SOLUTION: for what values of k will the remainder be the same when x^2 +kx +4 is divided by x-1 and x+1?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: for what values of k will the remainder be the same when x^2 +kx +4 is divided by x-1 and x+1?      Log On


   

Question 816802: for what values of k will the remainder be the same when x^2 +kx +4 is divided by x-1 and x+1?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
One way to solve this is to divide by x-1 and x+1, set the remainders equal and solve for k. (I hope you've learned synthetic division because that is how I will divide. Long division will get the same result.)
1  |   1   k    4
----       1    k+1
      -------------
       1   k+1  k+5

-1 |   1   k    4
----       -1   -k+1
      -------------
       1   k-1  -k+5
k+5 = -k+5
Add k:
2k + 5 = 5
Subtract 5:
2k = 0
Divide by 2:
k = 0

Another way to solve this uses two key facts:
  • The Remainder Theorem tells us that the remainder is equal to the y-coordinate for the x-value that makes the divisor zero.
  • x%5E2%2Bkx%2B4 is a parabola when graphed. Parabolas are symmetric about the line through the vertex.
From the first fact and from the fact that the remainders are equal, we get that y-coordinate for x=1 (which makes (x-1) zero) and the y-coordinate for x = -1 (which makes (x+1) zero) are the same. From the symmetry of a parabola we get that x=1 and x=-1 can have the same y-coordinates only if they are equally distant from the vertex. Halfway between 1 and -1 is zero. So the x-coordinate has to be zero in order for x=1 and x=-1 to have the same y-coordinates.

And finally, in the general form for parabolas, ax%5E2%2Bbx%2Bc, the x-coordinate of the vertex is given by -b%2F2a. Since we have determined that this x-coordinate must be zero:
0+=+-b%2F2a
Since a fraction is equal to zero only if the numerator is zero, the only way this equation can be true is if b = 0. And since the "b" of x%5E2%2Bkx%2B4 is "k", k must be zero.