When we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example: [latex]\left(x+4\right)\left(x+2\right)=x^{2}+2x+4x+8[/latex].

We can apply what we have learned about factoring out a common monomial  to return a four term polynomial to the product of two binomials. Why would we even want to do this?

Because it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:

[latex]\begin{array}{l}\left(x+4\right)\left(x+2\right)\\=x^{2}+2x+4x+8\\=x^2+6x+8\end{array}[/latex]

Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don’t all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.

Notice that when you factor a two term polynomial, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials. As we noted before, this is an important middle step in learning how to factor a three term polynomial.

This process is called the grouping technique. Broken down into individual steps, here’s how to do it (you can also follow this process in the example below).

• Group the terms of the polynomial into pairs that share a GCF.
• Find the greatest common factor and then use the distributive property to pull out the GCF
• Look for the common binomial between the factored terms
• Factor the common binomial out of the groups, the other factors will make the other binomial

Let’s try factoring a few more four-term polynomials. Note how there is a now a constant in front of the [latex]x^2[/latex] term. We will just consider this another factor when we are finding the GCF.

Another example follows that contains subtraction. Note how we choose a positive GCF from each group of terms, and the negative signs stay.

The video that follows provides another example of factoring by grouping.

In the next example, we will have a GCF that is negative.  It is important to pay attention to what happens to the resulting binomial when the GCF is negative.

In the following video we present another example of factoring by grouping when one of the GCF is negative.

Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

In the example above, each pair can be factored, but then there is no common factor between the pairs!

Factor Trinomials Part I

In the last section we introduced the technique of factoring by grouping as a means to be able to factor a trinomial. Now we will actually get to the work of starting with a three term polynomial, and rewriting it as a four term polynomial so it can be factored.

We will start with factoring trinomials of the form [latex]x^{2}+bx+c[/latex] that don’t have a coefficient in front of the [latex]x^2[/latex] term.

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

 [latex]\left(x+2\right)\left(x+5\right)=x^2+5x+2x+10=x^2+7x+10[/latex]

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

[latex](x^{2}+5x)+(2x+10)[/latex]

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

Then pull out the common binomial factor: [latex]\left(x+5\right)\left(x+2\right)[/latex]

What would have happened if we had rewritten [latex]7x[/latex] as [latex]6x+x[/latex]?

[latex](x^{2}+6x)+(x+10)[/latex]

Factor each pair: [latex]x\left(x+6\right)+1\left(x+10\right)[/latex]

Then we don’t have a common factor of [latex]\left(x+5\right)[/latex] like we did before. 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.

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

For example, to factor [latex]x^{2}+7x+10[/latex], 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 [latex]7x[/latex] as [latex]2x+5x[/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[/latex] as [latex]5x+2x[/latex]. Both will work.

Let’s factor the trinomial [latex]x^{2}+5x+6[/latex]. 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
[latex]1\cdot6=6[/latex]	[latex]1+6=7[/latex]
[latex]2\cdot3=6[/latex]	[latex]2+3=5[/latex]

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

Note that if you 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.

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

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 1, and that is when [latex]r=4[/latex], and [latex]s=−3[/latex]. Let’s use these to factor our original trinomial.

In the above example, you 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.

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.

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 the three problems you have seen so far.

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]

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 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.

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: [latex]x^2+bx+c[/latex] by finding r and s, 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]

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

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

Factor Trinomials Part II

The next goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored.  Additionally, we will explore one special case to look out for at the end of this page.

Not all trinomials look like [latex]x^{2}+5x+6[/latex], where the coefficient in front of the [latex]x^{2}[/latex] 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
[latex]2x^{2}+10x+12[/latex]	[latex]2(x^{2}+5x+6)[/latex]	[latex]2\left(x+2\right)\left(x+3\right)[/latex]
[latex]−5a^{2}−15a−10[/latex]	[latex]−5(a^{2}+3a+2)[/latex]	[latex]−5\left(a+2\right)\left(a+1\right)[/latex]
[latex]c^{3}–8c^{2}+15c[/latex]	[latex]c\left(c^{2}–8c+15\right)[/latex]	[latex]c\left(c–5\right)\left(c–3\right)[/latex]
[latex]y^{4}–9y^{3}–10y^{2}[/latex]	[latex]y^{2}\left(y^{2}–9y–10\right)[/latex]	[latex]y^{2}\left(y–10\right)\left(y+1\right)[/latex]

Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.

The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.

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

However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.

This is almost the same as factoring trinomials in the form [latex]x^{2}+bx+c[/latex], as in this form [latex]a=1[/latex]. Now you are looking for two factors whose product is [latex]a\cdot{c}[/latex], and whose sum is b.

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

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

In the following video, we present another example of factoring a trinomial using grouping.  In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[/latex], for instance. Can you think of 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.

In some situations, a 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.

Note that the answer above can also be written as [latex]\left(−h+3\right)\left(4h+1\right)[/latex] or [latex]\left(h–3\right)\left(−4h–1\right)[/latex] if you multiply [latex]−1[/latex] times one of the other factors.

In the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[/latex] using the grouping technique.

Difference of Squares

We would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms.  Let’s start from the product of two binomials to see the pattern.

Given the product of two binomials: [latex]\left(x-2\right)\left(x+2\right)[/latex], if we multiply them together, we lose the middle term that we are used to seeing as a result.

Multiply:

[latex]\begin{array}{l}\left(x-2\right)\left(x+2\right)\\\text{}\\=x^2-2x+2x-2^2\\\text{}\\=x^2-2^2\\\text{}\\=x^2-4\end{array}[/latex]

 The polynomial [latex]x^2-4[/latex] is called a difference of squares because teach term can be written as something squared.  A difference of squares will always factor in the following way:

Let’s factor [latex]x^{2}–4[/latex] by writing it as a trinomial, [latex]x^{2}+0x–4[/latex].  This is similar in format to the trinomials we have been factoring so far, so let’s use the same method.

Find the factors of [latex]a\cdot{c}[/latex] whose sum is b, in this case, 0:

Factors of [latex]−4[/latex]	Sum of the factors
[latex]1\cdot-4=−4[/latex]	[latex]1-4=−3[/latex]
[latex]2\cdot−2=−4[/latex]	[latex]2-2=0[/latex]
[latex]-1\cdot4=−4[/latex]	[latex]-1+4=3[/latex]

2, and -2 have a sum of 0. You can use these to factor [latex]x^{2}–4[/latex].

Since order doesn’t matter with multiplication, the answer can also be written as [latex]\left(x+2\right)\left(x–2\right)[/latex].

You can check the answer by multiplying [latex]\left(x–2\right)\left(x+2\right)=x^{2}+2x–2x–4=x^{2}–4[/latex].

The following video show two more examples of factoring a difference of squares.

Summary

When a trinomial is in the form of [latex]ax^{2}+bx+c[/latex], where a is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form [latex]ax^{2}+bx+c[/latex], find two integers, r and s, whose sum is b and whose product is ac. Then rewrite the trinomial as [latex]ax^{2}+rx+sx+c[/latex] and use grouping and the distributive property to factor the polynomial.

When [latex]ax^{2}[/latex] is negative, you can factor [latex]−1[/latex] out of the whole trinomial before continuing.

A difference of squares [latex]a^2-b^2[/latex] will factor in this way: [latex]\left(a+b\right)\left(a-b\right)[/latex].

Summary

Trinomials in the form [latex]x^{2}+bx+c[/latex] can be factored by finding two integers, r and s, whose sum is b and whose product is c. Rewrite the trinomial as [latex]x^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial.