How do I factor (x^3 + 8) ?
Any help would be greatly appreciated!
Thanks in advance!
Let $\displaystyle f(x)=x^{3}+8$
Since $\displaystyle f(-2)=0$, by factor theorem, $\displaystyle (x+2)$ is a factor.
So $\displaystyle x^{3}+8=(x+2)(ax^{2}+bx+c)$
Comparing coeff. of $\displaystyle x^{3}$, $\displaystyle a=1$
Comparing constants, $\displaystyle c=4$
Comparing coeff. of $\displaystyle x^{2}$, $\displaystyle b+2a=0$
$\displaystyle b=-2$
Note: you can find the quadratic factor by long division also.
Thus
$\displaystyle f(x)=(x+2)(x^{2}-2x+4)$