## Factoring a Trinomial with a Leading Coefficient of 1

### Learning Outcomes

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

Now that we have shown you how to factor by grouping, We are going to show you how to factor a trinomial whose leading coefficient is $1$.  That is, trinomials of the form $x^2+bx+c$.  Polynomials whose leading coefficients are 1 can be factored using the grouping method that we showed you in the previous section.  We will begin by showing a few examples using the grouping method and then we will show you a shortcut that will make factoring trinomials with a leading coefficient of 1 even easier!

The following is a summary of the method, then we will show some examples of how to use it.

### Factoring Trinomials in the form $x^{2}+bx+c$

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

$\begin{array}{l}r\cdot{s}=c\\\text{ and }\\r+s=b\end{array}$

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

Special Note:  Does the summary above look familiar?  It should!  How we factor polynomials of the form $x^{2}+bx+c$ using grouping is exactly the same as for polynomials of the form $ax^{2}+bx+c$.  The only difference is that, previously, we were looking for two integers, r and s, whose sum is b and whose product is ac.  Notice that if our polynomial is of the form  $x^{2}+bx+c$, then $a=1$, making $ac=1\cdot{c}=c$.  As a result, we can skip the step of multiplying a by c!

Let’s factor the trinomial $x^{2}+5x+6$. In this polynomial, the b part of the middle term is  $5$ and the c term is  $6$. A chart will help us organize possibilities. On the left, list all possible factors of the c term, $6$; on the right you’ll find the sums.

Factors whose product is  $6$ Sum of the factors
$1\cdot6=6$ $1+6=7$
$2\cdot3=6$ $2+3=5$

There are only two possible factor combinations, $1$ and $6$, and $2$ and $3$. You can see that $2+3=5$. So $2x+3x=5x$, giving us the correct middle term.

### Example

Factor $x^{2}+5x+6$.

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

In the following video, we present another example of how to use grouping to factor a quadratic polynomial.

Finally, let’s take a look at the trinomial $x^{2}+x–12$. In this trinomial, the c term is $−12$. So look at all of the combinations of factors whose product is $−12$. Then see which of these combinations will give you the correct middle term, where b is $1$.

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

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

### Example

Factor $x^{2}+x–12$.

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

### 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.

While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.

### Tips for Finding Values that Work

When factoring a trinomial in the form $x^{2}+bx+c$, 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. Either r or s will be negative, but not both.

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 $|r|>|s|$, 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 $|r|>|s|$, then r is negative and s is positive.

After you have factored a number of trinomials in the form $x^{2}+bx+c$, 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 $x^{2}+7x+10$ $x^{2}+5x+6$ $x^{2}+x-12$
r and s values $r=+5,s=+2$ $r=+2,s=+3$ $r=+4,s=–3$
Factored form $\left(x+5\right)\left(x+2\right)$ $\left(x+2\right)\left(x+3\right)$ $(x+4)(x–3)$

## 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 $x^{2}+bx+c$ (where the coefficient in front of $x^{2}$ is $1$), if you can identify the correct r and s values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let’s look at some examples where we use this idea.

Shortcut This Way

In the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: $x^2+bx+c$ by finding r and s, then placing them in two binomial factors like this:

$\left(x+r\right)\left(x+s\right)\text{ OR }\left(x+s\right)\left(x+r\right)$

### Example

Factor: $y^2+6y-27$

We will show one more example so you can gain more experience.

### Example

Factor: $-m^2+16m-48$

In the following video, we present two more examples of factoring a trinomial with a leading coefficient of 1.

### Try It

To summarize our process, consider the following steps:

### How To: Given a trinomial in the form ${x}^{2}+bx+c$, factor it

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

We will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.  We will show that when c is negative, either p or q will be negative.

### Example

Factor $x^{2}+x–12$.

### Think About It

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

In our last example, we will show how to factor a trinomial whose b term is negative.

### Example

Factor ${x}^{2}-7x+6$.

In the last example, the b term was negative and the c term was positive. This will always mean that if it can be factored, p and q 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 $1$ that cannot be factored as a product of binomials?

Use the textbox below to write your ideas.

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