How do u factor this completely? x3 − 2x2 − 4x + 8?

Question

x3 − 2x2 − 4x + 8

Answer 1

    X^3 -2X^2 -4X + 8.  First and for most,  We look for the factors of X^3 (which are x, x^2 & X^3) and + 8 (which are +1, -1, +2, -2, +4, -4, +8, -8)

      If X +2 is a factor, then X+2 = 0 then X= -2 . we substitute for X =-2 in the expression

i.e. (-2)^3 -2(-2)^2 - 4(-2) +8 =0 therefore X+2 is factor.

Similarly, If X -2 is a factor, then X - 2= 0 and X = 2, substitute for X = 2 in the expression



   i.e. (2)^3 -2(2)^2 -4(2) +8 = 0, therefore X - 2 is factor

  X^3 -2X^2 -4X +8   =  (X+2) (X-2) (X-2) are the factors  

Answer 2

2 is a factor of this polynomial, so take x - 2 as a factor and write 3 times because the degree is 3 and then multiply by suitable terms.

After factorization the answer will be (x - 2) ( x - 2 ) ( x + 2)

Answer 3

You use group factoring.

Factor a x^2 out of x^3 and 2x^2 and you have

x^2 * (x-2)

Now factor a -4 out of th eother 2 and you have

x^2 *(x-2)-4(x-2)

now you have:

(x-2)(x^2 -4)

But we're not DONE!

Do you see the x^2 -4 ? That is a difference of squares, so you have

The differenc of suqares is: (x-2)(x+2)

Don't forget the other (x-2)

you have:

(X-2)(X+2)(X-2)

Which can be broken down into:

(x+2)(x-2)^2

Multiply it out and you'll see it's right!

Answer 4

This one's easy, because both

x^3 - 2x^2 and

-4x + 8

are obviously divisible by (x - 2), so you can just perform that division to get it down to factoring a quadratic equation:

x^3 - 2x^2 - 4x + 8 = (x - 2) (x^2 -4)

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

because (x^2 - 4) is the difference of two squares.

Answer 5

(x^2 -4) (x-2)

then (x^2-4):

(x-2)(x+2)

Answer 6

(x+2)*(x-2)^2

when you put x=2 ; you find it as a solution of x ; separate it and find the remaining terms which comes out to be x^2-4 ,

Answer 7

(x^2 +4) (x+2)