Question 816802
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.)
<pre>
1  |   1   k    4
----       1    k+1
      -------------
       1   k+1  k+5

-1 |   1   k    4
----       -1   -k+1
      -------------
       1   k-1  -k+5
</pre>k+5 = -k+5
Add k:
2k + 5 = 5
Subtract 5:
2k = 0
Divide by 2:
k = 0<br>
Another way to solve this uses two key facts:<ul><li>The Remainder Theorem tells us that the remainder is equal to the y-coordinate for the x-value that makes the divisor zero.</li><li>{{{x^2+kx+4}}} is a parabola when graphed. Parabolas are symmetric about the line through the vertex.</li></ul>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.<br>
And finally, in the general form for parabolas, {{{ax^2+bx+c}}}, the x-coordinate of the vertex is given by {{{-b/2a}}}. Since we have determined that this x-coordinate must be zero:
{{{0 = -b/2a}}}
Since a fraction is equal to zero <i>only</i> if the numerator is zero, the only way this equation can be true is if b = 0. And since the "b" of {{{x^2+kx+4}}} is "k", k must be zero.