Here goes...

Let's start with the equation 20x^2 -44x -15 =0, and were just going to forget about the =0 part for now.

The part were going to concern ourselves with, 20x^2 -44x -15, is referred to as a

**POLYNOMIAL**.

**POLY** means "many" or "several", "more than one".

**NOMIAL** means "number". So polynomial means "many numbers".

Each part of the polynomial is called a

**TERM**. The parts or term's are [20x^2], and [-44x], and [-15].

Each term has a

**COEFFICIENT** which is the number part, [

**20**x^2], [

**-44x**], [

**-15**]

Each term may or may not have a

**VARIABLE**. The variable is the letter part, [20

**x**^2], [-44

**x**].

If a term has a coefficient but no variable, then that term is called a

**CONSTANT**.

In our particular polynomial, [-15] is a constant.

Taking this, 20x^2-44x-15, and turning it into this, (10x+3)(2x-5), is called

**FACTORING**.

Factoring is when a term or polynomial is broken down into smaller parts,

__but only under the operation of __**multiplication**.

By the way, this (10x+3), and this (2x-5), are each called a

**BINOMIAL**.

**BI** means "two", and you know what "nomial" means.

To turn our binomials, (10x+3)(2x-5), back into our polynomial, 20x^2-44x-15, we can use the

**F.O.I.L.** method.

**FIRST** we recombine the first terms(

**10x**+3)(

**2x**-5) for

**20x^2**.

**OUTSIDE** terms are recombined(

**10x**+3)(2x

**-5**) for 20x^2

**-50x**.

**INSIDE** terms (10x

**+3**)(

**2x**-5) for 20x^2-50x

**+6x**.

**LAST** terms (10x

**+3**)(2x

**-5**) for 20x^2-50x+6x

**-15**.

And 20x^2-50x+6x-15=20x^2-44x-15.

Originally Posted by

**somfooleishfool**
I find the textbook to be lacking in assumed knowledge. I found there is a whole other step needed in quadratics when a is not = 1 in ax^2. This for some reason was not explained in the textbook but I found it in my lecture notes.

Let me thank you for not making me explain that part. You posted while I was still writing this.

As an exercise, can you explain why setting the

**polynomial** to zero and

**factoring** it out into

**binomials** provides potential solutions to the equation in the

**constant's** of the

**binomial's**?