How do you evaluate #(5y - 7) ( 5y ^ { 2} - y - 1)#?

2 Answers
Dec 10, 2017

As no identities apply on this, you have to do it by distributive property
#5y(5y^2-y-1) - 7(5y^2-y-1)#
#[25y^3-5y^2-5y]-[35y^2-7y-7]#
#25y^3-5y^2-5y-35y^2+7y+7#
Regroup
#25y^3-5y^2-35y^2-5y+7y+7#
You get
#25y^3-40y^2+2y+7#

Dec 10, 2017

#color(white)color(green)(25y^3-40y^2+2y+7)#

Explanation:

#(5y-7)(5y^2-y-1)#

#color(white)(aaaaaaaaaaaaa)##5y^2-y-1#
#color(white)(aaaaaaaaaaa)## xx underline(5y-7)#
#color(white)(aaaaaaaaaaaaa)##25y^3-5y^2-5y#
#color(white)(aaaaaaaaaaaaaaaaa)##-35y^2+7y+7#
#color(white)(aaaaaaaaaaaaa)##overline(25y^3-40y^2+2y+7)#

#color(white)(aaaaaaaaaaaaa)##color(green)(25y^3-40y^2+2y+7)#