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

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

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

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.

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 $\displaystyle (x-2)$ is a factor of f(x) by the factor theorem.

$\displaystyle f(x) = (x-2)(Ax^2+Bx+C)$ 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

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

$\displaystyle x^3+2x^2-4x-8=x^2(x+2)-4(x+2)=(x+2)(x^2-4)$ . So,

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

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

i.e. the solutions are $\displaystyle x=2$ (simple) and $\displaystyle x=-2$ (double) .

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.