# 4.1 Quadratic concepts -- factoring  (Page 2/2)

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$\left(x+a\right)\left(x-a\right)={x}^{2}-{a}^{2}$

You can run this formula in reverse whenever you are subtracting two perfect squares . For instance, if we see ${x}^{2}-\text{25}$ , we recognize that both ${x}^{2}$ and 25 are perfect squares. We can therefore factor it as $\left(x+5\right)\left(x-5\right)$ . Other examples include:

• ${x}^{2}-\text{64}$ $=\left(x+8\right)\left(x-8\right)$
• $\text{16}{y}^{2}-\text{49}$ $=\left(4y+7\right)\left(4y-7\right)$
• ${2x}^{2}-\text{18}$ $=2\left({x}^{2}-9\right)$ $=2\left(x+3\right)\left(x-3\right)$

And so on. Note that, in the last example, we begin by pulling out a 2, and we are then left with two perfect squares. This is an example of the rule that you should always begin by pulling out common factors before you try anything else!

It is also important to note that you cannot factor the sum of two squares. ${x}^{2}+4$ is a perfectly good function, but it cannot be factored.

## Brute force, old-fashioned, bare-knuckle, no-holds-barred factoring

In this case, the multiplication that we are reversing is just FOIL. For instance, consider:

$\left(x+3\right)\left(x+7\right)={x}^{2}+3x+7x+\text{21}={x}^{2}+\text{10}x+\text{21}$

What happened? The 3 and 7 added to yield the middle term (10), and multiplied to yield the final term $\left(\text{21}\right)$ . We can generalize this as: $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+\text{ab}$ .

The point is, if you are given a problem such as ${x}^{2}+\text{10}x+\text{21}$ to factor, you look for two numbers that add up to 10, and multiply to 21. And how do you find them? There are a lot of pairs of numbers that add up to 10, but relatively few that multiply to 21. So you start by looking for factors of 21.

Below is a series of examples. Each example showcases a different aspect of the factoring process, so I would encourage you not to skip over any of them: try each problem yourself, then take a look at what I did.

If you are uncomfortable with factoring, the best practice you can get is to multiply things out . In each case, look at the final answer I arrive at, and multiply it with FOIL. See that you get the problem I started with. Then look back at the steps I took and see how they led me to that answer. The steps will make a lot more sense if you have done the multiplication already.

Factor ${x}^{2}+\text{11}x+\text{18}$

$\left(x+\text{__}\right)\left(x+\text{__}\right)$

What multiplies to 18? $1\cdot \text{18}$ , or $2\cdot 9$ , or $3\cdot 6$ .

Which of those adds to 11? $2+9$ .

$\left(x+2\right)\left(x+9\right)$

Start by listing all factors of the third term. Then see which ones add to give you the middle term you want.

Factor ${x}^{2}-\text{13}x+\text{12}$

$\left(x+\text{__}\right)\left(x+\text{__}\right)$

What multiplies to 12? $1\cdot \text{12}$ , or $2\cdot 6$ , or $3\cdot 4$

Which of those adds to 13? $1+\text{12}$

$\left(x-1\right)\left(x-\text{12}\right)$

If the middle term is negative, it doesn’t change much: it just makes both numbers negative. If this had been ${x}^{2}+\text{13}x+\text{12}$ , the process would have been the same, and the answer would have been $\left(x+1\right)\left(x+\text{12}\right)$ .

Factor ${x}^{2}+\text{12}x+\text{24}$

$\left(x+\text{__}\right)\left(x+\text{__}\right)$

What multiplies to 24? $1\cdot \text{24}$ , or $2\cdot \text{12}$ , or $3\cdot 8$ , or $4\cdot 6$

Which of those adds to 12? None of them.

It can’t be factored. It is “prime.”

Some things can’t be factored. Many students spend a long time fighting with such problems, but it really doesn’t have to take long. Try all the possibilities, and if none of them works, it can’t be factored.

Factor ${x}^{2}+2x-\text{15}$

$\left(x+\text{__}\right)\left(x+\text{__}\right)$

What multiplies to 15? $1\cdot \text{15}$ , or $3\cdot 5$

Which of those subtracts to 2? 5–3

$\left(x+5\right)\left(x-3\right)$

If the last term is negative, that changes things! In order to multiply to –15, the two numbers will have to have different signs—one negative, one positive—which means they will subtract to give the middle term. Note that if the middle term were negative, that wouldn’t change the process: the final answer would be reversed, $\left(x+5\right)\left(x-3\right)$ . This fits the rule that we saw earlier—changing the sign of the middle term changes the answer a bit, but not the process.

Factor ${2x}^{2}+\text{24}x+\text{72}$

$2\left({x}^{2}+\text{12}x+\text{36}\right)$

$2{\left(x+6\right)}^{2}$

Never forget, always start by looking for common factors to pull out. Then look to see if it fits one of our formulae. Only after trying all that do you begin the FOIL approach.

Factor ${3x}^{2}+\text{14}x+\text{16}$

$\left(3x+\text{__}\right)\left(x+\text{__}\right)$

What multiplies to 16? $1\cdot \text{16}$ , or $2\cdot 8$ , or $4\cdot 4$

Which of those adds to 14 after tripling one number ? $8+3\cdot 2$

$\left(3x+8\right)\left(x+2\right)$

If the ${x}^{2}$ has a coefficient, and if you can’t pull it out, the problem is trickier. In this case, we know that the factored form will look like $\left(3x+\text{__}\right)\left(x+\text{__}\right)$ so we can see that, when we multiply it back, one of those numbers—the one on the right—will be tripled, before they add up to the middle term! So you have to check the number pairs to see if any work that way.

There are two different ways to check your answer after factoring: multiplying back, and trying numbers.

1. Problem : Factor $\text{40}{x}^{3}-\text{250}x$
• $\text{10}x\left(4x-\text{25}\right)$ First, pull out the common factor
• $\text{10}x\left(2x+5\right)\left(2x-5\right)$ Difference between two squares
2. So, does $\text{40}{x}^{3}-\text{250}x=\text{10}x\left(2x+5\right)\left(2x-5\right)$ ? First let’s check by multiplying back.
• $\text{10}x\left(2x+5\right)\left(2x-5\right)$
• $=\left(\text{20}{x}^{2}+\text{50}x\right)\left(2x-5\right)$ Distributive property
• $=\text{40}{x}^{3}-\text{100}{x}^{2}+\text{100}{x}^{2}-\text{250}x$ FOIL
3. Check by trying a number. This should work for any number. I’ll use $x=7$ and a calculator.
• $\text{40}{x}^{3}-\text{250}x\stackrel{?}{=}\text{10}x\left(2x+5\right)\left(2x-5\right)$
• $\text{40}{\left(7\right)}^{3}-\text{250}\left(7\right)\stackrel{?}{=}\text{10}\left(7\right)\left(2\cdot 7+5\right)\left(2\cdot 7-5\right)$

I stress these methods of checking answers, not just because checking answers is a generally good idea, but because they reinforce key concepts. The first method reinforces the idea that factoring is multiplication done backward . The second method reinforces the idea of algebraic generalizations.

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