# Factoring Problem

1. Feb 18, 2006

### zombeast

I'm in calculus, but we're having this test on some curve sketing soon and I'm doing some practice problems that the teacher gave us. We're not allowed to use a calculator in this calculus class (except for very rare instances) so we have to know how to sketch graphs with no problem!

I'm wondering I have factored this correctly. If not, I'm wondering if someone can point me in the right direction.

Orginal Equation:
$$x^3 - 3x^2-6x+8$$

My Factors
$$(x+2)(x^2-2x+4) - 3x(x+2)$$

Turns into ...

$$(1-3x)(x+2)(x^2 -2x + 4)$$

If anyone could help me out I would appreciate it.

Thanks!

2. Feb 18, 2006

### 0rthodontist

You can always multiply them back out to see if it gets you where you started.

3. Feb 18, 2006

### HallsofIvy

Staff Emeritus
Your original polynomial has degree 3 yet you have 4 factors involving x!

Hint: $$(x+2)(x^2-2x+4) - 3x(x+2)$$ is correct but
ab- ca= a(b- c) NOT ab(1-c).

4. Feb 19, 2006

### zombeast

Ok, thanks for the help thus far. I was able to remember (through the help of another math friend) that if all the coefficients of the polynomial add up to zero, it is divisble by x-1 and x-1 is one of the factors. Therefore I had to take x-1 and divide it into $$x^3 - 3x^2 -6x+8$$ and I was given $$x^2-2x-8$$. From that I could get the other two factors $$(x-4)(x+2)$$

Thanks for looking at it though.

5. Feb 19, 2006

### benorin

Your factored expression is a 4th degree polynomial while your original expression is a 3rd degree polynomial.

As HallsofIvy said, $$(x+2)(x^2-2x+4) - 3x(x+2)$$ is correct, and the factored form should be

(x+2)(2nd degree poly).