Use synthetic division to divide f(x) by x - k for the given value of k. Then express f(x) in the form f(x)=(x-k)q(x)+r for the given value of k.

f(x) = -5x4 + x3 + 2x2 + 3x - 1; k = 1

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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step-by-step explanation:

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answer: i'm great at algebra

~coco

step-by-step explanation: