How to Factor Trinomials
- Compare the trinomial to the general form \(ax^2+bx+c\) and identify \(a\), \(b\), and \(c\).
- Find two numbers that add up to \(b\) and multiply to \(ac\).
- 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 \({\purple ac}={\yellow QR}{\green ST}\).
