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Factoring Quadratics - Part One

Reference > Mathematics > Algebra > Factoring
 

One very common process which is necessary in Algebra is factoring quadratics. The ability to factor quadratics gives us the ability to solve many second degree equations.

In this reading, we're going to look at the very simplest kind of quadratic factoring problem: factoring a quadratic with no leading coefficient.

What does that mean? It simply means that the x2term has no number in front of it.

Example 1: Factor x2 + 6x + 8.

Solution: There are two very important numbers we need to look at in this problem: +6 and +8. The first step in factoring this expression is to find:

Two numbers that add to +6, and multiply to +8.

___ + ___ = +6

___ x ___ = +8

Have you found them? They are: +2 and +4.

2 + 4 = 6
2 x 4 = 8

Once you've found these two numbers, you pretty much have the answer:

x2 + 6x + 8 = (x + 2)(x + 4)

Example 2: Factor x2 - 2x - 15

Solution: The two important numbers here are -2 and -15. We want two numbers that add to -2, and multiply to 15.

Did you find them? Hint: one of them is negative!

-5 + 3 = -2
-5 x 3 = -15

So the answer is (x - 5)(x + 3).

Example 3: Factor x2 - 8x + 12

Solution: We want two numbers that add to -8, and multiply to +12. In this case, both of the numbers are negative:

-6 + -2 = -8
-6 x -2 = +12

So the answer is (x - 6)(x - 2).

Questions

1.
Factor x2 + 7x + 12
2.
Factor x2 - 10x - 75
3.
Factor x2 + 2x + 1
4.
Factor x2 - x - 12
5.
Factor x2 + 12x + 20
6.
Factor x2 - 9x + 18
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Factoring with GCFFactoring with GCF
Factoring Quadratics - Part TwoFactoring Quadratics - Part Two
 

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