prerequisite calculus homework?

Question

I had an equation and after I worked it out, it came down to a tri-nomial (I think that's what it's called). But I can't seem to factor it, and I've been at it for quite some time trying to find what x equals. If it was any other typical x^2 I would use the quadratic formula but I can't with this. so how do I solve for x if this won't factor?

I have this so far:
Solve: (x+2)^2(x-1) + (x+2)(x-1)^2 = 0

(x+2)^2(x-1) + (x+2)(x-1)^2 = 0
(x^2 + 4x + 4)(x-1) + (x+2)(x^2 - 2x + 1) = 0
(x^3 + 4x^2 + 4x - x^2 - 4x -4) + (x^3 - 2x + x + 2x^2 - 4x + 2) = 0
(x^3 + 3x^2 - 4) + (x^3 - 3x + 2) = 0
2x^3 + 3x^2 - 3x - 2 = 0

and someone told me if I considered it to be like (a^2)b + a(b^2) = 0, but that doesn't really help or make sense either.
thanks in advance.

Answer 1

2x^3 + 3x^2 - 3x - 2 = 0

this is correct, but you're missing the last little step. the above equation can be factored. Unfortunately that factoring is really a pain, so do it the easy way!

(x+2)^2(x-1) + (x+2)(x-1)^2 = 0

Now factor out an (x+2)(x-1) from the whole thing so that you get

(x+2)(x-1)*((x+2)+(x-1)) = 0

Now add up the terms inside the brackets, and you get

((x+2)+(x-1)) = 2x+1

and now the equation is

(x+2)(x-1)(2x+1) = 0

And there you go!

Answer 2

Let p(x)= 2x^3+3x^2-3x-2
Find p(1)=2(1)^3 + 3(1)^2 –3(1)-2
= 2(1)+3(1)-3-2
=2+3-3-2
=5-5
=0
So by remainder theorem x-1 is a factor of 2x^3+3x^2-3x-2=0
On dividing p(x) by x-1 we get 2x^2+5x+2
On factoring 2x^2+5x+2=0
2x^2+x+4x+2=0
x(2x+1)+2(2x+1)=0
(2x+1)(x+2)=0
So factors of 2x^3+3x^2-3x-2=0 are (x-1) (2x+1) (x+2)=0

Answer 3

Solve: (x+2)^2(x-1) + (x+2)(x-1)^2 = 0 *is a times sqr is squareroot

x^2+4x^2+2x-x+2*x^2+1=0
6x^2+x+2*1=0
6x^2+x+2=0
this is as far as i get till i can call my friend so ill finish (if we can) later
srry