I want to know how to solve and check my answer

## (4x^3+8x^2-x-2)/(x+2) =

# 2 Answers

One method is "long division" . Using this method, you compare the leading terms of your numerator and denominator.

How many times does x (from x + 2) go into 4x^{3} ? 4x^{2} times (x * 4x^{2} = 4x^{3}). So we also multiply the 2 (from x + 2) by 4x^{2} (to get 8x^{2}), and now we subtract it from our numerator, and we are
left with (-x-2)/(x + 2)

Now, how many times does x go into -x ? -1 times. Multiply and subtract:

-x - 2 - (-1)(x + 2) = 0

So our result is 4x^{2} - 1

**__4x ^{2} ______
-1 _________________**

x + 2 | 4x^{3} + 8x^{2} - x - 2

4x^{3} + 8x^{2}

---------------------

-x - 2

-x - 2

--------

0

A way to check your answer, which is correct from Kevin S., is to multiply both the numerator and denominator by the original denominator (x+2). As shown:

(4x^2 - 1) / 1 * (x + 2) / (x + 2) (doing this step is effectively multiplying your answer by "1")

(4x^2 - 1)*(x + 2) / (x+2)

(4x^3 + 8x^2 - x - 2) / (x + 2) (FOIL)

**Since we got the same answer, the solution checks out. **

## Comments

Synthetic Division is another method.

In Synthetic Division, we "assume" that the denominator, x + 2, defines a zero; that is, x + 2 = 0 ==> x = -2

We then will set up our "division" problem, using only the coefficients and our "zero":

-2 | 4 8 -1 -2

|____________

We bring down our first coefficient unchanged:

-2 | 4 8 -1 -2

|____________

4

Then we multiply our zero by the number we brought down, and place in our problem:

-2 | 4 8 -1 -2

|___-8_________

4 0

We repeat for each term:

-2 | 4 8 -1 -2

|___-8__0 _______ <=== -2 * 0 = 0, so that gets inserted

4 0 -1

-2 | 4 8 -1 -2

|____-8__0 __2_____

4 0 -1 0

So our result would be 4x

^{2}- 1, which is the same result we got above.I'm not sure how synthetic division would work if you didn't have a linear divisior (denominator), such as the problem (4x

^{4}+ 8x^{3}+ 2x^{2}- 8x - 8)/ (x^{2}- 3x + 2), but long division would work in this case.Comment