Factoring Help

[hockeychick4018]

I have to factor the following equations.

f(x)=x^3+2x^2-51x+108

f(x)=2x^3-15x^2+34x-21

f(x)=5x^3-x^2-18x+8

f(x)=8x^6+125

I attempted to do grouping on the majority of them, but the answers didn't come out correctly. For example, I tried...
x^3+2x^2-51x+108
(x^3+2x^2)-(51x+108)
x^2(x+2)-3(17x-36)

Obviously, that isn't correct. Could you please show me how to do one of the first 3, so I can do the others. Also, can you show me how to do the last one; I don't know how to even go about it. Thank you!

 

[Loren]

The first one is not factorable. Possible typo?

I don't see that grouping works for the second one. Try long or synthetic division. For instance, divide the given polynomial by x-1. I think that works. Then factor the remaining trinomial. You should get (x-1)(x+a)(2x+b) where you will discover the values of a and b.

 

[mmm4444bot]

f(x) = 8x^6 + 125

Hello Hockey Chick:

Have you seen the factorization pattern for a Sum of Cubes?

A[sup:3vihavqy]3[/sup:3vihavqy] + B[sup:3vihavqy]3[/sup:3vihavqy] = (A + B) * (A[sup:3vihavqy]2[/sup:3vihavqy] - A * B + B[sup:3vihavqy]2[/sup:3vihavqy])

In the last exercise that you posted (quoted above), each quantity is a cube.

8 is 2[sup:3vihavqy]3[/sup:3vihavqy]

x[sup:3vihavqy]6[/sup:3vihavqy] is (x[sup:3vihavqy]2[/sup:3vihavqy])[sup:3vihavqy]3[/sup:3vihavqy]

125 is 5[sup:3vihavqy]3[/sup:3vihavqy]

Can you use this information to rewrite 8x[sup:3vihavqy]6[/sup:3vihavqy] + 125 as a sum of two cubed quantities?

If so, then factor the sum of these two cubes using the pattern above.

If not, then show us what you tried, and we'll help you get there.

f(x)=5x^3-x^2-18x+8

(x+2) is a factor of this particular f(x) that I found using synthetic division. Dividing f(x) by (x+2) yields 5x^2-11x+4.

The zeros of this quadratic expression are irrational numbers containing ?41, so f(x) does not factor nicely. You can get the other two factors by using the Quadratic Formula.

Cheers,

~ Mark