__Factoring Quadratic Trinomials__

So, what *is* a quadratic trinomial is? Well, a
trinomial is a polynomial with 3 terms. A quadratic
trinomial has the form:

**ax ^{2} + bx + c**, in
which a, b, and c are constants.

So, how can we factor that? I will show an example. For now, we will only work with quadratic trinomials in which a = 1, because they are much easier.

x^{2} + 5x + 6

To do this, we actually factor the expression into two **binomials**,
instead of a monomial and a polynomial. Can you figure it out?
Nah...I will just tell you: it is (x + 2)(x + 3)

*Keep in mind that the (x+2) and (x+3) can be switched because of the commutative property.*

To look for a pattern, we will multiply the binomials out, except in a vertical form (because I feel like it).

x + 3

__ * x + 2__

2x + 6

__x ^{2 }+ 3x
__

So it checks out. But how on earth did I know it would? (I'm a psychic - just kidding).

You can see that the two constants (2 and 3) I put in the binomials multiplied to give 6, or "c", and they added to the coefficient of x, or "b." So the rule is:

If the x^{2} coefficient is 1,

Find two numbers (say h and r) that multiply to give c and add to give b.
Then make two binomials in the form (x + h) and (x + r).

Yes, it's a little messy, but I'll do an example.

x^{2} + 7x + 12

First list some factors of 12:

12, 1 6, 2 3,
4 -4, -3 (remember you can multiply with negatives
too).

Do any of these pairs have a sum of 7? Yes, 3 and 4.

So we write x^{2} + 7x + 12 = (x +
3)(x + 4)

Let's do another example, except this time a little harder.

x^{2} + 4x - 21

First list factors of **-21** (we take the sign in front of
it to determine whether it's pos. or neg.)

-21, 1 -7, 3 -3,
7 -1, 21

Do any have a sum of 4? Yes, -3 and 7.

So we write (x - 3)(x + 7).

Some trinomials cannot be factored using integers, example is x^{2}
+ 4x + 17.

If you want a little practice, try these:

1. x^{2} + 4x - 5

2. x^{2} - 3x + 2

3. x^{2} - 6x - 7

4. x^{2} + 4x + 4

Answers

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