How do you divide #(a^3-19a^2+97a-63)-:(a-9)#?

1 Answer
Jan 8, 2017

#(a^3-19a^2+97a-63)/(a-9) = a^2-10a+7#

Explanation:

One way is to factor by grouping, splitting the terms so that each binomial is a multiple of #(a-9)# as follows:

#a^3-19a^2+97a-63 = (a^3-9a^2)-(10a^2-90a)+(7a-63)#

#color(white)(a^3-19a^2+97a-63) = a^2(a-9)-10a(a-9)+7(a-9)#

#color(white)(a^3-19a^2+97a-63) = (a^2-10a+7)(a-9)#

Hence:

#(a^3-19a^2+97a-63)/(a-9) = a^2-10a+7#