# f(x) = X^3 + 2x^2 -4x -8: For zeroes of polynomial functions

• Aug 7th 2011, 12:27 PM
Ingersoll
f(x) = X^3 + 2x^2 -4x -8: For zeroes of polynomial functions
This seems pretty simple but the book gave an unexpected answer

f(x)= x^3 +2x^2 -4x -8

0= the same as above

$x^2(x+2)$ and $-4(x+2)$

$x^2 -4= 0$ and $x+2= 0$

So that: +/- 4 and -2

This gives me (-4,0), (-2,0), and (4,0)

However the book gives the answer as {-2,2} without previously giving a similar answer to similar problems. Any guidance is much appreciated. Thanks.
• Aug 7th 2011, 12:30 PM
e^(i*pi)
Re: f(x) = X^3 + 2x^2 -4x -8: For zeroes of polynomial functions
By inspection f(2) = 2^3 + 2(4) - 4 - 8 = 0. Thus $(x-2)$ is a factor of f(x) by the factor theorem.

$f(x) = (x-2)(Ax^2+Bx+C)$ where A, B and C are constants to be found. You can either use long division or comparing coefficients to find them.

Once you have the quadratic you need to check if that will factor too
• Aug 7th 2011, 12:36 PM
FernandoRevilla
Re: f(x) = X^3 + 2x^2 -4x -8: For zeroes of polynomial functions
$x^3+2x^2-4x-8=x^2(x+2)-4(x+2)=(x+2)(x^2-4)$ . So,

$x^3+2x^2-4x-8=0\Leftrightarrow (x+2)(x^2-4)=0\Leftrightarrow( x+2=0)\vee (x^2-4=0)\Leftrightarrow$

$x=-2\vee x=\pm 2$ .

i.e. the solutions are $x=2$ (simple) and $x=-2$ (double) .
• Aug 7th 2011, 01:04 PM
Ingersoll
Re: f(x) = X^3 + 2x^2 -4x -8: For zeroes of polynomial functions
Cool. Got it. It's always the dumb mistakes... Anyway, thanks to both of you very much.