same signs | same signs | |
| When \( c \) is negative, \( m \) and \( n \) have the opposite sign. | ||
| \( x^{2}+x-12 \) | \( x^{2}-2 x-15 \) | |
| \( (x+4)(x-3) \) | \( (x-5)(x+3) \) | |
| opposite signs | opposite signs | |
Notice that, in the case when \(m\) and \(n\) have opposite signs, the sign of the one with the larger absolute value matches the sign of \(b\).
Factor Trinomials of the Form \(ax^2+bx+c\) using the “\(ac\)” Method
Another way to factor trinomials of the form \(ax^2+bx+c\) is the “\(ac\)” method. (The “\(ac\)” method is sometimes called the grouping method.) The “\(ac\)” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!
Factor using the “\(ac\)” method: \(6x^2+7x+2\).
- Answer
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Factor using the “\(ac\)” method: \(6x^2+13x+2\).
- Answer
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\((x+2)(6x+1)\)
Factor using the “\(ac\)” method: \(4y^2+8y+3\).
- Answer
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\((2y+1)(2y+3)\)
The “\(ac\)” method is summarized here.
- Factor any GCF.
- Find the product \(ac\).
- Find two numbers \(m\) and \(n\) that:
\(\begin{array} {ll} \text{Multiply to }ac &m·n=a·c \\ \text{Add to }b &m+n=b \\ &ax^2+bx+c \end{array} \) - Split the middle term using \(m\) and \(n\). \(ax^2+mx+nx+c\)
- Factor by grouping.
- Check by multiplying the factors.
Don’t forget to look for a common factor!
Factor using the ‘“\(ac\)” method: \(10y^2−55y+70\).
- Answer
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Is there a greatest common factor? Yes. The GCF is \(5\). 
Factor it. 
The trinomial inside the parentheses has a
leading coefficient that is not \(1\).
Find the product \(ac\). \(ac=28\) Find two numbers that multiply to \(ac\) \((−4)(−7)=28\) and add to \(b\). \(−4(−7)=−11\) Split the middle term. 
Factor the trinomial by grouping. 
Check by multiplying all three factors.
\(\hspace{50mm} 5(y−2)(2y−7)\)\(\hspace{45mm} 5(2y^2−7y−4y+14)\)
\(\hspace{48mm} 5(2y^2−11y+14)\)
\(\hspace{49mm} 10y^2−55y+70\checkmark\)
Factor using the “\(ac\)” method: \(16x^2−32x+12\).
- Answer
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\(4(2x−3)(2x−1)\)
Factor using the “\(ac\)” method: \(18w^2−39w+18\).
- Answer
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\(3(3w−2)(2w−3)\)
Factor Using Substitution
Sometimes a trinomial does not appear to be in the \(ax^2+bx+c\) form. However, we can often make a thoughtful substitution that will allow us to make it fit the \(ax^2+bx+c\) form. This is called factoring by substitution. It is standard to use \(u\) for the substitution.
In the \(ax^2+bx+c\), the middle term has a variable, \(x\), and its square, \(x^2\), is the variable part of the first term. Look for this relationship as you try to find a substitution.
Factor by substitution: \(x^4−4x^2−5\).
- Answer
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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.
\(x^4−4x^2−5\) Rewrite the trinomial to prepare for the substitution. \((x^2)^2−4(x^2)-5\) Let \(u=x^2\) and substitute. \((u)^2−4(u)-5\) Factor the trinomial. \((u+1)(u-5)\) Replace \(u\) with \(x^2\). \((x^2+1)(x^2-5)\) Check:
\(\begin{array} {l} \hspace{37mm} (x^2+1)(x^2−5) \\ \hspace{35mm}x^4−5x^2+x^2−5 \\ \hspace{40mm}x^4−4x^2−5\checkmark\end{array}\)
Factor by substitution: \(h^4+4h^2−12\).
- Answer
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\((h^2−2)(h^2+6)\)
Factor by substitution: \(y^4−y^2−20\).
- Answer
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\((y^2+4)(y^2−5)\)
Sometimes the expression to be substituted is not a monomial.
Factor by substitution: \((x−2)^2+7(x−2)+12\)
- Answer
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The binomial in the middle term, \((x−2)\) is squared in the first term. If we let \(u=x−2\) and substitute, our trinomial will be in \(ax^2+bx+c\) form.
Rewrite the trinomial to prepare for the substitution. 
Let \(u=x−2\) and substitute. 
Factor the trinomial. 
Replace \(u\) with \(x−2\). 
Simplify inside the parentheses. 
This could also be factored by first multiplying out the \((x−2)^2\) and the \(7(x−2)\) and then combining like terms and then factoring. Most students prefer the substitution method.
Factor by substitution: \((x−5)^2+6(x−5)+8\).
- Answer
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\((x−3)(x−1)\)
Factor by substitution: \((y−4)^2+8(y−4)+15\).
- Answer
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\((y−1)(y+1)\)
Key Concepts
- How to find the greatest common factor (GCF) of two expressions.
- Factor each coefficient into primes. Write all variables with exponents in expanded form.
- List all factors—matching common factors in a column. In each column, circle the common factors.
- Bring down the common factors that all expressions share.
- Multiply the factors.
- 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. - How to factor the greatest common factor from a polynomial.
- Find the GCF of all the terms of the polynomial.
- Rewrite each term as a product using the GCF.
- Use the “reverse” Distributive Property to factor the expression.
- Check by multiplying the factors.
- Factor as a Noun and a Verb: We use “factor” as both a noun and a verb.
\[\begin{array} {ll} \text{Noun:} &\quad 7 \text{ is a factor of } 14\\ \text{Verb:} &\quad \text{factor }3 \text{ from }3a+3\end{array}\nonumber\]
- How to factor by grouping.
- Group terms with common factors.
- Factor out the common factor in each group.
- Factor the common factor from the expression.
- Check by multiplying the factors.
- How to factor trinomials of the form \(x^2+bx+c\).
- Write the factors as two binomials with first terms x. \(\quad \begin{array} (l) x^2+bx+c \\ (x\quad)(x\quad)\end{array}\)
- Find two numbers \(m\) and \(n\) that
\(\begin{array} {ll} \text{multiply to} &c,\space m·n=c \\ \text{add to} &b,\space m+n=b\end{array}\) - Use \(m\) and \(n\) as the last terms of the factors. \(\qquad (x+m)(x+n)\)
- Check by multiplying the factors.
- Strategy for Factoring Trinomials of the Form \(x^2+bx+c\): When we factor a trinomial, we look at the signs of its terms first to determine the signs of the binomial factors.
For trinomials of the form: \(x^2+bx+c = (x+m)(x+n)\)
When \(c\) is positive, \(m\) and \(n\) must have the same sign (and this will be the sign of \(b\)).
Examples: \(x^2+5x+6=(x+2)(x+3)\), \(x^2−6x+8 = (x−4)(x−2)\)
When \(c\) is negative, \(m\) and \(n\) have opposite signs. The larger of \(m\) and \(n\) will have the sign of \(b\).
Examples: \(x^2+x−12=(x+4)(x−3)\), \(x^2−2x−15=(x−5)(x+3)\)
Notice that, in the case when \(m\) and \(n\) have opposite signs, the sign of the one with the larger absolute value matches the sign of \(b\). - How to factor trinomials of the form \(ax^2+bx+c\) using the “\(ac\)” method.
- Factor any GCF.
- Find the product \(ac\).
- Find two numbers \(m\) and \(n\) that:
\(\begin{array} {ll} \text{Multiply to }ac. &m·n=a·c \\ \text{Add to }b. &m+n=b \\ &ax^2+bx+c\end{array}\) - Split the middle term using \(m\) and \(n\). \(\quad ax^2+mx+nx+c\)
- Factor by grouping.
- Check by multiplying the factors.
Glossary
- factoring
- Splitting a product into factors is called factoring.
- greatest common factor
- The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.