Question

Find the value of a, of b and of c in the identity \[x^{3}+3x ^{2}-2x+16-c(x+2) = ax ^{2}(x-1)+b(x-2)^{2}(x-1)\]

Answer 1

i don't see a snap way to do this
sometimes it is easy by picking a nice value for \(x\) but i can't think of one here

Answer 2

you can either multiply all this junk out and equate like coefficients
or you can replace \(x\) by say 0, 1, 2 and get three equations in \(a, b,c\)

Answer 3

I tried the method of comparing coefficients but I go an extremely weird answer and along the way had 6=0 :/

Answer 4

put \(x=1\) first

Answer 5

you get
\[1+3-2+16-3c=0\]

Answer 6

oh right, okay

Answer 7

i get
\[c=6\] and then you can find exactly what the left side is