Answer:
the key to this problem is in recognizing that q(x) is a quotient, and r is a remainder.
Since f(x) is a higher degree polynomial than (x-k), If we divide (x-k) into f(x), we'll get a non-fractional quotient and a fractional remainder:
f(x)/(x-k) = q(x) + r/(x-k). Multiplying both sides by (x-k), and applying a little algebra, we have
f(x) = (x-k)q(x) + r
So, all we have to do to get f(x) in the desired form is divide it by (x-k) and then multiply the result again by (x-k). Sounds strange, right? But that's all there is to it! Let's give it a shot:
k= -1, so (x-k) = (x+1).
Now, we find (either with long or synthetic division) f(x)/(x+1) = (2x3+x2+x-6)/(x+1).
f(x)/(x+1) = 2x2 - x + 2 - 8/(x+1).
we can see that q(x) = 2x2-x+2, and
r/(x+1) = -8/(x+1).
Now, multiply both sides by (x+1):
f(x) = (x+1)[2x2-x+2-8/(x+1)]
=(x+1)(2x2-x+2) - (x+1)*8/(x+1). Notice I distributed (x+1) to q(x) and r independently.
=(x+1)(2x2-x+2) - 8.
This is in the form (x-k)q(x) + r.
Step-by-step explanation:
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