##$$$ 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…

\[\yellow (3x^3+2)(4x^3-3)\]

Leading Coefficient of 1

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

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 \({\purple ac}={\yellow QR}{\green ST}\).

Printable Worksheets

Online Practice