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:

off the internet