¨f(x)=x^3+2x^2-5x-6?

Question

can someone give me the steps to the solution of this equation. In the book the solution is done through dividing the equation by (x+1). Why?

thank u

Answer 1

The possible integer roots of your expression are - 6, - 3, - 2, -1, 1, 2, 3, and 6. By substituting each into the expression and evaluating to see if the result is 0, you may find one or more roots.
1^3 + 2*1^2 - 5*1 - 6 = - 8 (1 is not a root)

(-1)^3 + 2(-1)^2 - 5(-1) - 6 =
-1 + 2 + 5 - 6 = 0, so - 1 is a root, meaning a factor is (x + 1)
Dividing the original cubic by (x + 1) will yield a quadratic which may be factorable even if there are no more integer roots:
. . . . x^2 + x - 6
x + 1)x^3 + 2x^2 - 5x - 6
. . . . x^3 + 1x^2
. . . . . . . . . . x^2 - 5x - 6
. . . . . . . . . . x^2 + 1x
. . . . . . . . . . . . . - 6x - 6
. . . . . . . . . . . . . - 6x - 6

x^2 + x - 6 then factors into
(x + 3)(x - 2)

f(x) = (x + 1)(x + 3)(x - 2)

You could have found all of these factors by trying roots one-at-a-time, but that's usually not the case. A cubic having even one integer root is fairly rare.

Answer 2

If you can divide by x+1 and have no remainder when your done, then x+1 is a factor of your equation.
By the way, you can also check if x+1 goes into the equation by plugging in -1 for x and see if zero is the answer.

Answer 3

it's usually a great term to start dividing a cubic equation
set it up like long division
you will generate x² + 2x - 6 terms from this