2.06 Factoring trinomials

Sometimes the common factor that we take out when factoring is not just a single algebraic term. Consider this example: $a\left(x+1\right)+2\left(x+1\right)$a(x+1)+2(x+1). Let's consider the similar, but simpler, expression of $ay+2y$ay+2y, where $y=\left(x+1\right)$y=(x+1). This we can factor into $y\left(a+2\right)$y(a+2). So that must mean $a\left(x+1\right)+2\left(x+1\right)$a(x+1)+2(x+1) can be factored into $\left(x+1\right)\left(a+2\right)$(x+1)(a+2).

Can you see that $\left(x+1\right)$(x+1) here is a common factor of both terms even though it is a binomial in parentheses? This is the key idea with factoring by grouping.

 

Factoring by grouping

The above technique is very useful in certain situations when factoring four terms. Let's take a look at the example $14-7v+4u-2uv$14−7v+4u−2uv. There is no way we could factor that by finding an GCF of all four terms! But let's have a look at what factors there are in each term anyway:

Can you see that $14$14 and -$7v$7v have an GCF of $7$7, while $4u$4u and $2uv$2uv have an GCF of $2u$2u? That means we can group these four terms into two pairs and factor them separately. Let's see what happens!

$14-7v+4u-2uv$14−7v+4u−2uv	$=$=	$\left(14-7v\right)+\left(4u-2uv\right)$(14−7v)+(4u−2uv)
	$=$=	$7\left(2-v\right)+2u\left(2-v\right)$7(2−v)+2u(2−v)

Wow, this is very similar to the example we had before, where we can take out a parenthesized term to factor! Let's try that technique!

$7\left(2-v\right)+2u\left(2-v\right)=\left(2-v\right)\left(7+2u\right)$7(2−v)+2u(2−v)=(2−v)(7+2u)

Isn't that amazing? We managed to factor four terms!

 

Factoring with negatives

Sometimes we need to factor out a negative to make the two parentheses the same. Suppose we factor by grouping and get $3\left(y-1\right)-10y\left(1-y\right)$3(y−1)−10y(1−y). We don't have the exactly same factor as we have $\left(y-1\right)$(y−1) and $\left(1-y\right)$(1−y), so we think that it can't be factored. However, notice that $y-1$y−1 and $1-y$1−y look quite similar don't they? In fact, $1-y$1−y is equal to $-1\times\left(y-1\right)$−1×(y−1)! Let's see if we can use this to rewrite our expression.

$3\left(y-1\right)-10y\left(1-y\right)$3(y−1)−10y(1−y)	$=$=	$3\left(y-1\right)-10y\times\left(-1\right)\times\left(y-1\right)$3(y−1)−10y×(−1)×(y−1)
	$=$=	$3\left(y-1\right)+10y\left(y-1\right)$3(y−1)+10y(y−1)
This can then be factored using our usual approach:
	$=$=	$\left(y-1\right)\left(3+10y\right)$(y−1)(3+10y)

So when you see two binomials that are exactly the same but with inverted signs, you can use $-1$−1 to help you out!

 

Practice questions

Question 1

Question 2

Question 3

 

Factoring trinomials with a leading coefficient of 1

We have already learned how to distribute the product of two binomials, the result of which is called a quadratic. Now we are going to reverse the process of distributing and look at factoring back to the original binomial product. We are going to start with quadratics of the $x^2+bx+c$x2+bx+c, so with a leading coefficient of $1$1.

Recall that if we distribute $\left(x+m\right)\left(x+n\right)$(x+m)(x+n) and combine like terms, we found that:

 $\left(x+m\right)\left(x+n\right)=x^2+\left(m+n\right)x+mn$(x+m)(x+n)=x2+(m+n)x+mn

Using the fact above, we can factor expressions of the form, $x^2+bx+c$x2+bx+c.

1. Find two numbers, $m$m and $n$n, that have a sum of $b$b and a product of $c$c

• In other words, we are trying to find $m$m and $n$n, where $\left(m+n\right)=b$(m+n)=b and $mn=c$mn=c 
• It's usually a lot easier to look at $c$c and start listing out its factors first rather than first finding the numbers that add to make $b$b.

2. The factored form will be $\left(x+m\right)\left(x+n\right)$(x+m)(x+n)

 

 

Worked examples

Question 4

Factor $x^2+15x+56$x2+15x+56

Think: We are looking for two number that have a sum of 15 and a product of 56. What does this tell us about the signs of the two numbers?

Do:

$m+n=15$m+n=15 and $m\times n=56$m×n=56

$m$m and $n$n must both then be positive if they have both a positive product and sum.

If we look at the factors of $56$56, we get these pairs:

$1$1, $56$56

$2$2, $28$28

$4$4, $14$14

$7$7, $8$8

The only pair with a sum of $15$15 is the last pair, so $m$m and $n$n must equal $7$7 and $8$8.

So $x^2+15x+56=\left(x+7\right)\left(x+8\right)$x2+15x+56=(x+7)(x+8)

Reflect: Coming up with the list of factors requires a good knowledge of our times tables and possibly a calculator for very large numbers.

 

Question 5

Factor $x^2+2x-3$x2+2x−3 using the strategy or organizational tool of your choice.

Think: There are many ways to organize our thinking. Let's use the box method for this example. It is important to notice that there are two different factor pairs for $-3$−3 as it is a negative number.

Do:

$-3$−3 has the factor pairs $1$1 & $-3$−3 and $-1$−1 & $3$3. Let's try the first pair first:

	$x$x	$1$1
$x$x	$x^2$x2	$x$x
$-3$−3	$-3x$−3x	$-3$−3

Unfortunately, we get a sum of $-2x$−2x, not just $2x$2x like we wanted. We will have to try the other pair, $-1$−1 and $3$3.

	$x$x	$-1$−1
$x$x	$x^2$x2	$-x$−x
$3$3	$3x$3x	$-3$−3

This gives us the correct sum of $2x$2x, so the two numbers are $-1$−1 and $3$3, making the factored form $\left(x-1\right)\left(x+3\right)$(x−1)(x+3).

 

Practice questions

Question 6

Question 7

Question 8

 

Factoring trinomials with a leading coefficient not equal to 1

So far, the trinomials we've dealt with have had a coefficient of $1$1 on the $x^2$x2 term or if the coefficient is not $1$1, then we could factor out that coefficient from the whole quadratic to give a trinomial with a leading coefficient of $1$1.

eg. $2x^2-4x+6=2\left(x^2-2x+3\right)$2x2−4x+6=2(x2−2x+3).

 

But how do we factor quadratics that can't be simplified in this way? First let's have a look at how this type of quadratic distributes from a product of two binomials:

Now we are more familiar with these quadratics let's have a look at one of many methods you can use below.

Split method

The Split Method uses a similar idea we had with quadratics with leading coefficient of $1$1 where we think about sums and products, but slightly different.

 

Worked example

question 9

Factor $5x^2+11x-12$5x2+11x−12 using the Split method.

Think: What two numbers have a product of $5\times\left(-12\right)=60$5×(−12)=60 and a sum of $11$11? What do we know about their signs?

Do:

We want the sum of of $m$m & $n$n to be $11$11, and the product to be $5\times\left(-12\right)=-60$5×(−12)=−60. 

By listing factors, we find the two numbers to be $-4$−4 & $15$15, so we need split $11x$11x into $-4x+15x$−4x+15x:

$5x^2+11x-12$5x2+11x−12	$=$=	$5x^2-4x+15x-12$5x2−4x+15x−12
	$=$=	$x\left(5x-4\right)+3\left(5x-4\right)$x(5x−4)+3(5x−4)
	$=$=	$\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3)

Reflect: As always, we can distribute $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3) to double check that we get back the question of $5x^2+11x-12$5x2+11x−12. Once we get lots of practice, we may be able to factor expressions like this just by looking at them!

 

Practice questions

Question 10

Question 11