ok last factoring question!?

Question

Factor:

x^3 – 2x^2 – 4x – 8

Answer 1

Factor by grouping. Group the first two and the last two terms together, and then pull out the greatest common factor: (I think you may have the problem written down wrong...one of those minus should probably be a plus)

(x^3 - 2x^2) + (-4x - 8)
x^2(x - 2) -4 (x + 2) (Factor out greatest common factors of x^2 and -4)

In the example you gave, that's all. But let's assume that the problem was instead:

(x^3 - 2x^2) + (- 4x + 8)
x^2 (x - 2) -4 (x - 2) (Pull out the common x^2 & -4)
(x-2)(x^2-4) (Notice both have a x-2 term...pull that out as a common factor)
(x-2)(x-2)(x+2) (Factor x^2 - 4 as a difference of squares)

So your answer to that one would be (x-2)(x-2)(x+2) or (x+2)(x-2)^2

Answer 2

it is factorable
factor it by using GRP(grouping

first, get the common multiple of the first & the second term, and the common multiple of the third and fourth term

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

notice that the terms in the parehesis are alike, that means that you are doing it correctly

then combine the common mulitples then multiply it to the remaining multiple

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

this can still be factorized using DOTS (difference of two squares)
(x^2-4)=(x+2)(x-2)

therefor the answer is
(x+2)(x-2)(x-2)


hope it may help you


^_^

Answer 3

the first term is the cube of x and the last cube of 2.

lets develop (x-2)^3 = x^3 -6x^2 +12x-8

so x^3 -2x^2-4x-8 = (x-2)^3+4x^2-16x= (x-2)^3+4(x^2-4)

= (x-2)^3 +4 (x-2)(x+2) we factorize (x-2)

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

(x-2) (x^2+12)

Answer 4

you cantfactor out an x if the 8 dont haveone