Question

(3w^3+7w^2-4w+3)/(w+3) ? @dmezzullo

Answer 1

Factor it out. The way these problems work is the term on the bottom should cancel out a term on the top. So you should find (w + 3) on top when you factor it out.

Now let's get on to factoring the top:

3w^3 + 7w^2 - 4w + 3

The highest power is 3, so that means either there's a w^2 in your terms or you have 3 terms. At least one term is negative and probably only one is because there are two + symbols.

What we want to do now is look for a binomial expression (that means one that goes up to w^2 instead of w^3) and we want to factor out that (w+3) from what we have now.

So we're looking for the format:

(Aw^2 + Bw - C)(w+3)

How do we get that? Well we know that 3 has to multiply through the first expression and we know that we need 3w^3 to show up. So A = 3, because then we get:

(3w^2)(w) = 3w^3

as the first term. So far, so good.

Now we need to solve for B. We have

3w^3 + 9w^2 + Bw^2 + 3Bw - Cw - 3C

So those two w^2 terms need to add up to 7. We already have 9w^2 so we now need to subtract 2, so B = -2.

Substituting:

3w^3 + 7w^2 - 6w - Cw - 3C

Okay, now our first terms are great. Let's figure out C:

-6w - Cw - 3C = -4w - 3

Therefore C = -2:

-6w + 2w - 6 = -4w - 3
-4w - 6 = -4w - 3

Doesn't work! Is this unsolvable? Let's ask WolframAlpha:

http://www.wolframalpha.com/input/?i=%283w%5E3+%2B+7w%5E2+-+4w+%2B+3%29%2F%28w%2B3%29

They say it factors to (3w^2 - 2w - 2)(w+3) - 3, which is what we arrived at.

Answer 2

Thankss :)