6.3: Factoring Trinomials with Leading Coefficient of 1

section 6.3 Learning Objectives

6.3: Factoring Trinomials with Leading Coefficient of 1

Factor trinomials of the type $x^2 + bx + c$ by grouping
Factor trinomials of the type $x^2+bx+c$ using the Product and Sum method
Rewrite and factor trinomials of the above type by first factoring out the GCF
Factor two-variable trinomials of the type $x^2 + bxy + cy^2$

Factor Trinomials Part I – Using Grouping

In the last section we introduced the technique of factoring by grouping as a means to be able to factor a four-term polynomial. However, many of the polynomials we encountered were actually trinomials in disguise. Now we will get to the work of starting with a three-term polynomial, and rewriting it as a four term polynomial so it can be factored using our grouping process.

We will start with factoring trinomials of the form $x^{2}+bx+c$ , and hence with a leading coefficient of 1.

Remember that when $\left(x+2\right)$ and $\left(x+5\right)$ , are multiplied, the result is a four term polynomial and then it is simplified into a trinomial:

$\left(x+2\right)\left(x+5\right)=x^2+5x+2x+10=x^2+7x+10$

Factoring is the reverse of multiplying, so let’s go in reverse and factor the trinomial $x^{2}+7x+10$ . The individual terms $x^{2}$ , $7x$ , and 10 share no common factors. If we rewrite the middle term as the sum of the two terms $7x=5x+2x$ then we can use the grouping technique:

$(x^{2}+5x)+(2x+10)$

Factor each pair: $\begin{array}{l}x\underbrace{\left(x+5\right)}+2\underbrace{\left(x+5\right)}\\\text{common binomial factor}\end{array}$

Then pull out the common binomial factor: $\left(x+5\right)\left(x+2\right)$

But how would we know to rewrite $7x$ as $5x+2x$ ? What would have happened if we had rewritten $7x$ as $6x+x$ ?

$(x^{2}+6x)+(x+10)$

Factor each pair: $x\left(x+6\right)+1\left(x+10\right)$

Then we don’t have a common binomial! There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.

Shakespeare quote: "Though this be madness, yet there is method in it."

Method to the Madness

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 $\left(x+s\right)$ .

For example, to factor $x^{2}+7x+10$ , you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).

Look at factor pairs of 10: 1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite $7x$ as $2x+5x$ , and continue factoring as in the example above. Note that you can also rewrite $7x$ as $5x+2x$ . Both will work.

With all terms positive in the last example, we did not really need to consider any negative factors. 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 1

Factor $x^{2}+x–12$ .

Show Solution

Rewrite the trinomial using the values from the chart above. Use values

r = 4

$r=4$ and

s = - 3

$s=−3$ .

$x^{2}+4x−3x–12$

Group pairs of terms.

$\left(x^{2}+4x\right)+\left(−3x–12\right)$

Factor x out of the first group.

$x\left(x+4\right)+\left(-3x-12\right)$

Factor −3 out of the second group.

$x\left(x+4\right)–3\left(x+4\right)$

Factor out $\left(x+4\right)$ .

$\left(x+4\right)\left(x-3\right)$

Answer

$x^2+x-12=\left(x+4\right)\left(x-3\right)$

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.

In the following video, we present another example of how to use grouping to factor a trinomial of this form.

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

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 includes some of the problems you have seen so far.

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	$\left(x+5\right)\left(x+2\right)$	$\left(x+2\right)\left(x+3\right)$	$(x+4)(x–3)$

Factoring Trinomials II – The Shortcut (Product and Sum Method)

Notice that in all of the examples, 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.

Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.

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:

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

This technique is sometime referred to as the “Product and Sum Method.”

Let us revisit the problem $x^2+7x+10$ . We determined that the factors $2$ and $5$ multiply to $10$ while simultaneously adding to $7$ . We proceeded by rewriting the problem and transforming it into a grouping problem. Alternatively, we could have jumped to the final answer immediately using Product and Sum, factoring the polynomial as

$x^2+7x+10=(x+2)(x+5)$ .

Note that the order does not matter, and we could have also written $x^2+7x+10=(x+5)(x+2)$ .

We give several more examples demonstrating the Product and Sum Method.

Example 2

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

Show Solution

Find r and s:

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

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

In this case:

$\begin{array}{l}r=-3\\s=9\end{array}$

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

$\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)\left(\,\,\,\,\,\,\,\,\,\,\,\,\right)$

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

$\left(y\,\,\,\,\,\,\,\,\right)\left(y\,\,\,\,\,\,\,\,\right)$

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

$\left(y-3\right)\left(y+9\right)$

Note how we kept the sign on each of the values. We also could have reversed the order and written $(y+9)(y-3)$ instead due to the commutative property of multiplication.

One nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.

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

Answer

$y^2+6y-27=\left(y-3\right)\left(y+9\right)$

Let us try another.

Example 3

Factor: $x^2-x-12$

Show Solution

We have $b=-1$ and $c=-12$ . Instead of listing from possible pair of factors for $-12$ , let us utilize the tips shared earlier. Since both b and c are negative, we know r and s must have opposite signs (one positive and one negative), and that the absolute of the negative value must be the larger of the two.

Factors whose product is -12	Sum of the factors
$1\cdot{-12}=-12$	$1-12=-11$
$2\cdot(-6)=-12$	$-2-6=-4$
$3\cdot(-4)=-12$	$3-4=-1$

Hence, we should use $r=3$ and $s=-4$ .

We know the factored form will look like

$(x\hspace{.25in})(x\hspace{.25in})$

Inserting the values we found for r and s gives

$(x+3)(x-4)$

Answer

$x^2-x-12=(x+3)(x-4)$

In the following video, we present two more examples of factoring a trinomial using the shortcut presented here.

Another situation we sometime encounter is a trinomial with two variables of the form $x^2+bxy+cy^2$ . Applying the same logic behind beginning each parenthetical expression with $x$ and $x$ to produce the leading term of $x^2$ , we will similarly end each expression with $y$ . We will then be able to deal with the coefficients using the same approach as previous examples.

Example 4

Factor: $x^2-12xy+20y^2$

Show Answer

To ensure this can FOIL to produce the original expression, we will factor it as

$(x\hspace{.3in}y)(x\hspace{.3in}y)$ .

We simply need to fill in the gaps with the corrects numbers and operations, which we can use the normal Product and Sum approach for. So, we are looking for two numbers that multiply to $20$ and add to $-12$ . Note that in order to multiply to a positive but add to a negative, this will require two negative factors.

Factors whose product is 20	Sum of the factors
$-1\cdot{-20}=20$	$-1-20=-21$
$-2\cdot(-10)=20$	$-2-10=-12$
$-4\cdot(-5)=20$	$-4-5=-9$

We conclude that the correct pair is $-2$ and $-12$ . Filling in the blanks in our above factored form gives us

$(x-2y)(x-10y)$ .

It may not be obvious that this would match the original polynomial. Let us check the answer by performing the FOIL.

$(x-2y)(x-10y)=x^2-10xy-2xy+20y^2=x^2-12xy+20y^2$

This matches the original problem and therefore the factoring was correct.

Answer

$x^2-12xy+20y^2=(x-2y)(x-10y)$

Its Okay To Be Prime

Many students become frustrated when searching for the proper pair of factors, only to discover that no such pair existed! Remember, not every polynomial can be factored.

ExAMPLE 5

Factor, if possible: $x^2-7x+18$

Show Solution

When looking at the coefficient, we may be tempted to think that we can somehow use 9 and 2 to get the desired results. But this stresses how crucial it is to consider the signs as well. Let’s list all the possible pairs of factors.

Factors whose product is 18	Sum of the factors
$-1\cdot{-18}=18$	$-1-18=-19$
$-2\cdot{-9}=18$	$-2-9=-11$
$-3\cdot -6=18$	$-3-6=-9$
$1\cdot{18}=18$	$1+18=19$
$2\cdot{9}=18$	$2+9=11$
$3\cdot 6=18$	$3+6=9$

So, there is no pair of numbers that multiply to 18 and add to -7, meaning this polynomial cannot be factored.

Answer

Cannot be factored (Prime)

Factor Trinomials Part III – Factoring Out A GCF First

Not all trinomials look like $x^{2}+5x+6$ , where the coefficient in front of the $x^{2}$ term is 1. In these cases, your first step should be to look for common factors for the three terms.

Trinomial	Factor out Common Factor	Factored
$2x^{2}+10x+12$	$2(x^{2}+5x+6)$	$2\left(x+2\right)\left(x+3\right)$
$−5a^{2}−15a−10$	$−5(a^{2}+3a+2)$	$−5\left(a+2\right)\left(a+1\right)$
$c^{3}–8c^{2}+15c$	$c\left(c^{2}–8c+15\right)$	$c\left(c–5\right)\left(c–3\right)$
$y^{4}–9y^{3}–10y^{2}$	$y^{2}\left(y^{2}–9y–10\right)$	$y^{2}\left(y–10\right)\left(y+1\right)$

Notice that once you have identified and pulled out the common factor, if the resulting trinomial has a leading coefficient of 1, you can factor the remaining trinomial as usual.

Let’s begin with an example where the leading coefficient is $-1$ . Recall that in Section 6.1, we discussed that it could be useful to factor out the negative in this case. Here, it is imperative in order to directly apply the shortcut we have introduced.

Example 6

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

Show Solution

There is a negative in front of the squared term, so we will factor out a negative one from the whole trinomial first. Remember, this boils down to changing the sign of all the terms:

$-m^2+16m-48=-1\left(m^2-16m+48\right)$

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

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

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

$\begin{array}{l}r=-4\\s=-12\end{array}$

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

$\left(m-4\right)\left(m-12\right)$

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

$-1\left(m-4\right)\left(m-12\right)$

Answer

$-m^2+16m-48=-\left(m-4\right)\left(m-12\right)$

In the next example, we must first factor out a more substantial GCF.

Example 7

Factor $3x^{3}–3x^{2}–90x$ .

Show Solution

Since

3 x

$3x$ is a common factor for the three terms, factor it out first.

$3x\left(x^{2}–x–30\right)$

Now you can factor the trinomial $x^{2}–x–30$ . To find r and s, identify two numbers whose product is $−30$ and whose sum is $−1$ .

The pair of factors is $−6$ and $5$ . Using our shortcut, $x^2-x-30$ factors as

$(x-6)(x+5)$

However, do not forget the GCF of $3x$ that we factored out. This needs to remain in front, implying that the original expression factors as

$3x\left(x–6\right)\left(x+5\right)$

Answer

$3x^3-3x^2-90x=3x\left(x–6\right)\left(x+5\right)$

The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF.

The general form of trinomials with a leading coefficient of a is $ax^{2}+bx+c$ . Sometimes the factor of a can be factored out as you saw above; this happens when a is in common to all three terms. The remaining trinomial that still needs factoring (if possible) will then be simpler, with the leading term only being an $x^{2}$ term, instead of an $ax^{2}$ term.

But what do we do if a is not 1 and yet is not a common factor? We explore this in the next section.

Module 6: Factoring Polynomials