Algebra help please-- factoring polynomials?

Question

Hello. please help me, thanks for your help every one.
***Factor each polynomial.

x^3+8x^2+15x
x^3+10x^2+24x
3x^3-300x
x^3+3x^2-x-3
x^3+x^2+x+1

thank you... bless u!!!!!!!!
also, show the work if u can for a few. thx.

Answer 1

1) x^3+8x^2+15x factor out the x
x (x^2 + 8x + 15) then you get
x (x + 5) (x + 3)

2) x^3+10x^2+24x
x (x^2 + 10x + 24)
x (x + 6) (x + 4)

3) 3x^3-300x factor out 3x
3x (x^2 - 100)
3x (x - 10) (x + 10)

4) x^3+3x^2-x-3 you have to factor by grouping
(x^3+3x^2)(-x-3) factor out x^2 from the first part then factor out a -1 from the second part
x^2 (x + 3) - 1 (x + 3) then put the parts together
(x^2 - 1) (x + 3) factor first part
(x - 1) (x + 1) (x +3)

5) (x^3+x^2)(+x+1)
x^2 (x +1) +1 (x+1)
(x^2 + 1) (x + 1)

hope that helps you

Answer 2

Notice that in x^3 + 8x^2 + 15x, you have at least one "x" in each term. So factor that out, which gives you x(x^2 + 8x + 15). When it comes to factoring a quadratic (something in the form of (x^2 + bx + c)) into something in the form of (x+m)(x+n), there are a couple of ways to go about it. The simplest is the "Reverse FOIL". Notice that (x+m)(x+n) is x^2 + (m+n)x + mn when you multiply it out. So in this case, we need to find an m and n such that (m+n)=8 and mn=15. Think of the factors of 15: 1, 3, 5, 15. Of these, 3 and 5 fit both conditions. So the x^2 + 8x +15 can be rewritten as (x+5)(x+3). This means the original polynomial can be factored into x(x+5)(x+3).

For the second problem, use the same approach. x^3 + 10x^2 +24x = x(x^2 + 10x + 24), which you can eventually factor out to x(x+6)(x+4).

For 3x^3 - 300x, notice that we can factor out both an x and a 3. So this is 3x(x^2 - 100). We have a minus sign on the end and no "x" term, so we should see if we can factor this into (x+m)(x-m). (Notice that when you multiply out (x+m)(x-m), you get x^2 + mx - mx -m^2, or x^2 - m^2). It turns out that we can factor this to 3x(x + 10)(x - 10).

The last two are a little tricky, because off-hand it looks like you can't factor one thing out of each of the 4 terms. But notice the third one can be rewritten as (x^3 + 3x^2) - (x+3). We see the "x" and "3" showing up. You can factor an x^2 out of the first term and get (x^2)(x+3) - (x+3). Now you have two terms that both have an "x+3" in them, so you can factor again: (x+3)(x^2 - 1). This is just using the distributive rule in reverse. Finally, you can factor x^2 - 1 into (x+1)(x-1), so the answer is (x+3)(x+1)(x-1).

For the last problem, use the same trick. (x^3 + x^2) + (x + 1) = x^2(x + 1) + (x+1) = (x^2 + 1)(x + 1). We can't factor x^2 +1 any further, so the answer stays.

Answer 3

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