6.2 Factor Trinomials

Learning objectives

Learning objectives

• Factor trinomials of the form ${x}^{2}+bx+c$

• Factor trinomials of the form $a{x}^{2}+bx+c$ using trial and error

• Factor trinomials of the form $a{x}^{2}+bx+c$ using the ‘ac’ method

• Factor using substitution

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 $(x+n).$

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 6.9

Let’s summarize the steps we used to find 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 6.10

Solution

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 $-11.$

${y}^{2}-11y+28$

Write the factors as two binomials with first terms $y.$ · $(y\,)(y\,)$

Find two numbers that: multiply to 28 and add to −11.

Factors of $28$ · Sum of factors $-1,-28$$-2,-14$$-4,-7$ · $-1+(-28)=-29$$-2+(-14)=-16$$\,-4+(-7)={-11}^{*}$

Use $-4,-7$ as the last terms of the binomials. · $(y-4)(y-7)$

Check: $\,\begin{array}{c} (y-4)(y-7) \\ {y}^{2}-7y-4y+28 \\ {y}^{2}-11y+28\mathrm{✓} \\ \end{array}$

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 6.11

Solution

$2x+{x}^{2}-48$

First we put the terms in decreasing degree order. · ${x}^{2}+2x-48$ Factors will be two binomials with first terms $x.$ · $(x\,)(x\,)$

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}^{*}$

$\mathrm{Use}\,-6,8\,\mathrm{as the last terms of the binomials.}$ · $(x-6)(x+8)$

Check: $\,\begin{array}{c} (x-6)(x+8) \\ {x}^{2}-6q+8q-48 \\ {x}^{2}+2x-48\mathrm{✓} \\ \end{array}$

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, $x\cdot{}x.$ 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 6.12

Solution

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.

${r}^{2}-8rs-9{s}^{2}$

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.

Factors of $-9$ · Sum of factors $\,1,-9$ · $\,-1+9=8$$-1,9$ · $1+(-9)=-{8}^{*}$$\,3,-3$ · $3+(-3)=0$

$\mathrm{Use}\,1,-9\,\mathrm{as coefficients of the last terms.}$ · $(r+s)(r-9s)$

Check: $\,\begin{array}{c} (r-9s)(r+s) \\ {r}^{2}+rs-9rs-9{s}^{2} \\ {r}^{2}-8rs-9{s}^{2}\mathrm{✓} \\ \end{array}$

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 6.13

Solution

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.

${u}^{2}-9uv-12{v}^{2}$

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.

Factors of $-12$ · 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.

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

Factor Trinomials of the form ax${}^{2}$ + 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 6.14

Solution

Is there a greatest common factor? · $4{x}^{3}+16{x}^{2}-20x$ Yes, $\mathrm{GCF}=4x.$ 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 thatmultiply to −5 and add to 4. · $4x(x-1)(x+5)$

Factors of $-5$ · Sum of factors $-1,5$$\,1,-5$ · $\,-1+5={4}^{*}$$1+(-5)=-4$

• Check:$\,\begin{array}{c} 4x(x-1)(x+5) \\ 4x({x}^{2}+5x-x-5) \\ 4x({x}^{2}+4x-5) \\ 4{x}^{3}+16{x}^{2}-20x\mathrm{✓} \\ \end{array}$

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.

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.

Check: Does $1x\cdot{}3x=3{x}^{2}?$

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.

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

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

Our result of the factoring is:

Example 6.15

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

Example 6.16

Solution

Consider all the combinations of factors.

Possible factors · Product $(b-1)(6b-5)$ · $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)$ · $6{b}^{2}-17b+5$

The correct factors are those whose productis the original trinomial. · $(2b-1)(3b-5)$

Check by multiplying:$\,\begin{array}{c} (2b-1)(3b-5) \\ 6{b}^{2}-10b-3b+5 \\ 6{b}^{2}-13b+5\mathrm{✓} \\ \end{array}$

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 6.17

Solution

Consider all the combinations of factors.

The correct factors are those whose product is the original trinomial. · $(2x-3y)(9x-5y)$

Check by multiplying: $\,\begin{array}{c} (2x-3y)(9x-5y) \\ 18{x}^{2}-10xy-27xy+15{y}^{2} \\ 18{x}^{2}-37xy+15{y}^{2}\mathrm{✓} \\ \end{array}$

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

Example 6.18

Solution

Consider all the combinations.

The correct factors are those whose productis the original trinomial. Remember to includethe factor $-5{y}^{2}.$ · $-5{y}^{2}(y+4)(2y+3)$

Check by multiplying: $\,\begin{array}{c} -5{y}^{2}(y+4)(2y+3) \\ -5{y}^{2}(2{y}^{2}+8y+3y+12) \\ -10{y}^{4}-55{y}^{3}-60{y}^{2}\mathrm{✓} \\ \end{array}$

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 6.19

The “ac” method is summarized here.

Don’t forget to look for a common factor!

Example 6.20

Solution

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 6.21

Solution

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

Sometimes the expression to be substituted is not a monomial.

Example 6.22

Solution

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.

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.

Section 6.2 Exercises

Section 6.2 Exercises

Factor Trinomials of the Form ${x}^{2}+bx+c$

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

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

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

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

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

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

Factor Using Substitution

In the following exercises, factor using substitution.

Mixed Practice

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

Writing Exercises

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

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

1. Factor: ${q}^{2}+10q+24.$

2. Factor: ${t}^{2}+14t+24.$

3. Factor: ${u}^{2}-9u+18.$

4. Factor: ${y}^{2}-16y+63.$

5. Factor: $9m+{m}^{2}+18.$

1. Factor: $-7n+12+{n}^{2}.$

2. Factor: ${a}^{2}-11ab+10{b}^{2}.$

3. Factor: ${m}^{2}-13mn+12{n}^{2}.$

4. Factor: ${x}^{2}-7xy-10{y}^{2}.$

5. Factor: ${p}^{2}+15pq+20{q}^{2}.$

1. Factor completely: $5{x}^{3}+15{x}^{2}-20x.$

2. Factor completely: $6{y}^{3}+18{y}^{2}-60y.$

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

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

5. Factor completely using trial and error: $8{x}^{2}-14x+3.$

1. Factor completely using trial and error: $10{y}^{2}-37y+7.$

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

3. Factor completely using trial and error: $30{x}^{2}-53xy-21{y}^{2}.$

4. Factor completely using trial and error: $15{n}^{3}-85{n}^{2}+100n.$

5. Factor completely using trial and error: $56{q}^{3}+320{q}^{2}-96q.$

1. Factor using the ‘ac’ method: $6{x}^{2}+13x+2.$

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

3. Factor using the ‘ac’ method: $16{x}^{2}-32x+12.$

4. Factor using the ‘ac’ method: $18{w}^{2}-39w+18.$

5. Factor by substitution: ${h}^{4}+4{h}^{2}-12.$

1. Factor by substitution: ${y}^{4}-{y}^{2}-20.$

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

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

4. ${p}^{2}+11p+30$

5. ${n}^{2}+19n+48$

1. ${a}^{2}+25a+100$

2. ${x}^{2}-8x+12$

3. ${y}^{2}-18y+45$

4. ${x}^{2}-8x+7$

5. $5p-6+{p}^{2}$

1. $8-6x+{x}^{2}$

2. ${x}^{2}-12-11x$

3. ${x}^{2}-2xy-80{y}^{2}$

4. ${m}^{2}-64mn-65{n}^{2}$

5. ${a}^{2}+5ab-24{b}^{2}$

1. ${x}^{2}-3xy-14{y}^{2}$

2. ${m}^{2}-5mn+30{n}^{2}$

3. ${p}^{3}-8{p}^{2}-20p$

4. $3{m}^{3}-21{m}^{2}+30m$

5. $5{x}^{4}+10{x}^{3}-75{x}^{2}$

1. $2{t}^{2}+7t+5$

2. $11{x}^{2}+34x+3$

3. $4{w}^{2}-5w+1$

4. $4{q}^{2}-7q-2$

5. $6{p}^{2}-19pq+10{q}^{2}$

1. $4{a}^{2}+17ab-15{b}^{2}$

2. $-16{x}^{2}-32x-16$

3. $-30{q}^{3}-140{q}^{2}-80q$

4. $5{n}^{2}+21n+4$

5. $4{k}^{2}-16k+15$

1. $6{y}^{2}+y-15$

2. $2{n}^{2}-27n-45$

3. $60{y}^{2}+290y-50$

4. $48{z}^{3}-102{z}^{2}-45z$

5. $16{s}^{2}+40s+24$

1. $48{y}^{2}+12y-36$

2. ${x}^{4}-6{x}^{2}-7$

3. ${x}^{4}-3{x}^{2}-28$

4. ${(x-3)}^{2}-5(x-3)-36$

5. ${(3y-2)}^{2}-(3y-2)-2$

1. ${u}^{2}-12u+36$

2. ${r}^{2}-20rs+64{s}^{2}$

3. $12{y}^{2}-29y+14$

4. $6{n}^{2}+5n-4$

5. $13{z}^{2}+39z-26$

1. $3{p}^{2}+21p$

2. $6{r}^{2}+30r+36$

3. $24{n}^{2}+20n+4$

4. ${x}^{4}-4{x}^{2}-12$

5. ${(x+3)}^{2}-9(x+3)-36$

1. 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.

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