MX Algebra Factor Trinomials

Section 6.2Factor Trinomials

Definition

Before you get started, take this readiness quiz.

1

Find all the factors of 72.
If you missed this problem, review Example 2.

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Definition
2

Find the product: \((3y+4)(2y+5).\)
If you missed this problem, review Example 4.

\(6{y}^{2}+23y+20\)

Definition
3

Simplify: \(-9(6);\) \(-9(-6).\)
If you missed this problem, review Example 7.

−54, 54

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.

Figure shows the equation open parentheses x plus 2 close parentheses open parentheses x plus 3 close parentheses equals x squared plus 5 x plus 6. The left side of the equation is labeled factors and the right is labeled product. An arrow pointing right is labeled multiply. An arrow pointing left is labeled factor.

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 \((x+n).\)

Table 1
The image displays the algebraic expression (x + m)(x + n), which represents the product of two binomials. This is a common form in algebra, often encountered when expanding or factoring quadratic equations.
Foil to find the product.A mathematical expression showing x squared plus mx plus nx plus mn.
Factor the GCF from the middle terms.An algebraic expression is shown: x² + (m + n)x + mn. This is a quadratic expression with variables x, m, and n, demonstrating the expanded form of (x+m)(x+n).
Our trinomial is of the form \({x}^{2}+bx+c.\)This image illustrates the relationship between the standard quadratic form x^2 + bx + c and its factored form components x^2 + (m + n)x + mn, highlighting b = m + n and c = mn.

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.

Example 1How to Factor a Trinomial of the form \({x}^{2}+bx+c\)

Factor: \({x}^{2}+11x+24.\)

Find two numbers that multiply to 24 and add to 11.

Step 1 is to write the factors of x squared plus 11x plus 24 as two binomials with first terms x. Write two sets of parentheses and put x as the first term. Step 2 is to find two numbers m and n that multiply to c, m times n is c and add to b, m plus n is b. So, find two numbers that multiply to 24 and add to 11. Factors of 24 are 1 and 24, 2 and 12, 3 and 8, 4 and 6. Sum of factors: 1 plus 24 is 25, 2 plus 12 is 14, 3 plus 8 is 11 and 4 plus 6 is 10. Step 3 is to use m and n, in this case, 3 and 8, as the last terms of the binomials. So we get open parentheses x plus 3 close parentheses open parentheses x plus 8 close parentheses Step 4 is to check by multiplying the factors to get the original polynomial.
Try It #1

Factor: \({q}^{2}+10q+24.\)

\((q+4)(q+6)\)

Did you get it?
Try It #2

Factor: \({t}^{2}+14t+24.\)

\((t+2)(t+12)\)

Did you get it?

Let’s summarize the steps we used to find the factors.

Factor trinomials of the form \({x}^{2}+bx+c.\)
  • Write the factors as two binomials with first terms x. \(\,\begin{array}{l}{x}^{2}+bx+c \\ (x\,)(x\,)\end{array}\)
  • 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. \(\,(x+m)(x+n)\)
  • 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.

Example 2

Factor: \({y}^{2}-11y+28.\)

Since the last term is positive and the middle term is negative, look for two negative numbers that multiply to 28 and add to −11.

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 \(-11.\)


Table 2
\({y}^{2}-11y+28\)
Write the factors as two binomials with first terms \(y.\)\((y\,)(y\,)\)
Find two numbers that: multiply to 28 and add to −11.
Table 3
Factors of \(28\)Sum of factors
\(\text{-}1,-28\)

\(\text{-}2,-14\)

\(\text{-}4,-7\)		\(\text{-}1+(\text{-}28)=-29\)

\(\text{-}2+(\text{-}14)=-16\)

\(\,\text{-}4+(\text{-}7)={-11}^{*}\)
Table 4
Use \(-4,-7\) as the last terms of the binomials.\((y-4)(y-7)\)
Check:
\(\,\begin{array}{l} \\ \\ (y-4)(y-7) \\ {y}^{2}-7y-4y+28 \\ {y}^{2}-11y+28✓\end{array}\)
Try It #3

Factor: \({u}^{2}-9u+18.\)

\((u-3)(u-6)\)

Did you get it?
Try It #4

Factor: \({y}^{2}-16y+63.\)

\((y-7)(y-9)\)

Did you get it?

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.

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.

Example 3

Factor: \(2x+{x}^{2}-48.\)

Rewrite the terms in descending degree order before looking for m and n.



Table 5
\(2x+{x}^{2}-48\)
First we put the terms in decreasing degree order.\({x}^{2}+2x-48\)
Factors will be two binomials with first terms \(x.\)\((x\,)(x\,)\)
Table 6
Factors of \(-48\)Sum of factors
\(-1,48\)
\(-2,24\)
\(-3,16\)
\(-4,12\)
\(-6,8\)		\(-1+48=47\)
\(-2+24=22\)
\(-3+16=13\)
\(-4+12=8\)
\(\,-6+8={2}^{*}\)
Table 7
\(\text{Use}\,-6,8\,\text{as the last terms of the binomials.}\)\((x-6)(x+8)\)
Check:
\(\,\begin{array}{l} \\ \\ (x-6)(x+8) \\ {x}^{2}-6q+8q-48 \\ {x}^{2}+2x-48✓\end{array}\)
Try It #5

Factor: \(9m+{m}^{2}+18.\)

\((m+3)(m+6)\)

Did you get it?
Try It #6

Factor: \(-7n+12+{n}^{2}.\)

\((n-3)(n-4)\)

Did you get it?

Sometimes you’ll need to factor trinomials of the form \({x}^{2}+bxy+c{y}^{2}\) with two variables, such as \({x}^{2}+12xy+36{y}^{2}.\) 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.

Example 4

Factor: \({r}^{2}-8rs-9{s}^{2}.\)

Since the last term is negative, look for two numbers with opposite signs that multiply to −9 and add to −8.

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.


Table 8
\({r}^{2}-8rs-9{s}^{2}\)
Note that the first terms are \(r,\) last terms contain \(s.\)\((r\,s)(r\,s)\)
Find the numbers that multiply to −9 and add to −8.
Table 9
Factors of \(-9\)Sum of factors
\(\,1,-9\)\(\,-1+9=8\)
\(-1,9\)\(1+(-9)=\text{-}{8}^{*}\)
\(\,3,-3\)\(3+(-3)=0\)
Table 10
\(\text{Use}\,1,-9\,\text{as coefficients of the last terms.}\)\((r+s)(r-9s)\)
Check:
\(\,\begin{array}{l}(r-9s)(r+s) \\ {r}^{2}+rs-9rs-9{s}^{2} \\ {r}^{2}-8rs-9{s}^{2}✓\end{array}\)
Try It #7

Factor: \({a}^{2}-11ab+10{b}^{2}.\)

\((a-b)(a-10b)\)

Did you get it?
Try It #8

Factor: \({m}^{2}-13mn+12{n}^{2}.\)

\((m-n)(m-12n)\)

Did you get it?

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Example 5

Factor: \({u}^{2}-9uv-12{v}^{2}.\)

List every factor pair of −12 and check each pair's sum against −9.

We need u in the first term of each binomial and v in the second term. The last term of the trinomial is negative, so the factors must have opposite signs.

Table 11
\({u}^{2}-9uv-12{v}^{2}\)
Note that the first terms are \(u,\) last terms contain \(v.\)\((u\,v)(u\,v)\)
Find the numbers that multiply to −12 and add to −9.
Table 12
Factors of \(\text{-}12\)Sum of factors
\(\,1,-12\)
\(-1,12\)
\(\,2,-6\)
\(-2,6\)
\(\,3,-4\)
\(-3,4\)		\(1+(-12)=-11\)
\(\,-1+12=11\)
\(\,2+(-6)=-4\)
\(\,-2+6=4\)
\(\,3+(-4)=-1\)
\(\,-3+4=1\)

Note there are no factor pairs that give us \(-9\) as a sum. The trinomial is prime.

Try It #9

Factor: \({x}^{2}-7xy-10{y}^{2}.\)

prime

Did you get it?
Try It #10

Factor: \({p}^{2}+15pq+20{q}^{2}.\)

prime

Did you get it?

Let’s summarize the method we just developed to factor trinomials of the form \({x}^{2}+bx+c.\)

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.

\[\begin{array}{l}\begin{array}{l}{x}^{2}+bx+c \\ (x+m)(x+n)\end{array} \\ \text{When}\,\text{c}\,\text{is positive,}\,\text{m}\,\text{and}\,\text{n}\,\text{have the same sign.} \\ b\,\text{positive}\,b\,\text{negative} \\ m,n\,\text{positive}\,m,n\,\text{negative} \\ {x}^{2}+5x+6\,{x}^{2}-6x+8 \\ (x+2)(x+3)\,(x-4)(x-2) \\ \text{same signs}\,\text{same signs} \\ \text{When}\,\text{c}\,\text{is negative,}\,\text{m}\,\text{and}\,\text{n}\,\text{have opposite signs.} \\ {x}^{2}+x-12\,{x}^{2}-2x-15 \\ (x+4)(x-3)\,(x-5)(x+3) \\ \text{opposite signs}\,\text{opposite signs}\end{array}\]

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 ax2 + bx + c using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form \(a{x}^{2}+bx+c.\)

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

Example 6

Factor completely: \(4{x}^{3}+16{x}^{2}-20x.\)

Check for a GCF first and factor it out before dealing with the trinomial.



Table 13
Is there a greatest common factor?\(4{x}^{3}+16{x}^{2}-20x\)
Yes, \(\text{GCF}=4x.\) Factor it.\(4x({x}^{2}+4x-5)\)
Binomial, trinomial, or more than three terms?
It is a trinomial. So “undo FOIL.”\(4x(x\,)(x\,)\)
Use a table like the one shown to find two numbers that
multiply to −5 and add to 4.		\(4x(x-1)(x+5)\)
Table 14
Factors of \(\text{-}5\)Sum of factors
\(\text{-}1,5\)
\(\,1,-5\)		\(\,\text{-}1+5={4}^{*}\)
\(1+(\text{-}5)=-4\)
Table 15
Check:
\(\,\begin{array}{l} \\ \\ 4x(x-1)(x+5) \\ 4x({x}^{2}+5x-x-5) \\ 4x({x}^{2}+4x-5) \\ 4{x}^{3}+16{x}^{2}-20x✓\end{array}\)
Try It #11

Factor completely: \(5{x}^{3}+15{x}^{2}-20x.\)

\(5x(x-1)(x+4)\)

Did you get it?
Try It #12

Factor completely: \(6{y}^{3}+18{y}^{2}-60y.\)

\(6y(y-2)(y+5)\)

Did you get it?

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial \(3{x}^{2}+5x+2.\)

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

\[\begin{array}{l} \\ \\ 3{x}^{2}+5x+2 \\ (\,)(\,)\end{array}\]

We know the first terms of the binomial factors will multiply to give us \(3{x}^{2}.\) The only factors of \(3{x}^{2}\) are \(1x,3x.\) We can place them in the binomials.

The polynomial is 3x squared plus 5x plus 2. There are two pairs of parentheses, with the first terms in them being x and 3x.

Check: Does \(1x·3x=3{x}^{2}?\)

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.

Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses.

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

Figure shows the polynomial 3x squared plus 5x plus 2 and two possible pairs of factors. One is open parentheses x plus 1 close parentheses open parentheses 3x plus 2 close parentheses. The other is open parentheses x plus 2 close parentheses open parentheses 3x plus 1 close parentheses. In each case, arrows are shown pairing the first term of the first factor with the last term of the second factor and the first term of the second factor with the last term of the first factor.

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

\[\begin{array}{l}(x+1)(3x+2) \\ 3{x}^{2}+2x+3x+2 \\ 3{x}^{2}+5x+2✓\end{array}\]

Our result of the factoring is:

\[\begin{array}{l} \\ 3{x}^{2}+5x+2 \\ (x+1)(3x+2)\end{array}\]

Example 7How to Factor a Trinomial Using Trial and Error

Factor completely using trial and error: \(3{y}^{2}+22y+7.\)

List the factor pairs of the first term's coefficient and of the last term, then test combinations.

Step 1 is to write the trinomial in descending order. The trinomial 3 y squared plus 22y plus 7 is already in descending order. Step 2 is to factor the GCF. Here, there is none. Step 3 is Find all the factor pairs of the first term. The only factors here are 1y and 3y. Since there is only one pair, we can put each as the first term in the parentheses. Step 4 is to find all the factor pairs of the third term. Here, the only pair is 1 and 7. Step 5 is to test all the possible combinations of the factors until the correct product is found. For possible factors open parentheses y plus 1 close parentheses open parentheses 37 plus 7 close parentheses, the product is 3 y squared plus 10y plus 7. For the possible factors open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses, the product is 3 y squared plus 22y plus 7, which is the correct product. Hence, the correct factors are open parentheses y plus 7 close parentheses open parentheses 3y plus 1 close parentheses. Step 6 is to check by multiplying.
Try It #13

Factor completely using trial and error: \(2{a}^{2}+5a+3.\)

\((a+1)(2a+3)\)

Did you get it?
Try It #14

Factor completely using trial and error: \(4{b}^{2}+5b+1.\)

\((b+1)(4b+1)\)

Did you get it?
Factor trinomials of the form \(a{x}^{2}+bx+c\) using trial and error.
  • Write the trinomial in descending order of degrees as needed.
  • Factor any GCF.
  • Find all the factor pairs of the first term.
  • Find all the factor pairs of the third term.
  • Test all the possible combinations of the factors until the correct product is found.
  • Check by multiplying.

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Example 8

Factor completely using trial and error: \(6{b}^{2}-13b+5.\)

List the factor pairs of \(6{b}^{2}\) and of 5, using negative factors of 5 since the middle term is negative.

Table 16
The trinomial is already in descending order.The image displays the quadratic expression 6b^2 - 13b + 5.
Find the factors of the first term.A quadratic expression, 6b^2 - 13b + 5, is shown with potential factors for the 6b^2 term listed below it: 1b * 6b and 2b * 3b, in red text.
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.		The image shows the quadratic expression 6b^2 - 13b + 5, with potential factors for the first term (1b*6b, 2b*3b) and the constant term (-1, -5) listed below, indicating the process of factoring.

Consider all the combinations of factors.

Table 17
\(6{b}^{2}-13b+5\)
Possible factorsProduct
\((b-1)(6b-5)\)\(6{b}^{2}-11b+5\)
\((b-5)(6b-1)\)\(6{b}^{2}-31b+5\)
\((2b-1)(3b-5)\)\(6{b}^{2}-13b+{5}^{*}\)
\((2b-5)(3b-1)\)\(6{b}^{2}-17b+5\)
Table 18
The correct factors are those whose product
is the original trinomial.		\((2b-1)(3b-5)\)
Check by multiplying:
\(\,\begin{array}{l} \\ \\ (2b-1)(3b-5) \\ 6{b}^{2}-10b-3b+5 \\ 6{b}^{2}-13b+5✓\end{array}\)
Try It #15

Factor completely using trial and error: \(8{x}^{2}-14x+3.\)

\((2x-3)(4x-1)\)

Did you get it?
Try It #16

Factor completely using trial and error: \(10{y}^{2}-37y+7.\)

\((2y-7)(5y-1)\)

Did you get it?

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 9

Factor completely using trial and error: \(18{x}^{2}-37xy+15{y}^{2}.\)

List the factor pairs of the first and last terms, choosing negative factors since the middle term is negative and the last term is positive.

Table 19
The trinomial is already in descending order.The image shows the algebraic expression 18x^2 - 37xy + 15y^2.
Find the factors of the first term.Steps to factor the trinomial 18x^2 - 37xy + 15y^2 are shown, with initial factor pairs for the 18x^2 term listed as 1x*18x, 2x*9x, and 3x*6x.
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.		An image displaying the algebraic expression 18x^2 - 37xy + 15y^2, with potential factors for the first and last terms shown in red below it, indicating steps for factoring the trinomial.

Consider all the combinations of factors.

This table shows the possible factors and corresponding products of the trinomial 18 x squared minus 37xy plus 15 y squared. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: open parentheses x minus 1y close parentheses open parentheses 18x minus 15y close parentheses, highlighted. Factor, open parentheses x minus 15y close parentheses open parentheses 18x minus 1y close parentheses; product: 18 x squared minus 271xy plus 15 y squared. Factor open parentheses x minus 3y close parentheses open parentheses 18x minus 5 y close parentheses; product: 18 x squared minus 59xy plus 15 y squared. Factor: open parentheses x minus 5y close parentheses open parentheses 18x minus 3y close parentheses highlighted. Factor: open parentheses 2x minus 1y close parentheses open parentheses 9x minus 15y close parentheses highlighted. Factor: open parentheses 2x minus 15y close parentheses open parentheses 9x minus 1y close parentheses; product 18 x squared minus 137 xy plus 15y squared. Factor: open parentheses 2x minus 3y close parentheses open parentheses 9x minus 5y close parentheses; product: 18 x squared minus 37xy plus 15 y squared, which is the original trinomial. Factor: open parentheses 2x minus 57 close parentheses open parentheses 9x minus 3y close parentheses highlighted. Factor: open parentheses 3x minus 1y close parentheses open parentheses 6x minus 15y close parentheses highlighted. Factor: open parentheses 3x minus 15y close parentheses highlighted open parentheses 6x minus 1y close parentheses. Factor: open parentheses 3x minus 3y close parentheses highlighted open parentheses 6x minus 5y.
Table 20
The correct factors are those whose product is the original trinomial.\((2x-3y)(9x-5y)\)
Check by multiplying:
\(\,\begin{array}{l} \\ \\ (2x-3y)(9x-5y) \\ 18{x}^{2}-10xy-27xy+15{y}^{2} \\ 18{x}^{2}-37xy+15{y}^{2}✓\end{array}\)
Try It #17

Factor completely using trial and error \(18{x}^{2}-3xy-10{y}^{2}.\)

\((3x+2y)(6x-5y)\)

Did you get it?
Try It #18

Factor completely using trial and error: \(30{x}^{2}-53xy-21{y}^{2}.\)

\((3x+y)(10x-21y)\)

Did you get it?

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

Example 10

Factor completely using trial and error: \(-10{y}^{4}-55{y}^{3}-60{y}^{2}.\)

Factor out the greatest common factor, including the negative sign, before factoring the remaining trinomial.

Table 21
A mathematical expression: negative ten y to the fourth power, minus fifty-five y cubed, minus sixty y squared.
Notice the greatest common factor, so factor it first.negative five y squared times open parenthesis two y squared plus eleven y plus twelve close parenthesis
Factor the trinomial.A mathematical expression -5y²(2y² + 11y + 12) illustrating steps to factor the quadratic trinomial. Red text shows factors for 2y² (y*2y) and factors for 12 (1*12, 2*6, 3*4).

Consider all the combinations.

This table shows the possible factors and product of the trinomial 2 y squared plus 11y plus 12. In some pairs of factors, when one factor contains two terms with a common factor, that factor is highlighted. In such cases, product is not an option because if trinomial has no common factors, then neither factor can contain a common factor. Factor: y plus 1, 2y plus 12 highlighted. Factor: y plus 12, 2y plus 1; product: 2 y squared plus 25y plus 12. Factor: y plus 2, 2y plus 6 highlighted. Factor: y plus 6, 2y plus 2 highlighted. Factor: y plus 3, 2y plus 4 highlighted. Factor: y plus 4, 2y plus 3; product: 2 y squared plus 11y plus 12. This is the original trinomial.
Table 22
The correct factors are those whose product
is the original trinomial. Remember to include
the factor \(\text{-}5{y}^{2}.\)		\(\text{-}5{y}^{2}(y+4)(2y+3)\)
Check by multiplying:
\(\,\begin{array}{l} \\ \\ \text{-}5{y}^{2}(y+4)(2y+3) \\ \text{-}5{y}^{2}(2{y}^{2}+8y+3y+12) \\ \text{-}10{y}^{4}-55{y}^{3}-60{y}^{2}✓\end{array}\)
Try It #19

Factor completely using trial and error: \(15{n}^{3}-85{n}^{2}+100n.\)

\(5n(n-4)(3n-5)\)

Did you get it?
Try It #20

Factor completely using trial and error: \(56{q}^{3}+320{q}^{2}-96q.\)

\(8q(q+6)(7q-2)\)

Did you get it?

Factor Trinomials of the Form \(a{x}^{2}+bx+c\) using the “ac” Method

Another way to factor trinomials of the form \(a{x}^{2}+bx+c\) 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!

Example 11How to Factor Trinomials using the “ac” Method

Factor using the ‘ac’ method: \(6{x}^{2}+7x+2.\)

Find the product ac, then find two numbers that multiply to ac and add to b.

Step 1 is to factor the GCF. There is none in 6 x squared plus 7x plus 2. Step 2 is to find the product of a and c. The product of 6 and 2 is 12. Step 3 is to find 2 numbers m and n such that mn is ac and m plus n is b. So we need to numbers that multiply to 12 and add to 7. Both factors must be positive. 3 times 4 is 12 and 3 plus 4 is 7. Step 4 is to split the middle term using m and n. So we rewrite 7 x as 3x plus 4x. It would give the same result if we used 4x plus 3x. Rewriting, we get 6 x squared plus 3x plus 4x plus 2. Notice that this is the same as the original polynomial. We just split the middle term to get a more useful form Step 5 is to factor by grouping. So, we get, 3x open parentheses 2x plus 1 close parentheses plus 2 open parentheses 2x plus 1 close parentheses. This is equal to 2x plus 1, 3x plus 2. Step 6 is to check by multiplying the factors.
Try It #21

Factor using the ‘ac’ method: \(6{x}^{2}+13x+2.\)

\((x+2)(6x+1)\)

Did you get it?
Try It #22

Factor using the ‘ac’ method: \(4{y}^{2}+8y+3.\)

\((2y+1)(2y+3)\)

Did you get it?

The “ac” method is summarized here.

Factor trinomials of the form \(a{x}^{2}+bx+c\) using the “ac” method.
  • Factor any GCF.
  • Find the product ac.
  • Find two numbers m and n that:
    \(\begin{array}{llllll}\text{Multiply to}\,ac & & & & & m·n=a·c \\ \text{Add to}\,b & & & & & m+n=b \\ & & & & & a{x}^{2}+bx+c\end{array}\)
  • Split the middle term using m and n. \(\,a{x}^{2}+mx+nx+c\)
  • Factor by grouping.
  • Check by multiplying the factors.

Don’t forget to look for a common factor!

Example 12

Factor using the ‘ac’ method: \(10{y}^{2}-55y+70.\)

Factor out the GCF first, then apply the "ac" method to what's left inside the parentheses.

Table 23
Is there a greatest common factor?
Yes. The GCF is 5.The image shows the quadratic expression 10y^2 - 55y + 70.
Factor it.The image shows the mathematical expression 5(2y^2 - 11y + 14).
The trinomial inside the parentheses has a
leading coefficient that is not 1.		The image displays two algebraic expressions: 'ax^2 + bx + c' in red, representing a general quadratic equation, and '5(2y^2 - 11y + 14)' in black, a factored quadratic expression.
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.Mathematical expression 5(2y^2 - 11y + 14) with arrows highlighting the -11y term, likely for breaking it down during factorization.
The algebraic expression 5(2y^2 - 7y - 4y + 14) is shown, with blue brackets indicating the grouping of terms (2y^2 - 7y) and (-4y + 14), typically a step in factorization.
Factor the trinomial by grouping.A mathematical expression featuring the difference of two products, 5(y(2y - 7)) - 2(2y - 7)), suitable for algebraic manipulation or solving.
A mathematical expression displays 5 multiplied by the quantity (y minus 2), which is then multiplied by the quantity (2y minus 7), representing a factored quadratic expression.
Check by multiplying all three factors.

\(\,\begin{array}{l}5(y-2)(2y-7) \\ 5(2{y}^{2}-7y-4y+14) \\ 5(2{y}^{2}-11y+14) \\ 10{y}^{2}-55y+70✓\end{array}\)
Try It #23

Factor using the ‘ac’ method: \(16{x}^{2}-32x+12.\)

\(4(2x-3)(2x-1)\)

Did you get it?
Try It #24

Factor using the ‘ac’ method: \(18{w}^{2}-39w+18.\)

\(3(3w-2)(2w-3)\)

Did you get it?

Factor Using Substitution

Sometimes a trinomial does not appear to be in the \(a{x}^{2}+bx+c\) form. However, we can often make a thoughtful substitution that will allow us to make it fit the \(a{x}^{2}+bx+c\) form. This is called factoring by substitution. It is standard to use u for the substitution.

In the \(a{x}^{2}+bx+c,\) 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.

Example 13

Factor by substitution: \({x}^{4}-4{x}^{2}-5.\)

Let \(u={x}^{2}\) so the expression becomes a factorable trinomial in u.

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 \(a{x}^{2}+bx+c\) form we need to factor it.

Table 24
A mathematical expression reads 'x to the power of 4 minus 4x squared minus 5'.
Rewrite the trinomial to prepare for the substitution.A mathematical expression showing a quadratic in terms of x squared: (x^2)^2 - 4(x^2) - 5. The 'x^2' terms are highlighted in red.
Let \(u={x}^{2}\) and substitute.The image shows the mathematical expression u squared minus 4u minus 5, with the 'u' characters in a reddish hue and the numbers and minus signs in grey/black.
Factor the trinomial.The mathematical expression (u+1)(u-5) is shown, representing the product of two binomials.
Replace u with \({x}^{2}.\)The image displays the mathematical expression (x^2 + 1)(x^2 - 5), which represents the product of two binomials involving x squared.
Check:

\(\,\begin{array}{l}({x}^{2}+1)({x}^{2}-5) \\ {x}^{4}-5{x}^{2}+{x}^{2}-5 \\ {x}^{4}-4{x}^{2}-5✓\end{array}\)
Try It #25

Factor by substitution: \({h}^{4}+4{h}^{2}-12.\)

\(({h}^{2}-2)({h}^{2}+6)\)

Did you get it?
Try It #26

Factor by substitution: \({y}^{4}-{y}^{2}-20.\)

\(({y}^{2}+4)({y}^{2}-5)\)

Did you get it?

Sometimes the expression to be substituted is not a monomial.

Example 14

Factor by substitution: \({(x-2)}^{2}+7(x-2)+12\)

Let \(u=x-2\) so the expression becomes a factorable trinomial in u.

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 \(a{x}^{2}+bx+c\) form.

Table 25
A mathematical expression is displayed, which reads as quantity x minus 2 squared, plus 7 times quantity x minus 2, plus 12. This is a quadratic expression in terms of (x-2).
Rewrite the trinomial to prepare for the substitution.A mathematical expression reads as 'open parenthesis x minus 2 close parenthesis squared plus 7 open parenthesis x minus 2 close parenthesis plus 12'.
Let \(u=x-2\) and substitute.A quadratic expression is displayed against a white background, reading 'u^2 + 7u + 12' with the 'u' characters in red.
Factor the trinomial.An algebraic expression showing the product of two binomials, (u+3) and (u+4).
Replace u with \(x-2.\)A mathematical expression featuring (x-2) plus 3 times (x-2) plus 4. The 'x-2' terms are highlighted in red, indicating a common factor within the polynomial expression.
Simplify inside the parentheses.The mathematical expression (x+1)(x+2) is displayed in black text on a white background.

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.

Try It #27

Factor by substitution: \({(x-5)}^{2}+6(x-5)+8.\)

\((x-3)(x-1)\)

Did you get it?
Try It #28

Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)

\((y-1)(y+1)\)

Did you get it?
Media

Access this online resource for additional instruction and practice with factoring.

Key Concepts

Section Exercises

Practice Makes Perfect

Factor Trinomials of the Form \({x}^{2}+bx+c\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bx+c.\)

4

\({p}^{2}+11p+30\)

\((p+5)(p+6)\)

5

\({w}^{2}+10w+21\)

6

\({n}^{2}+19n+48\)

\((n+3)(n+16)\)

7

\({b}^{2}+14b+48\)

8

\({a}^{2}+25a+100\)

\((a+5)(a+20)\)

9

\({u}^{2}+101u+100\)

10

\({x}^{2}-8x+12\)

\((x-2)(x-6)\)

11

\({q}^{2}-13q+36\)

12

\({y}^{2}-18y+45\)

\((y-3)(y-15)\)

13

\({m}^{2}-13m+30\)

14

\({x}^{2}-8x+7\)

\((x-1)(x-7)\)

15

\({y}^{2}-5y+6\)

16

\(5p-6+{p}^{2}\)

\((p-1)(p+6)\)

17

\(6n-7+{n}^{2}\)

18

\(8-6x+{x}^{2}\)

\((x-4)(x-2)\)

19

\(7x+{x}^{2}+6\)

20

\({x}^{2}-12-11x\)

\((x-12)(x+1)\)

21

\(-11-10x+{x}^{2}\)

In the following exercises, factor each trinomial of the form \({x}^{2}+bxy+c{y}^{2}.\) If the trinomial cannot be factored, answer “Prime.”

22

\({x}^{2}-2xy-80{y}^{2}\)

\((x+8y)(x-10y)\)

23

\({p}^{2}-8pq-65{q}^{2}\)

24

\({m}^{2}-64mn-65{n}^{2}\)

\((m+n)(m-65n)\)

25

\({p}^{2}-2pq-35{q}^{2}\)

26

\({a}^{2}+5ab-24{b}^{2}\)

\((a+8b)(a-3b)\)

27

\({r}^{2}+3rs-28{s}^{2}\)

28

\({x}^{2}-3xy-14{y}^{2}\)

Prime

29

\({u}^{2}-8uv-24{v}^{2}\)

30

\({m}^{2}-5mn+30{n}^{2}\)

Prime

31

\({c}^{2}-7cd+18{d}^{2}\)

Factor Trinomials of the Form \(a{x}^{2}+bx+c\) Using Trial and Error

In the following exercises, factor completely using trial and error.

32

\({p}^{3}-8{p}^{2}-20p\)

\(p(p-10)(p+2)\)

33

\({q}^{3}-5{q}^{2}-24q\)

34

\(3{m}^{3}-21{m}^{2}+30m\)

\(3m(m-5)(m-2)\)

35

\(11{n}^{3}-55{n}^{2}+44n\)

36

\(5{x}^{4}+10{x}^{3}-75{x}^{2}\)

\(5{x}^{2}(x-3)(x+5)\)

37

\(6{y}^{4}+12{y}^{3}-48{y}^{2}\)

38

\(2{t}^{2}+7t+5\)

\((2t+5)(t+1)\)

39

\(5{y}^{2}+16y+11\)

40

\(11{x}^{2}+34x+3\)

\((11x+1)(x+3)\)

41

\(7{b}^{2}+50b+7\)

42

\(4{w}^{2}-5w+1\)

\((4w-1)(w-1)\)

43

\(5{x}^{2}-17x+6\)

44

\(4{q}^{2}-7q-2\)

\((4q+1)(q-2)\)

45

\(10{y}^{2}-53y-11\)

46

\(6{p}^{2}-19pq+10{q}^{2}\)

\((2p-5q)(3p-2q)\)

47

\(21{m}^{2}-29mn+10{n}^{2}\)

48

\(4{a}^{2}+17ab-15{b}^{2}\)

\((4a-3b)(a+5b)\)

49

\(6{u}^{2}+5uv-14{v}^{2}\)

50

\(-16{x}^{2}-32x-16\)

\(-16(x+1)(x+1)\)

51

\(-81{a}^{2}+153a+18\)

52

\(-30{q}^{3}-140{q}^{2}-80q\)

\(-10q(3q+2)(q+4)\)

53

\(-5{y}^{3}-30{y}^{2}+35y\)

Factor Trinomials of the Form \(a{x}^{2}+bx+c\) using the ‘ac’ Method

In the following exercises, factor using the ‘ac’ method.

54

\(5{n}^{2}+21n+4\)

\((5n+1)(n+4)\)

55

\(8{w}^{2}+25w+3\)

56

\(4{k}^{2}-16k+15\)

\((2k-3)(2k-5)\)

57

\(5{s}^{2}-9s+4\)

58

\(6{y}^{2}+y-15\)

\((3y+5)(2y-3)\)

59

\(6{p}^{2}+p-22\)

60

\(2{n}^{2}-27n-45\)

\((2n+3)(n-15)\)

61

\(12{z}^{2}-41z-11\)

62

\(60{y}^{2}+290y-50\)

\(10(6y-1)(y+5)\)

63

\(6{u}^{2}-46u-16\)

64

\(48{z}^{3}-102{z}^{2}-45z\)

\(3z(8z+3)(2z-5)\)

65

\(90{n}^{3}+42{n}^{2}-216n\)

66

\(16{s}^{2}+40s+24\)

\(8(2s+3)(s+1)\)

67

\(24{p}^{2}+160p+96\)

68

\(48{y}^{2}+12y-36\)

\(12(4y-3)(y+1)\)

69

\(30{x}^{2}+105x-60\)

Factor Using Substitution

In the following exercises, factor using substitution.

70

\({x}^{4}-6{x}^{2}-7\)

\(({x}^{2}+1)({x}^{2}-7)\)

71

\({x}^{4}+2{x}^{2}-8\)

72

\({x}^{4}-3{x}^{2}-28\)

\(({x}^{2}-7)({x}^{2}+4)\)

73

\({x}^{4}-13{x}^{2}-30\)

74

\({(x-3)}^{2}-5(x-3)-36\)

\((x-12)(x+1)\)

75

\({(x-2)}^{2}-3(x-2)-54\)

76

\({(3y-2)}^{2}-(3y-2)-2\)

\((3y-4)(3y-1)\)

77

\({(5y-1)}^{2}-3(5y-1)-18\)

Mixed Practice

In the following exercises, factor each expression using any method.

78

\({u}^{2}-12u+36\)

\((u-6)(u-6)\)

79

\({x}^{2}-14x-32\)

80

\({r}^{2}-20rs+64{s}^{2}\)

\((r-4s)(r-16s)\)

81

\({q}^{2}-29qr-96{r}^{2}\)

82

\(12{y}^{2}-29y+14\)

\((4y-7)(3y-2)\)

83

\(12{x}^{2}+36y-24z\)

84

\(6{n}^{2}+5n-4\)

\((2n-1)(3n+4)\)

85

\(3{q}^{2}+6q+2\)

86

\(13{z}^{2}+39z-26\)

\(13({z}^{2}+3z-2)\)

87

\(5{r}^{2}+25r+30\)

88

\(3{p}^{2}+21p\)

\(3p(p+7)\)

89

\(7{x}^{2}-21x\)

90

\(6{r}^{2}+30r+36\)

\(6(r+2)(r+3)\)

91

\(18{m}^{2}+15m+3\)

92

\(24{n}^{2}+20n+4\)

\(4(2n+1)(3n+1)\)

93

\(4{a}^{2}+5a+2\)

94

\({x}^{4}-4{x}^{2}-12\)

\(({x}^{2}+2)({x}^{2}-6)\)

95

\({x}^{4}-7{x}^{2}-8\)

96

\({(x+3)}^{2}-9(x+3)-36\)

\((x-9)(x+6)\)

97

\({(x+2)}^{2}-25(x+2)-54\)

Writing Exercises

98

Many trinomials of the form \({x}^{2}+bx+c\) factor into the product of two binomials \((x+m)(x+n).\) Explain how you find the values of m and n.

Answers will vary.

99

Tommy factored \({x}^{2}-x-20\) as \((x+5)(x-4).\) Sara factored it as \((x+4)(x-5).\) Ernesto factored it as \((x-5)(x-4).\) Who is correct? Explain why the other two are wrong.

100

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form \(a{x}^{2}+bx+c.\)

Answers will vary.

101

How is the “ac” method similar to the “undo FOIL” method? How is it different?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 4 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: factor trinomials of the form x squared plus bx plus c, factor trinomials of the form a x squared plus b x plus c using trial and error, factor trinomials of the form a x squared plus bx plus c with using the “ac” method, factor using substitution.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?