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

1 Answer
Jun 23, 2018

#p^4+2p^3+2p^2-2p-3=(p-1)(p+1)(p^2+2p+3)#

Explanation:

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 #+-1# into the polynomial and seeing if the result is 0 because these substitutions are easy and can be done quickly. If #+-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+2p^3+2p^2-2p-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+2p^3+2p^2-2p-3=(p-1)(p^3+3p^2+5p+3)#. Now trying #p=-1# into the cubic polynomial gives 0 so we get #(p-1)(p^3+3p^2+5p+3)=(p-1)(p+1)(p^2+2p+3)#. Looking at the quadratic polynomial, we know it is irreducible because no factors of 3 add up to 2 and we are done.