Answer
$P(x)=(x-1)(x+1)(x+2)(x+4)$
Zeros: $ \ -4,\ -2, \ -1, \ 1 $
Work Step by Step
$P(x)=x^{4}+6x^{3}+7x^{2}-6x-8$
$P(-x)=x^{4}-6x^{3}+7x^{2}+6x-8$
Descart's rule of signs:
P(x) has 1 sign changes $\Rightarrow$ 1 positive zeros.
P(-x) has 3 sign changes $\Rightarrow$ 3 or 1 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4,\pm 8$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4,\pm 8$
Testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & 6 & 7 & -6 & -8\\\hline
& 1 & 7 & 14 & 8\\\hline
1& 7 & 14 & 8&|\ \ 0\end{array}$
$P(x)=(x-1)(x^{3}+7x^{2}+14x+8)$
Again, testing with synthetic division,
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1& 7 & 14 & 8\\\hline
& -1 & -6& -8\\\hline
1& 6 & 8 &|\ \ 0\end{array}$
$P(x)=(x-1)(x+1)(x^{2}+6x+8)$
... factor the trinomial (factors of 8 with sum 6 ... are 4 and 2)
$P(x)=(x-1)(x+1)(x+2)(x+4)$
Zeros: $ \ -4,\ -2, \ -1, \ 1 $
