# How do you factor completely p^4+2p^3+2p^2-2p-3?

${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3 = \left(p - 1\right) \left(p + 1\right) \left({p}^{2} + 2 p + 3\right)$
When factoring polynomials, a good strategy is to use the Rational Root Theorem. When applying this theorem, it is a good idea to try substituting $\pm 1$ into the polynomial and seeing if the result is 0 because these substitutions are easy and can be done quickly. If $\pm 1$ doesn't give 0, try substituting in other factors of the constant term divided by factors of the leading term.
Looking at ${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3$, trying $p = 1$ gives 0 so we know $p - 1$ is a factor. Using synthetic division (or polynomial long division), we see that ${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3 = \left(p - 1\right) \left({p}^{3} + 3 {p}^{2} + 5 p + 3\right)$. Now trying $p = - 1$ into the cubic polynomial gives 0 so we get $\left(p - 1\right) \left({p}^{3} + 3 {p}^{2} + 5 p + 3\right) = \left(p - 1\right) \left(p + 1\right) \left({p}^{2} + 2 p + 3\right)$. Looking at the quadratic polynomial, we know it is irreducible because no factors of 3 add up to 2 and we are done.