Solve 27-27p^2+p^3-p^5

$$27 - 27 p ^ { 2 } + p ^ { 3 } - p ^ { 5 }$$

Factor

$$\left(-p-1\right)\left(p-1\right)\left(p+3\right)\left(p^{2}-3p+9\right)$$

Solution Steps

Do the grouping 27-27p^{2}+p^{3}-p^{5}=\left(27-27p^{2}\right)+\left(p^{3}-p^{5}\right), and factor out 27 in the first and p^{3} in the second group.

$$27\left(-p^{2}+1\right)+p^{3}\left(-p^{2}+1\right)$$

Factor out common term -p^{2}+1 by using distributive property.

$$\left(-p^{2}+1\right)\left(27+p^{3}\right)$$

Consider -p^{2}+1. Rewrite -p^{2}+1 as 1^{2}-p^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(1-p\right)\left(1+p\right)$$

Reorder the terms.

$$\left(-p+1\right)\left(p+1\right)$$

Consider p^{3}+27. Rewrite p^{3}+27 as p^{3}+3^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(p+3\right)\left(p^{2}-3p+9\right)$$

Rewrite the complete factored expression. Polynomial p^{2}-3p+9 is not factored since it does not have any rational roots.

$$\left(-p+1\right)\left(p+1\right)\left(p+3\right)\left(p^{2}-3p+9\right)$$

Evaluate

$$27-27p^{2}+p^{3}-p^{5}$$