8.6.4: Factoring Trinomials

Learning Outcome

  • Apply an algorithm to rewrite a trinomial as a four term polynomial and factor
  • Use factoring by grouping to factor a trinomial
  • Factor trinomials of the form [latex]a{x}^{2}+bx+c[/latex]
  • Factoring perfect square trinomials

Key words

  • Prime: having only two factors; itself and 1
  • Perfect square trinomial: any trinomial that factors into two identical binomials that can be written as the square of that binomial

In the last section, we looked at factoring trinomials of the form [latex]a{x}^{2}+bx+c[/latex] by grouping. We can use a similar method to factor trinomials of the form [latex]ax^2+bx+c[/latex].  The only difference is that we have to include the leading coefficient [latex]a[/latex] in the process.  We need to find to numbers [latex]r[/latex] and [latex]s[/latex] that are factors of the product [latex]ac[/latex] and that sum to [latex]b[/latex].

The trinomial [latex]2{x}^{2}+5x+3[/latex] can be rewritten as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[/latex] and then factor each portion of the expression to obtain [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression.

Below is a summary of the steps we will use followed by an example demonstrating how to use the steps.

Factoring Trinomials in the form [latex]ax^{2}+bx+c[/latex]

To factor a trinomial in the form [latex]ax^{2}+bx+c[/latex], find two integers, r and s, whose sum is b and whose product is ac.

[latex]\begin{array}{l}r\cdot{s}=a\cdot{c}\\r+s=b\end{array}[/latex]

Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial.

The first step in this process is to figure out what two numbers to use to re-write the [latex]x[/latex]-term as the sum of two new terms. Making a table to keep track of our work is helpful. We are looking for two numbers with a product of [latex]ac=6[/latex] and a sum of [latex]b=5[/latex]

Factors of [latex]ac=2\cdot3=6[/latex] Sum of Factors
[latex]1,6[/latex] [latex]7[/latex]
[latex]-1,-6[/latex] [latex]-7[/latex]
[latex]2,3[/latex] [latex]5[/latex]
[latex]-2,-3[/latex] [latex]-5[/latex]

The pair [latex]r=2 \text{ and }s=3[/latex] will give the correct [latex]x[/latex]-coefficient of [latex]5[/latex], so we will rewrite it using the new factors:

[latex]2{x}^{2}+5x+3=2x^2+2x+3x+3[/latex]

Now we can group the polynomial into two binomials.

[latex]2x^2+2x+3x+3=(2x^2+2x)+(3x+3)[/latex]

Identify the GCF of each binomial.

[latex]2x[/latex] is the GCF of [latex](2x^2+2x)[/latex] and [latex]3[/latex] is the GCF of [latex](3x+3)[/latex]. We use these this to rewrite the polynomial:

[latex](2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)[/latex]

The GCF of our new polynomial is [latex](x+1)[/latex]. We factor this out as well:

[latex]2x(x+1)+3(x+1)=(x+1)(2x+3)[/latex].

Sometimes it helps visually to write the polynomial as [latex](x+1)2x+(x+1)3[/latex] before we factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.

 

Now let’s see how this strategy works for factoring [latex]6z^{2}+11z+4[/latex].

In this trinomial, [latex]a=6[/latex], [latex]b=11[/latex], and [latex]c=4[/latex]. According to the strategy, we need to find two factors, [latex]r[/latex] and [latex]s[/latex], whose sum is [latex]b=11[/latex] and whose product is [latex]a\cdot{c}=6\cdot4=24[/latex]. We can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since [latex]ac[/latex] is positive and [latex]b[/latex] is positive, we can be certain that the two factors we’re looking for are also positive numbers.)

Factors whose product is  [latex]ac=24[/latex] Sum of the factors
[latex]1\cdot24=24[/latex] [latex]1+24=25[/latex]
[latex]2\cdot12=24[/latex] [latex]2+12=14[/latex]
[latex]3\cdot8=24[/latex] [latex]3+8=11[/latex]
[latex]4\cdot6=24[/latex] [latex]4+6=10[/latex]

There is only one combination where the product is [latex]24[/latex] and the sum is [latex]11[/latex], and that is when [latex]r=3[/latex], and [latex]s=8[/latex]. Let’s use these values to factor the original trinomial.

Example

Factor [latex]6z^{2}+11z+4[/latex].

Solution

Rewrite the middle term, [latex]11z[/latex], as [latex]3z + 8z[/latex] (from the chart above):

[latex]6z^{2}+3z+8z+4[/latex]

Group pairs. Use grouping to consider the terms in pairs:

[latex]\left(6z^{2}+3z\right)+\left(8z+4\right)[/latex]

Factor [latex]3z[/latex] out of the first group and [latex]4[/latex] out of the second group:

[latex]3z\left(2z+1\right)+4\left(2z+1\right)[/latex]

Factor out [latex]\left(2z+1\right)[/latex]:

[latex]\left(2z+1\right)\left(3z+4\right)[/latex]

Answer

[latex]\left(2z+1\right)\left(3z+4\right)[/latex]

Try It

Factor [latex]10y^2+23y+12[/latex]

Now, let’s look at an example where [latex]c[/latex] is negative.

Example

Factor [latex]5{x}^{2}+7x - 6[/latex] by grouping.

Solution

We have a trinomial with [latex]a=5,b=7[/latex], and [latex]c=-6[/latex]. First, determine [latex]ac=-30[/latex]. We need to find two numbers with a product of [latex]-30[/latex] and a sum of [latex]7[/latex]. In the table, we list factors until we find a pair with the desired sum.

Factors of [latex]-30[/latex] Sum of Factors
[latex]1,-30[/latex] [latex]-29[/latex]
[latex]-1,30[/latex] [latex]29[/latex]
[latex]2,-15[/latex] [latex]-13[/latex]
[latex]-2,15[/latex] [latex]13[/latex]
[latex]3,-10[/latex] [latex]-7[/latex]
[latex]-3,10[/latex] [latex]7[/latex]

So [latex]r=-3[/latex] and [latex]s=10[/latex].

[latex]\begin{array}{cc}\;\;\;\;5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+rx+sx+c\hfill \\ =x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}\hfill \\ =\left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}\hfill \end{array}[/latex]

Answer

[latex]5{x}^{2}+7x - 6=\left(5x - 3\right)\left(x+2\right)[/latex]

 

We can check our work by multiplying to confirm that [latex]\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6[/latex].

Try It

Factor [latex]4x^2-8x-5[/latex]

We can summarize our process in the following way:

factor a trinomial in the form [latex]a{x}^{2}+bx+c[/latex]

  1. List factors of [latex]ac[/latex].
  2. Find [latex]r[/latex] and [latex]s[/latex], a pair of factors of [latex]ac[/latex] with a sum of [latex]b[/latex].
  3. Rewrite the original expression as [latex]a{x}^{2}+rx+sx+c[/latex].
  4. Pull out the GCF of [latex]a{x}^{2}+rx[/latex].
  5. Pull out the GCF of [latex]sx+c[/latex].
  6. Factor out the GCF of the expression.

Try It

The following video presents another example of factoring a trinomial using grouping.  In this example, the [latex]x[/latex]-term, [latex]b[/latex], is negative. Note how having a negative middle term and a positive constant term influences the options for r and s when factoring.

Example

Factor [latex]2{x}^{2}+9x+9[/latex].

Solution

Find two numbers [latex]r[/latex] and [latex]s[/latex] such that [latex]r\cdot{s}=18[/latex] and [latex]r + s = 9[/latex].

[latex]9[/latex] and [latex]18[/latex] are both positive, so we will only consider positive factors.

Factors of [latex]2\cdot9=18[/latex] Sum of Factors
[latex]1, 18[/latex] [latex]19[/latex]
[latex]3,6[/latex] [latex]9[/latex]

We can stop because we have found our factors.

Rewrite the original expression and group:

[latex]\;\;\;\;2x^2+3x+6x+9 \\= (2x^2+3x)+(6x+9)[/latex]

Factor out the GCF of each binomial and write as a product of two binomials:

[latex]\;\;\;\;(2x^2+3x)+(6x+9)\\=x(2x+3)+3(2x+3)\\=(x+3)(2x+3)[/latex]

Answer

[latex]2{x}^{2}+9x+9=(x+3)(2x+3)[/latex]

Try It

Factor [latex]12x^2+25x+7[/latex]

Here is an example where the constant term is negative.

Example

Factor [latex]6{x}^{2}+x - 1[/latex].

Solution

Determine [latex]ac=6(-1)=-6[/latex]. We need two factors of [latex]ac[/latex] that sum to [latex]b=1[/latex].

Factors of [latex]6\cdot-1=-6[/latex] Sum of Factors
[latex]-1,6[/latex] [latex]5[/latex]
[latex]1,-6[/latex] [latex]-5[/latex]
[latex]-2,3[/latex] [latex]1[/latex]

We can stop because we have found our factors.

Rewrite the original expression and group:

[latex]\;\;\;\;6{x}^{2}+x - 1\\=6x^2-2x+3x-1[/latex]

Factor out the GCF of each binomial and write as a product of two binomials:

[latex]\;\;\;\;(6x^2-2x)+(3x-1)\\=2x(3x-1)+1(3x-1)\\=(2x+1)(3x-1)[/latex]

Answer

[latex]6{x}^{2}+x - 1=(2x+1)(3x-1)[/latex]

Try It

Factor [latex]6x^2+53x-9[/latex]

Prime Trinomials

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Indeed, there are more trinomials that can’t be factored than trinomials that can be factored. Take the trinomial [latex]2z^{2}+35z+7[/latex], for instance. Are there two integers whose sum is [latex]b=35[/latex] and whose product is [latex]a\cdot{c}=2\cdot7=14[/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.  We will sometimes encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials.

Example

Factor [latex]7x^{2}-16x–5[/latex].

Solution

Find [latex]r, s[/latex] such that [latex]r\cdot{s}=-35\text{ and }r+s=-16[/latex]:

Factors of [latex]7\cdot{-5}=-35[/latex] Sum of Factors
[latex]-1, 35[/latex] [latex]34[/latex]
[latex]1, -35[/latex] [latex]-34[/latex]
[latex]-5, 7[/latex] [latex]2[/latex]
[latex]-7,5[/latex] [latex]-2[/latex]

We have exhausted all integer pairs, so the trinomial cannot be factored. None of the factors add up to [latex]-16[/latex].

Answer

[latex]7x^{2}-16x–5[/latex] is prime.

Try It

Factor [latex]5x^2-3x+1[/latex]

Factoring out a GCF

Consider [latex]2x^{2}+10x+12[/latex]. Since [latex]ac=2\cdot 12=24[/latex] and [latex]b=10[/latex], the values [latex]r=6[/latex] and [latex]s=4[/latex] can be used to rewrite the trinomial as [latex]2x^2+6x+4x+12[/latex]. Factoring each pair leads to [latex]2x(x+3)+4(x+3)=(2x+4)(x+3)[/latex].  But notice that this is not completely factored because the binomial [latex]2x+4[/latex] can be further factored by pulling out the greatest common factor of [latex]2[/latex] resulting in [latex]2(x+2)[/latex]. This means that [latex]2x^{2}+10x+12[/latex] factors to [latex]2(x+2)(x+3)[/latex].

We really should have noticed the GCF right away: [latex]2x^{2}+10x+12=2(x^2+5x+6)[/latex]. Pulling the GCF out first makes the rest of the factoring easier as we are dealing with smaller values of [latex]a, b[/latex] and [latex]c[/latex].  In this case,

[latex]\;\;\;\;2x^{2}+10x+12\\=2(x^2+5x+6)\\=2(x+3)(x+2)[/latex]

Exactly the same answer we got before.

Consequently, we should always start factoring by looking for a GCF for all terms in the trinomial. Once we have identified and pulled out the greatest common factor, we can factor the remaining trinomial as usual.

Example

Factor [latex]3x^{3}–3x^{2}–90x[/latex].

Solution

Since [latex]3x[/latex] is a common factor for the three terms, factor out the[latex]3x[/latex] first:

[latex]3x\left(x^{2}–x–30\right)[/latex]

Now we can factor the trinomial [latex]x^{2}–x–30[/latex]. To find r and s, identify two numbers whose product is [latex]ac=−30[/latex] and whose sum is [latex]b=−1[/latex].

The pair of factors is [latex]−6[/latex] and [latex]5[/latex]. So replace [latex]–x[/latex] with [latex]−6x+5x[/latex]:

[latex]3x\left(x^{2}–6x+5x–30\right)[/latex]

Use grouping to consider the terms in pairs:

[latex]3x\left[\left(x^{2}–6x\right)+\left(5x–30\right)\right][/latex]

Factor [latex]x[/latex] out of the first group and factor [latex]5[/latex] out of the second group:

[latex]3x\left[\left(x\left(x–6\right)\right)+5\left(x–6\right)\right][/latex]

Then factor out [latex]x–6[/latex]:

[latex]3x\left(x–6\right)\left(x+5\right)[/latex]

Answer

[latex]3x^{3}–3x^{2}–90x=3x\left(x–6\right)\left(x+5\right)[/latex]

Try It

Factor [latex]30x^4+27x^3-27x^2[/latex]

Factoring out a Negative

In some situations, [latex]a[/latex] is negative, as in [latex]−4h^{2}+11h+3[/latex]. It often makes sense to factor out [latex]−1[/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[/latex] from negative to positive, making the remaining trinomial easier to factor.

Example

Factor [latex]−4h^{2}+11h+3[/latex]

Solution

Factor [latex]−1[/latex] out of the trinomial. Notice that the signs of all three terms will change to their opposites:

[latex]−4h^{2}+11h+3[/latex]

[latex]=−1\left(4h^{2}–11h–3\right)[/latex]

To factor the trinomial, we need to figure out how to rewrite [latex]−11h[/latex]. The product [latex]rs=4\cdot−3=−12[/latex], and the sum [latex]r+s=−11[/latex].

[latex]r\cdot{s}=−12[/latex] [latex]r+s=−11[/latex]
[latex]−12\cdot1=−12[/latex] [latex]−12+1=−11[/latex]
[latex]−6\cdot2=−12[/latex] [latex]−6+2=−4[/latex]
[latex]−4\cdot3=−12[/latex] [latex]−4+3=−1[/latex]

Rewrite the middle term [latex]−11h[/latex] as [latex]−12h+1h[/latex]:

[latex]−1\left(4h^{2}–12h+1h–3\right)[/latex]

Group terms:

[latex]−1\left[\left(4h^{2}–12h\right)+\left(1h–3\right)\right][/latex]

Factor out [latex]4[/latex]h from the first pair. The only common factor of the second group is [latex]1[/latex], so we can write it as [latex]+1\left(h–3\right)[/latex] since [latex]+1\left(h–3\right)=\left(h–3\right)[/latex]. This helps with factoring in the next step.

[latex]−1\left[4h\left(h–3\right)+1\left(h–3\right)\right][/latex]

Factor out a common factor of [latex]\left(h–3\right)[/latex]. Notice we are left with [latex]\left(h–3\right)\left(4h+1\right)[/latex]; the [latex]+1[/latex] comes from the term [latex]+1\left(h–3\right)[/latex] in the previous step.

[latex]−1\left[\left(h–3\right)\left(4h+1\right)\right][/latex]

Answer

[latex]−4h^{2}+11h+3=−1\left(h–3\right)\left(4h+1\right)[/latex]

The following video shows an example where the leading term is negative.  We will see how, by factoring out the negative sign, factoring the trinomial becomes easier.

Try It

Factor [latex]-60x^5-35x^4+60x^3[/latex]

Try It

Factor [latex]-9x^3+24x^2-16x[/latex]

Notice that [latex]-9x^3+24x^2-16x[/latex] factors into [latex]-x(3x-4)(3x-4)[/latex], which can be written as [latex]-x(3x-4)^2[/latex].

[latex](3x-4)^2[/latex] is the square of the binomial [latex](3x-4)[/latex]. When [latex](3x-4)[/latex] is squared we get [latex]{(3x-4)}^{2}=(3x-4)(3x-4)=9x^2-24x+16[/latex]. The trinomial [latex]9x^2-24x+16[/latex] is known as a perfect square trinomial.

Example

Factor [latex]3x^4-18x^3+27x^2[/latex]

Solution

Look for a GCF first: [latex]3x^2[/latex]

[latex]\;\;\;\;3x^4-18x^3+27x^2[/latex]

[latex]=3x^2(x^2-6x+9)[/latex]

[latex]ac=9[/latex] and [latex]b=-6[/latex]:

[latex]ac=9[/latex] Sum = [latex]6[/latex]
9, 1 10
-9, -1  -10
-9, 1 -8
9, -1 8
3, 3 6
-3, -3 -6

Rewrite [latex]-6x[/latex] as [latex]-3x-3x[/latex]:

[latex]3x^2(x^2-3x-3x+9)[/latex]

Group and pull out the common factors:

[latex]3x^2\left [ x(x-3)-3(x-3)\right ] [/latex]

Pull out the common factor [latex](x-3)[/latex]:

[latex]3x^2(x-3)(x-3)[/latex]

Rewrite the binomials as a perfect square:

[latex]3x^2{(x-3)}^{2}[/latex]

Answer

[latex]3x^4-18x^3+27x^2=3x^2(x-3)^2[/latex]

Try It

Factor [latex]-25x^2+20x-4[/latex]