# ##\$ Factor Trinomials

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## How to Factor Trinomials

1. Compare the trinomial to the general form $$ax^2+bx+c$$ and identify $$a$$, $$b$$, and $$c$$.
2. Find two numbers that add up to $$b$$ and multiply to $$ac$$.
3. Rewrite the middle term $$bx$$ as two terms and factor by grouping.
• As shown below, this step can be skipped if the leading coefficient is 1.

## Examples

Factor the trinomial:

$x^2+2x-48$

What are the coefficients of the trinomial?

When I compare $$\blue x^2+2x-48$$ to $$ax^2+bx+c$$, I can see that…

$\blue a=1$

$\blue b=2$

$\blue c=-48$

Which two numbers add up to $$b$$ and multiply to $$ac$$?

I need to find two numbers that multiply to $$\blue -48$$ and add to $$\blue 2$$ because…

$\blue ac = (1)(-48)=-48$

$\blue b = 2$

The factor pairs for $$\blue -48$$ are…

$$\blue(-1)(48)$$ or $$\blue(1)(-48)$$

$$\blue(-2)(24)$$ or $$\blue(2)(-24)$$

$$\blue(-3)(16)$$ or $$\blue(3)(-16)$$

$$\blue(-4)(12)$$ or $$\blue(4)(-12)$$

$$\blue(-6)(8)$$ or $$\blue(6)(-8)$$

When I look at all of these pairs, the only two numbers that add up to $$\blue 2$$ are $$\blue -6+8$$.

What is the factored form of $$\blue x^2+2x-48$$?

Using the numbers I found in Step 2, I can rewrite the polynomial as…

$x^2{\blue -6x+8x}-48$

Then I can factor by grouping…

First, I will split the terms of the polynomial into two groups:

${\blue (}x^2-6x{\blue )}+{\blue (}8x-48{\blue )}$

Then I will factor out the greatest common factor from each group.

The GCF of the first group is $$\blue x$$. The GCF of the second group is $$\blue 8$$.

${\blue x(}x-6{\blue )}+{\blue 8(}x-6{\blue )}$

The expressions in the parentheses match so I can factor them out.

${\blue(}x-6{\blue )}{\blue (x+8)}$

The factored form of $$\blue x^2+2x-48$$ is…

$\blue (x-6)(x+8)$

Factor the trinomial:

$6x^2-37x+56$

What are the coefficients of the trinomial?

When I compare $$\green 6x^2-37x+56$$ to $$ax^2+bx+c$$, I can see that…

$\green a=6$

$\green b=-37$

$\green c=56$

Which two numbers add up to $$b$$ and multiply to $$ac$$?

I need to find two numbers that multiply to $$\green 336$$ and add to $$\green -37$$ because…

$\green ac = (6)(56)=336$

$\green b = -37$

The factor pairs for $$\green 336$$ are…

$$\green(1)(336)$$ or $$\green(-1)(-336)$$

$$\green(2)(168$$ or $$\green(-2)(-168)$$

$$\green(3)(112$$ or $$\green(-3)(-112)$$

$$\green(4)(84$$ or $$\green(-4)(-84)$$

$$\green(6)(56$$ or $$\green(-6)(-56)$$

$$\green(7)(48$$ or $$\green(-7)(-48)$$

$$\green(8)(42$$ or $$\green(-8)(-42)$$

$$\green(12)(28$$ or $$\green(-12)(-28)$$

$$\green(14)(24) or \(\green(-14)(-24)$$

$$\green(16)(21$$ or $$\green(-16)(-21)$$

When I look at all of these pairs, the only two numbers that add up to $$\green -37$$ are $$\green -16+(-21)$$.

What is the factored form of $$\green 6x^2-37x+56$$?

Using the numbers I found in Step 2, I can rewrite the polynomial as…

$6x^2{\green -16x-21x}+56$

Then I can factor by grouping…

First, I will split the terms of the polynomial into two groups:

${\green (}6x^2-16x{\green )}+{\green (}-21x+56{\blue )}$

Then I will factor out the greatest common factor from each group.

The GCF of the first group is $$\green 2x$$. The GCF of the second group is $$\green -7$$.

Technically, you could say the GCF of the second group is $$7$$, but I chose to make it negative so that the expressions in the parentheses match.

${\green 2x(}3x-8{\green )}{\green -7(}3x-8{\blue )}$

The expressions in the parentheses match so I can factor them out.

${\green(}3x-8{\green )}{\green (2x-7)}$

The factored form of $$\green 6x^2-37x+56$$ is…

$\green (3x-8)(2x-7)$

Factor the trinomial:

$12x^6-x^3-6$

What are the coefficients of the trinomial?

When I compare $$\yellow 12x^6-x^3-6$$ to $$ax^2+bx+c$$, I can see that…

$\yellow a=12$

$\yellow b=-1$

$\yellow c=-6$

I also notice that there is an $$x^6$$ and $$x^3$$ instead of the standard $$x^2$$ and $$x$$. But the factoring process will still work because…

$x^6=(x^3)^2$

Which two numbers add up to $$b$$ and multiply to $$ac$$?

I need to find two numbers that multiply to $$\yellow -72$$ and add to $$\yellow -1$$ because…

$\yellow ac = (12)(-6)=-72$

$\yellow b = -1$

The factor pairs for $$\yellow -72$$ are…

$$\yellow(-1)(72)$$ or $$\yellow(1)(-72)$$

$$\yellow(-2)(36)$$ or $$\yellow(2)(-36)$$

$$\yellow(-3)(24)$$ or $$\yellow(3)(-24)$$

$$\yellow(-4)(18)$$ or $$\yellow(4)(-18)$$

$$\yellow(-6)(12)$$ or $$\yellow(6)(-12)$$

$$\yellow(-8)(9)$$ or $$\yellow(8)(-9)$$

When I look at all of these pairs, the only two numbers that add up to $$\yellow -1$$ are $$\yellow 8+(-9)$$.

What is the factored form of $$\yellow 12x^6-x^3-6$$?

Using the numbers I found in Step 2, I can rewrite the polynomial as…

$12x^6{\yellow 8x^3-9x^3}-6$

Then I can factor by grouping…

First, I will split the terms of the polynomial into two groups:

${\yellow (}12x^6+8x^3{\yellow )}+{\yellow (}-9x^3-6{\yellow)}$

Then I will factor out the greatest common factor from each group.

The GCF of the first group is $$\yellow 4x^3$$. The GCF of the second group is $$\yellow -3$$.

${\yellow 4x^3(}3x^3+2{\yellow )}{\yellow -3(}3x^3+2{\yellow )}$

The expressions in the parentheses match so I can factor them out.

${\yellow(}3x^3+2{\yellow )}{\yellow (4x^3-3)}$

The factored form of $$\yellow 12x^6-x^3-6$$ is…

$\yellow (3x^3+2)(4x^3-3)$

In the blue example above, you may have noticed that the two numbers found in Step 2 ($$\blue -6$$ and $$\blue 8$$) ended up in the final answer $$\blue (x-6)(x+8)$$.

This will happen anytime the leading coefficient (the number with the $$x^2$$) is 1. So, if the leading coefficient is 1, then you can skip Step 3 and write your answer with the numbers you found in Step 2.

## Rewriting b

It doesn’t matter which order you rewrite b. It will change the GCFs of the groups, but it will not change the final answer. For example, if you were factoring $$8x^2-2x-45$$, you would find out in Steps 1 and 2, that the numbers $$-20$$ and $$18$$ multiply to $$ac$$ and add to $$b$$.

However, in Step 3, you could rewrite $$b$$ two different ways:

• $$8x^2-20x+18x-45$$
• $$8x^2+18x-20x-45$$

And the way you rewrite $$b$$ determines how the factoring by grouping will go.

Option 1: $8x^2-20x+18x-45$

$8x^2-20x+18x-45$

First, I will split the terms of the polynomial into two groups:

${\blue (}8x^2-20x{\blue )}+{\blue (}18x-45{\blue )}$

Then I will factor out the greatest common factor from each group.

The GCF of the first group is $$\blue 4x$$. The GCF of the second group is $$\blue 9$$.

${\blue 4x(}2x-5{\blue )}+{\blue 9(}2x-5{\blue )}$

The expressions in the parentheses match so I can factor them out.

${\blue(}2x-5{\blue )}{\blue (4x+9)}$

The factored form of $$\blue 8x^2-20x+18x-45$$ is…

$\blue (2x-5)(4x+9)$

Option 2: $8x^2+18x-20x-45$

$8x^2+18x-20x-45$

First, I will split the terms of the polynomial into two groups:

${\blue (}8x^2+18x{\blue )}+{\blue (}-20x-45{\blue )}$

Then I will factor out the greatest common factor from each group.

The GCF of the first group is $$\blue 2x$$. The GCF of the second group is $$\blue -5$$.

${\blue 2x(}4x+9{\blue )}{\blue -5(}4x+9{\blue )}$

The expressions in the parentheses match so I can factor them out.

${\blue(}4x+9{\blue )}{\blue (2x-5)}$

The factored form of $$\blue 8x^2+18x-20x-45$$ is…

$\blue (4x+9)(2x-5)$

## Why It Works

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I like using the FOIL method to multiply binomials, but you can also use the box method or the multiplication algorithm. $(Qx+S)(Tx+R)$

Multiply the FIRST terms: $({\red Qx})({\red Tx})={\red QTx^2}$

Multiply the OUTER terms: $({\yellow Qx})({\yellow R})={\yellow QRx}$

Multiply the INNER terms: $({\green S})({\green Tx})={\green STx}$

Multiply the LAST terms: $({\blue S})({\blue R})={\blue SR}$

The $$\yellow QRx$$ and $$\green STx$$ terms are like terms because they both have an $$x$$ and in a real problem the coefficients $$QR$$ and $$ST$$ will be numbers.

So, the expanded polynomial would look like this:

${\red QTx^2}+({\yellow QR}+{\green ST})x+{\blue SR}$

${\red QTx^2}+({\yellow QR}+{\green ST})x+{\blue SR}$

When we compare this polynomial to $$\purple ax^2+bx+c$$, we can see that…

${\purple a} = {\red QT}$

${\purple b} = {\yellow QR}+{\green ST}$

${\purple c} = {\blue SR}$

When we look for two numbers that multiply to $$\purple ac$$ and add to $$\purple b$$, we are basically looking for $$\yellow QR$$ and $$\green ST$$.

This is because $${\purple b} = {\yellow QR}+{\green ST}$$ and $${\purple ac}={\red QT}{\blue SR}$$. The letters for $$\purple ac$$ can be multiplied in any order so we could say that $${\purple ac}={\yellow QR}{\green ST}$$.