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.
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!
The key when factoring by grouping is to make sure that each pair of terms factors to give the same common factor.
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!
Factor the following expression by taking out the common factor:
$5\left(a+b\right)+v\left(a+b\right)$5(a+b)+v(a+b)
Factor the following expression:
$2x+xz-40y-20yz$2x+xz−40y−20yz
Factor $x^2+5x+8x+40$x2+5x+8x+40 by grouping in pairs.
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
A shortcut to distributing $\left(x+m\right)\left(x+n\right)$(x+m)(x+n) is to notice that the coefficient of the $x^2$x2 term will be $1$1, the coefficient of the $x$x term will be the sum of $m$m and $n$n, $\left(m+n\right)$(m+n), and the constant will be the product of $m$m and $n$n, $\left(mn\right)$(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
2. The factored form will be $\left(x+m\right)\left(x+n\right)$(x+m)(x+n)
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.
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).
Factor $x^2-2x-8$x2−2x−8.
Factor $44-15x+x^2$44−15x+x2.
Factor the expression completely by first taking out a common factor:
$3x^2-21x-54$3x2−21x−54
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.
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.
For a quadratic in the form $ax^2+bx+c$ax2+bx+c:
1. Find two numbers, $m$m and $n$n, that have a SUM of $b$b and a PRODUCT of $ac$ac (not just $c$c).
2. Rewrite the quadratic by splitting the linear term as $ax^2+mx+nx+c$ax2+mx+nx+c.
3. Use grouping in pairs (as seen earlier in the chapter) to factor the four-termed expression.
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!
Factor the trinomial:
$7x^2-75x+50$7x2−75x+50
Factor the following trinomial:
$6x^2+13x+6$6x2+13x+6
Determine the quotient of a polynomial of degree one and polynomial of degree two when divided by a polynomial of degree one and polynomial of degree two when the degree of the divisor does not exceed the degree of the dividend
Rewrite polynomial expressions of degree one and degree two in equivalent forms using the distributive property
Factor, if possible, trinomials with real factors in the form ax^2 + b^x + c, including perfect square trinomials of degree two