# Algebra problems

• Feb 18th 2008, 05:21 PM
p4l1ndr0m3
Algebra problems
I've got my head stuck around these 2 problems. I'm trying to factor these polynomials, but I just can seem to get them.

The first one is f(x) = x^3 + x^2 - 8x - 6

The 2nd one is f(x) = x^5 + 3x^4 - 4x^3 - 11x^2 - 3x + 2

I'd appreciate any tips, thanks!
• Feb 18th 2008, 05:35 PM
galactus
For the second one, try dividing by a possible root. Try x-2 and/or x+1

If it reduces down, then that's a root and you're on your way.

$\displaystyle \frac{x^{5}+3x^{4}-4x^{3}-11x^{2}-3x+2}{x-2}=x^{4}+5x^{3}+6x^{2}+x-1$

See?. That reduced down to a quartic. Try dividing that by another factor or some other tactic and whittle away at it.
• Feb 18th 2008, 05:41 PM
OzzMan
Is that the only way to go about factoring these?
• Feb 18th 2008, 05:42 PM
geton
For 1.

f(-3) = -27 + 9 + 24 - 6 = 0

So (x+3) is a factor of f(x).

By using long division,

f(x) = (x+3) (x^2 - 2x - 2)

Then do factories.

Same for 2.
• Feb 18th 2008, 05:53 PM
OzzMan
why exactly did you do f(-3) for the first one? where did the -3 come from exactly. Sorry I'm posting on this thread when its not mine but I'm just interested in this problem.
• Feb 18th 2008, 06:04 PM
p4l1ndr0m3
Quote:

Originally Posted by OzzMan
why exactly did you do f(-3) for the first one? where did the -3 come from exactly. Sorry I'm posting on this thread when its not mine but I'm just interested in this problem.

Feel free, this helps me too!
• Feb 18th 2008, 06:29 PM
geton
Well, it is not exactly f(-3). You’ve to take some values of x, like +1, -1, +2, -2………. After doing some exercise you can easily realize that what is the factor of f(x), without doing any rough.