Factor: 6x^3 + 23x^2 - 6x - 8 ..step by step, best answer will be rewarded.

.. 6x^3 + 23x^2 - 6x - 8
= 6x^3 + ( 24 - 1 ) x^2 - ( 2 + 4 ) x - 8
= 6x^3 - x^2 + 24x^2 - 2x - 4x - 8
= (6x^3 - x^2 - 2x) + (24x^2 - 4x - 8)
= x (6x^2 - x - 2) + 4 (6x^2 - x - 2)
= (x + 4) (6x^2 - x - 2)
= (x + 4) (2x + 1) (3x - 2)

Do I get a cookie ?

This is a tough one. The rational root theorem says that the factors of this polynomial are the quotients of the last and first terms. (See first reference below)
In this case: +/- 1, 2, 4, 8 / 1, 3, 6

Plug in the simple possibilities: +/- 1/1, 2/1, 4/1, 8/1 into the polynomial.

You find that -4 is a root. So one factor is (x+4).

Divide this into your ploynomial. You get 6x^2-x-2.

The factors of this are: (2x+1)(3x-2)


So the factored polynomial is (x+4)(2x+1)(3x-2)


Go to the second website referenced. It does the factoring symbolically. See below.

6x^3 + 23x^2 - 6x - 8 =
= 6*x^3 +23*x^2 -6*x -8
= 6*(x +4)*(x +0.5)*(x -2/3)
= (x+4)(2x+1)(3x-2)

possible rational roots are factors of 8 (±) over factors of 6. a little trial and error with synthetic division gets you
∙
∙ ∙ ∙ ∙ 6∙ ∙ ∙ 23∙ ∙ ∙ -6∙ ∙ ∙ -8
-4∙ ∙ ∙ ∙ ∙ ∙ -24∙ ∙ ∙ 4 ∙ ∙ ∙ 8
-------------------------------------
∙ ∙ ∙ ∙ 6∙ ∙ ∙ -1∙ ∙ ∙ -2∙ ∙ ∙ 0

so (x + 4) is a factor, and the depressed equation (polynomial) is
6x² - x - 2 which factors as
(2x + 1)(3x - 2)

so 6x³ + 23x² - 6x - 8 = (x + 4)(2x + 1)(3x - 2)

Let f(x) = 6x^3 + 23x^2 - 6x - 8

f(-4) = 0 => x + 4 is a factor

so, (x + 4)(6x^2 + bx - 2) = 6x^3 + 23x^2 - 6x - 8

now, expand to give:

6x^3 + (b + 24)x^2 + (4b - 2)x - 8 = 6x^3 + 23x^2 - 6x - 8

=> b + 24 = 23 and 4b - 2 = -6.........which both give b = -1

so, f(x) = (x + 4)(6x^2 - x - 2) = (x + 4)(3x - 2)(2x + 1)

:)>

why don't you try doing it and repost this with you work, then i might help