Remainder Theorem

how to avoid Polynomial Long Division when finding factors!!

"7 divided by 2 equals 3 with a remainder of 1"

Each part of the division has special names:

We can write as a sum like:

Polynomials:

Polynomial is an expression containing more than 2 algebraic terms.

We can also divide polynomials.

f(x) ÷ g(x) = q(x) with a remainder of r(x)

But it is better to write it as a sum like this: 

Example:

                             2x²-5x-1 divided by x-3

•  f(x) is 2x²-5x-1
• g(x) is x-3

After dividing we get the answer 2x+1, but there is a remainder 2.

• q(x) is 2x+1
• r(x) is 2

we can write:

                  f(x) = g(x)·q(x) + r(x)           

          2x²-5x-1 = (x-3)(2x+1) + 2

When you divide by a polynomial of degree 1 (such as "x-3") the remainder will have degree 0.

2. The Remainder Theorem

When you divide a polynomial f(x) by x-c you get:

f(x) = (x-c)·q(x) + r(x)

But r(x) is simply the constant r (remember) when you divide by (x-

c) the remainder is a constant)

f(x) = (x-c)·q(x) + r

Now put x equal to c:

f(c) = (c-c)·q(c) + r
f(c) = (0)·q(c) + r
f(c) = r

So we get this:

The Remainder Theorem:
When you divide a polynomial f(x) by x-c the remainder r will be f(c)

So if you want to know the remainder after dividing by x-c you don't need to do any division:

Just calculate f(c).

Example: 

2x²-5x-1 divided by x-3

We don't need to divide by (x-3) ... just calculate f(3):

2(3)²-5(3)-1 = 2x9-5x3-1 = 18-15-1 = 2

And that is the remainder we got from our calculations above.
We didn't need to do Long Division at all!

Example:

 Dividing by x-4

(Continuing our examplee)

What would the remainder be if we divided by "x-4" ?
"c" is 4, so let us check f(4):

2(4)²-5(4)-1 = 2x16-5x4-1 = 32-20-1 = 11