How do you use the remainder theorem to evaluate #f(a)=a^3+5a^2+10a+12# at a=-2?

Question

3297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-11T04:46:08

the remainder of the division #(a^3+5a^2+10a+12):(a+2)# is 4

We simply will substitute #a=-2# in #f(a)#:

#f(-2)=(-2)^3+5(-2)^2+10(-2)+12#

#=-8cancel+20cancel(-20)+12#

#=4#

that means the remainder of the division:

#(a^3+5a^2+10a+12):(a+2)#

is 4