x^3+2x^2-3x-6?

Question

Factor the Following polynomial completely.

x^3+2x^2-3x-6

I have forgotten how to do it. Please show work. Thanks

Answer 1

*Rule: with 4 terms - combine "like" terms.

(x^3 - 3x) + (2x^2 - 6)

First: factor both sets of parenthesis.

(x^3 - 3x) + (2x^2 - 6)
x(x^2 - 3) + 2(x^2 - 3)

Sec: the "inner" terms (in parenthesis) need to match. combine the inner terms (once) with the outer terms.

(x^2 - 3)(x+2), or (x+2)(x^2-3)

Answer 2

(x^3 + 2x^2 + 3x - 6) = (x - a million) (x^2 + 3x + 6) you do no longer choose long branch btw, only factorise. speedier and simpler! had to combat with my a point instructor to be waiting to do it this way.

Answer 3

first you check for divisors of -6 ( they are 1, -1, 2, -2, 3, -3, 6, -6)
as for (-2) you get 0 , this is a root
you divide the polynomial by (x=2) and get x^2-3, then you factorize it:
(x+2)(x^2 -3) = (x+2)(x+square root of 3)(x-square root of 3)

Answer 4

x^3+2x^2-3x-6=
x(x^2 -3) +2(x^2 -3) =
(x^2 -3)(x+2) = (x - √3)(x + √3)(x + 2)

Answer 5

x^3 + 2x^2 - 3x - 6
(x^3 + 2x^2) - 3(x + 2)
x^2(x + 2) - 3(x + 2)
(x^2 - 3)(x + 2)

Answer 6

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

Answer 7

X^2(X+2)-3(x+2)=(X^2-3)(X+2) if you multiple the answer it will give back the problem so u know your right

Answer 8

x^3 + 2x^2 -- 3x -- 6
= x^2(x + 2) -- 3(x + 2)
= (x + 2)(x^2 -- 3)
= (x + 2)(x + sqrt3)(x -- sqrt3)