Solve -p^2-p^8*p^6

$$- p ^ { 2 } - p ^ { 8 } \cdot p ^ { 6 }$$

Factor

$$p^{2}\left(p^{4}+1\right)\left(-p^{8}+p^{4}-1\right)$$

Solution Steps

Factor out p^{2}.

$$p^{2}\left(-1-p^{6}p^{6}\right)$$

Consider -1-p^{12}. Find one factor of the form kp^{m}+n, where kp^{m} divides the monomial with the highest power -p^{12} and n divides the constant factor -1. One such factor is p^{4}+1. Factor the polynomial by dividing it by this factor.

$$\left(p^{4}+1\right)\left(-p^{8}+p^{4}-1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: -p^{8}+p^{4}-1,p^{4}+1.

$$p^{2}\left(p^{4}+1\right)\left(-p^{8}+p^{4}-1\right)$$

Evaluate

$$p^{2}\left(-p^{12}-1\right)$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 8 and 6 to get 14.

$$-p^{2}-p^{14}$$