Question

how would you find the solutions to
x^3-x^2-4x+4=0?

Answer 1

You could guess a root. In this case x=1 works, so you can factor x-1 out. Then divide and solve the quadratic the usual way.

Answer 2

i know the answers... there -2,1 and 2. i know the first step is to factor x^3-x^2-4x+4 when you factor that you get (x-2)(x-1)(x-2)

Answer 3

i wanna know how to get x-2)(x-1)(x-2)

Answer 4

how do you factor it

Answer 5

Well, there's a formula, like for quadratics. But believe it or not, it's usual to just guess at some solution x and use long division.

Answer 6

how will that help me factor?

Answer 7

Do grouping first.

(x^3-x^2) (4x+4)
then look what their common factor is.

Answer 8

i know "x" is

Answer 9

If you have a solution x1 then x-x1 divides the original polynomial. You divide the original polynomial by x-x1 and go on from there.

Can you divide polynomials?

Answer 10

yes i can.
http://www.wolframalpha.com/input/?i=x^3-x^2-4x%2B4%3D0#socialwelcome
these are the steps, i just wanna know how to factor...

Answer 11

You mean how to factor after you know x=1 is a root or how to find x=1 in the first place?

Answer 12

u might have to scroll down next to answer and click on step by step solution

Answer 13

i just wanna know how to factor x^3-x^2-4x+4

Answer 14

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

Answer 15

is this it? If you are factoring wit those numbers, i think you use grouping, right?

Answer 16

thanks but ur factoring (x^3 - x^2) (4x +4)
i need x^3-x^2-4x+4

Answer 17

you group them

Answer 18

then you find gcf

Answer 19

OH! Okay

I think i got it now
(x^3-x^2)(-4x+4)
x^2(x-1)4(x-1)
x^2+4 and x-1
x^2+4=0
x^2=-4
square root it and that's the answer which is +2i and -2i
x-1 x=0,+ and -2i

Answer 20

you copied the wrong equation hs.

Anyway writing it as \

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

etc.

Answer 21

thanks