How do you divide #(3a^3+17a^2+12a-5)/(a+5)#?
1 Answer
Explanation:
First, let's split the top polynomial up into "multiples" of the bottom polynomial. To see what I mean, let's first try to take care of the
We can see that
#3a^2(a+5) = 3a^3+15a^2# . So, let's separate this from our polynomial:
#3a^3 + 17a^2 + 12a - 5#
#(3a^3+15a^2) + 2a^2+12a-5#
#3a^2(a+5) + 2a^2+12a-5#
See how this "gets rid of" the
We can see that
#2a(a+5) = 2a^2 + 10a# .
#3a^2(a+5) + (2a^2+10a) + 2a-5#
#3a^2(a+5) + 2a(a+5)+ 2a-5#
The next highest term to deal with is the
We can see that
#2(a+5) = 2a+10# .
#3a^2(a+5) + 2a(a+5)+ (2a+10) -15#
#3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15#
We can't do anything about the
Finally, let's divide everything by
#(3a^3+17a^2+12a-5)/(a+5) #
#= (3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15)/(a+5)#
#= 3a^2+2a+2 - 15/(a+5)#
Final Answer