# What are the roots of x^3 + 9 x^2 + 6x -16 ?

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We need to solve x^3 + 9 x^2 + 6x -16

x^3 + 9 x^2 + 6x -16

=>x^3 - x^2 + 10x^2 + 16x – 10x – 16 =0

=>x^3 – x^2 + 10x^2 – 10x + 16x -16 =0

=>x^2 (x-1) + 10x(x-1) + 16 (x-1) =0

=>(x -1) (x^2 + 10x + 16) =0

=>(x-1)(x^2 + 8x + 2x +16) =0

=>(x -1)[x (x+8) +2 (x+8)] =0

=>(x -1) (x+2) (x +8) =0

For x – 1 = 0, x = 1

For x +2 =0, x = -2

For x +8 =0, x = -8

**Therefore the roots of x^3 + 9 x^2 + 6x -16 are (-8, -2, 1).**

To find the roots of x^3+9x^2+6x-16 = 0

The sum of the coefficients , 1+9+6-16 = 0.

Therefore x= 1 is a root.

So x-1 is a factor.

Therefore x^3+9x^2+6x-16 = (x-1)(x^2+kx+16)

We equate the coefficients of x on both sides:

6= 16-k.

So k = 16-6 = 10.

Or k = 10

Therefore x^2+kx+16 = x^2+10x+16 = (x+8)(x+2).

So x^3+9x^2+6x-16 = 0 = (x-1)(x+8)(x+2).

So the other roots are given by x+8 = 0, or x+2 = 0.

x+8 = 0 gives x= -8.

x+2 = 0 gives x= -2.

So the x =1, x=-8 and x= -2 are the roots of the given equation.