8.6.3: Factoring Trinomials with Leading Coefficients of 1

Learning Outcomes

  • Factor a trinomial with a leading coefficient of [latex]1[/latex]
  • Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[/latex]

Key words

  • Leading term: the term with the highest degree
  • Leading coefficient: the number in the leading term

Factoring Trinomials

Now that we know how to factor by grouping, we are going to factor a trinomial whose leading coefficient is [latex]1[/latex].  In particular, we will look at trinomials of the form [latex]x^2+bx+c[/latex], where [latex]b, c[/latex] are real numbers.  Polynomials whose leading coefficients are 1 can be factored using the grouping method that we used in the previous section.  For example, [latex]x^2+5x+4[/latex] can be written as [latex]x^2+x+4x+4[/latex], which can then be factored using grouping: [latex]x^2+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)[/latex].  The question is, how do we know how to split up the [latex]x[/latex]-term?  If we multiply the generic binomials [latex](x+r)(x+s)[/latex], we get [latex]x(x+s)+r(x+s)=x^2+sx+rx+rs=x^2+(r+s)x+rs[/latex].

If we now compare this to the generic trinomial [latex]x^2+bx+c[/latex], we can equate the [latex]x[/latex]-terms so that [latex]r+s=b[/latex] and we can equate the constants so that [latex]c=rs[/latex].  In other words, we split up the [latex]x[/latex]-term [latex]bx[/latex] into [latex]rx+sx[/latex] where [latex]r+s=b[/latex] and [latex]rs=c[/latex].

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

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

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

Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\left(x+r\right)[/latex] and [latex]\left(x+s\right)[/latex].

Let’s factor the trinomial [latex]x^{2}+5x+6[/latex]. In this polynomial, [latex]b=5[/latex] and [latex]c=6[/latex]. We look for two numbers that add to 5 and multiply to 6. A chart will help us organize possibilities. On the left, list all possible factors of the [latex]c[/latex]-term, [latex]6[/latex]; on the right we’ll find the sums.

Factors whose product is  [latex]6[/latex] Sum of the factors
[latex]1\cdot6=6[/latex] [latex]1+6=7[/latex]
[latex]2\cdot3=6[/latex] [latex]2+3=5[/latex]
[latex]-1\cdot -6=6[/latex] [latex]-1+(-6)=-7[/latex]
[latex]-2\cdot -3=6[/latex] [latex]-2+(-3)=-5[/latex]

There are only two possible factor combinations with positive factors, [latex]1[/latex] and [latex]6[/latex], or [latex]2[/latex] and [latex]3[/latex], but there are also two possible factor combinations with negative factors, [latex]-1[/latex] and [latex]-6[/latex], or [latex]-2[/latex] and [latex]-3[/latex] . We can see that [latex]2+3=5[/latex]. So [latex]2x+3x=5x[/latex], giving us the correct [latex]x[/latex]-term.

Example

Factor [latex]x^{2}+5x+6[/latex].

Solution

From above, we know we need to split [latex]5x[/latex] into [latex]2x+3x[/latex]:

[latex]x^{2}+2x+3x+6[/latex]

Group the pairs of terms:

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

Factor x out of the first pair of terms:

[latex]x\left(x+2\right)+\left(3x+6\right)[/latex]

Factor [latex]3[/latex] out of the second pair of terms:

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

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

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

Answer

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

Note that if we wrote [latex]x^{2}+5x+6[/latex] as [latex]x^{2}+3x+2x+6[/latex] and grouped the pairs as [latex]\left(x^{2}+3x\right)+\left(2x+6\right)[/latex]; then factored, [latex]x\left(x+3\right)+2\left(x+3\right)[/latex], and factored out [latex]x+3[/latex], the answer would be [latex]\left(x+3\right)\left(x+2\right)[/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.

The following video presents another example of how to use grouping to factor a quadratic polynomial.

Let’s take a look at the trinomial [latex]x^{2}+x–12[/latex]. In this trinomial, the c term is [latex]−12[/latex]. So we look at all of the combinations of factors whose product is [latex]−12[/latex]. Then see which of these combinations will give us the correct [latex]x[/latex]-term, where b is [latex]1[/latex].

Factors whose product is [latex]−12[/latex] Sum of the factors
[latex]1\cdot−12=−12[/latex] [latex]1+−12=−11[/latex]
[latex]2\cdot−6=−12[/latex] [latex]2+−6=−4[/latex]
[latex]3\cdot−4=−12[/latex] [latex]3+−4=−1[/latex]
[latex]4\cdot−3=−12[/latex] [latex]4+−3=1[/latex]
[latex]6\cdot−2=−12[/latex] [latex]6+−2=4[/latex]
[latex]12\cdot−1=−12[/latex] [latex]12+−1=11[/latex]

There is only one combination where the product is [latex]−12[/latex] and the sum is [latex]1[/latex], and that is when [latex]r=4[/latex], and [latex]s=−3[/latex]. Let’s use these to factor our original trinomial.

Example

Factor [latex]x^{2}+x -12[/latex].

Solution

Rewrite the trinomial using the correct values from the chart above. Use values [latex]r=4[/latex] and [latex]s=−3[/latex].

[latex]x^{2}+4x−3x–12[/latex]

Group pairs of terms:

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

Factor [latex]x[/latex] out of the first group:

[latex]x\left(x+4\right)+\left(-3x-12\right)[/latex]

Factor [latex]−3[/latex]  out of the second group:

[latex]x\left(x+4\right)–3\left(x+4\right)[/latex]

Factor out [latex]\left(x+4\right)[/latex]:

[latex]\left(x+4\right)\left(x-3\right)[/latex]

Answer

[latex]\left(x+4\right)\left(x-3\right)[/latex]

In this example, we could also rewrite [latex]x^{2}+x-12[/latex] as [latex]x^{2}– 3x+4x–12[/latex] first. Then factor [latex]x\left(x – 3\right)+4\left(x–3\right)[/latex], and factor out [latex]\left(x–3\right)[/latex] getting [latex]\left(x–3\right)\left(x+4\right)[/latex]. Since multiplication is commutative, this is the same answer.

Try It

1. Factor [latex]x^2+7x-8[/latex]

 

2. Factor [latex]x^2-7x+10[/latex]

 

3. Factor [latex]x^2+6x+9[/latex]

 

4. Factor [latex]x^2-5x-24[/latex]

 

Factoring Tips

Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored. In this case the trinomial is said to be prime.

Here are some tips that can ease the way.

Tips for Finding Values that Work

When factoring a trinomial in the form [latex]x^{2}+bx+c[/latex], consider the following tips.

Look at the c term first.

  • If the c term is a positive number, then the factors of c will both be positive or both be negative. In other words, r and s will have the same sign.
  • If the c term is a negative number, then one factor of c will be positive, and one factor of c will be negative. In other words, r or s will have opposite signs.

Look at the b term second.

  • If the c term is positive and the b term is positive, then both r and s are positive.
  • If the c term is positive and the b term is negative, then both r and s are negative.
  • If the c term is negative and the b term is positive, then the factor that is positive will have the greater absolute value. That is, if [latex]|r|>|s|[/latex], then r is positive and s is negative.
  • If the c term is negative and the b term is negative, then the factor that is negative will have the greater absolute value. That is, if [latex]|r|>|s|[/latex], then r is negative and s is positive.

After you have factored a number of trinomials in the form [latex]x^{2}+bx+c[/latex], you may notice that the numbers you identify for r and s end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews three examples.

Trinomial [latex]x^{2}+7x+10[/latex] [latex]x^{2}+5x+6[/latex] [latex]x^{2}+x-12[/latex]
r and s values [latex]r=+5,s=+2[/latex] [latex]r=+2,s=+3[/latex] [latex]r=+4,s=–3[/latex]
Factored form [latex]\left(x+5\right)\left(x+2\right)[/latex] [latex]\left(x+2\right)\left(x+3\right)[/latex] [latex](x+4)(x–3)[/latex]

This is only true when the leading coefficient of the trinomial is 1.

The Shortcut

Notice that in each of the examples above, the r and s values are repeated in the factored form of the trinomial. So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[/latex] (where the coefficient in front of [latex]x^{2}[/latex] is [latex]1[/latex]), if we can identify the correct r and s values, we can effectively skip the grouping steps and go straight to the factored form. The idea is that we can build factors for a trinomial in this form: [latex]x^2+bx+c[/latex] by finding [latex]r[/latex] and [latex]s[/latex], then placing them in two binomial factors like this:

[latex]\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)[/latex]

Example

Factor: [latex]y^2+6y-27[/latex]

Solution

Find[latex]r[/latex] and [latex]s[/latex]:

Factors whose product is -27 Sum of the factors
[latex]1\cdot{-27}=-27[/latex] [latex]1-27=-26[/latex]
[latex]3\cdot{-9}=-27[/latex] [latex]3-9=-6[/latex]
[latex]-3\cdot{9}=-27[/latex]  [latex]-3+9=6[/latex]

Instead of rewriting the middle term, we will use the values of[latex]r[/latex] and [latex]s[/latex] that give the product and sum that we need.

In this case:

[latex]\begin{array}{l}r=-3\\s=9\end{array}[/latex]

It helps to start by writing two empty sets of parentheses:

[latex]\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)[/latex]

The squared term is y, so we will place a y in each set of parentheses:

[latex]\left(y\,\,\,\,\,\,\,\,\right)\left(y\,\,\,\,\,\,\,\,\right)[/latex]

Now we can fill in the rest of each binomial with the values we found for r and s.

[latex]\left(y-3\right)\left(y+9\right)[/latex]

Note how we kept the sign on each of the values.

The nice thing about factoring is we can check our work by multiplying the binomials together to get back to the original trinomial.

[latex]\begin{array}{l}\left(y-3\right)\left(y+9\right)\\=y^2+9y-3y-27\\=y^2+6y-27\end{array}[/latex]

Answer

[latex]\left(y-3\right)\left(y+9\right)[/latex]

In the next example, the leading coefficient is negative.

Example

Factor: [latex]-m^2+16m-48[/latex]

Solution

There is a negative in front of the squared term, so we will start by factoring out a [latex]-1[/latex] from the whole trinomial. Remember, this boils down to changing the sign of all the terms:

[latex]-m^2+16m-48=-1\left(m^2-16m+48\right)[/latex]

Now we can factor [latex]\left(m^2-16m+48\right)[/latex] by finding [latex]r[/latex] and [latex]s[/latex]. Note that [latex]b[/latex] is negative, and [latex]c[/latex] is positive so we are probably looking for two negative numbers:

Factors whose product is 48 Sum of the factors
[latex]-1\cdot{-48}=48[/latex] [latex]-1-48=-49[/latex]
[latex]-2\cdot{-12}=48[/latex] [latex]-2-12=-14[/latex]
[latex]-3\cdot{-16}=-48[/latex] [latex]-3-16=-19[/latex]
[latex]-4\cdot{-12}=-48[/latex] [latex]-4-12=-16[/latex]

There are more factors whose product is [latex]48[/latex], but we have found the ones that sum to  [latex]-16[/latex], so we can stop.

[latex]\begin{array}{l}r=-4\\s=-12\end{array}[/latex]

Now we can fill in each binomial with the values we found for r and s, make sure to use the correct variable!

[latex]\left(m-4\right)\left(m-12\right)[/latex]

We are not done yet, remember that we factored out a negative sign in the first step. We need to remember to include that.

 [latex]-1\left(m-4\right)\left(m-12\right)[/latex]

Answer

[latex]-m^2+16m-48=-1\left(m-4\right)\left(m-12\right)[/latex]

The following video presents two more examples of factoring a trinomial with a leading coefficient of 1.

Try It

To summarize our process, consider the following steps:

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

  1. List factors of [latex]c[/latex].
  2. Find [latex]r[/latex] and [latex]s[/latex], a pair of factors of [latex]c[/latex] with a sum of [latex]b[/latex].
  3. Write the factored expression [latex]\left(x+p\right)\left(x+q\right)[/latex].

Example

Factor [latex]x^{2}+x–12[/latex].

Solution

Consider all the combinations of numbers whose product is [latex]-12[/latex] and list their sum.

Factors whose product is [latex]−12[/latex] Sum of the factors
[latex]1\cdot−12=−12[/latex] [latex]1+−12=−11[/latex]
[latex]2\cdot−6=−12[/latex] [latex]2+−6=−4[/latex]
[latex]3\cdot−4=−12[/latex] [latex]3+−4=−1[/latex]
[latex]4\cdot−3=−12[/latex] [latex]4+−3=1[/latex]
[latex]6\cdot−2=−12[/latex] [latex]6+−2=4[/latex]
[latex]12\cdot−1=−12[/latex] [latex]12+−1=11[/latex]

Choose the values whose sum is [latex]+1[/latex]:  [latex]r=4[/latex] and [latex]s=−3[/latex], and place them into a product of binomials.

 

[latex]\left(x+4\right)\left(x-3\right)[/latex]

Think About It

Which property of multiplication can be used to describe why [latex]\left(x+4\right)\left(x-3\right) =\left(x-3\right)\left(x+4\right)[/latex]. Use the textbox below to write down your ideas before you look at the answer.

Try It

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

 

 

Example

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

Solution

List the factors of [latex]6[/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.

Factors of [latex]6[/latex] Sum of Factors
[latex]1,6[/latex] [latex]7[/latex]
[latex]2, 3[/latex] [latex]5[/latex]
[latex]-1, -6[/latex] [latex]-7[/latex]
[latex]-2, -3[/latex] [latex]-5[/latex]

Choose the pair that sum to [latex]-7[/latex], which is [latex]-1, -6[/latex]

Write the pair as constant terms in a product of binomials.

[latex]\left(x-1\right)\left(x-6\right)[/latex]

In the last example, [latex]b[/latex] was negative and [latex]c[/latex] was positive. This will always mean that if it can be factored, [latex]r[/latex] and [latex]s[/latex] will both be negative.

Think About It

Can every trinomial be factored as a product of binomials?

Mathematicians often use a counterexample to prove or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient [latex]1[/latex] that cannot be factored as a product of binomials?

Use the textbox below to write your ideas.

Try It

The more examples you work through, the more likely you are to spot patterns. This will lead to faster factoring, which is useful in later courses.