## Finite Math and Applied Calculus (6th Edition)

$y^{5}+4y^{4} +4y^{3}-y$
If we factor out y from the first parentheses, $y(y^{2}+2y+1)(y^{2}+2y-1)=...,$ we may recognize a special product, a difference of squares: ... $I^{2}-II^{2}=(I+II)(I-II)$ $=y[(y^{2}+2y)+1]\cdot[(y^{2}+2y)-1]=y[(y^{2}+2y)^{2}-1^{2}]$ ... first term in the brackets is a square of a sum ... ... $(I+II)^{2}= I^{2}+2\cdot I\cdot II+II^{2}$, $=y[(y^{2})^{2}+2(y^{2})(2y)+(2y)^{2}-1]$ $= y(y^{4}+4y^{3}+4y^{2}-1)$ $=y^{5}+4y^{4} +4y^{3}-y$