Express f(x) in the form f(x)=(x-k)q(x)+r

Question

Express f(x) in the form f(x)=(x-k)q(x)+r

Answer 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.

 

 

 

Answer 2

q(x) is the quotient when F(x) is divided by x + 1 and r is the remainder.

 

Dividing synthetically:

 

-1⌋       2       1         1       -6

 

          _____   -2____1___   -2

            2       -1        2       -8

 

q(x) = 2x² - x + 2 and r = -8

 

So, F(x) = (x + 1)(2x² - x + 2) - 8