Question 232400
If ( x + 3) is a factor of {{{x^3 - 2kx + k^2}}} , then (x+3) will divide into {{{x^3 - 2kx + k^2}}} evenly. And synthetic division is probably the easiest way to find out if (x+3) divides evenly:
<pre>
-3 |  1   0   -2k   k^2
----     -3    9  -27+6k               
     -----------------------
      1  -3  9-2k   k^2+6k-27
</pre>
When a division works out evenly, the remainder is zero. So the value(s) of k that makes {{{k^2+6k-27}}} zero will be our answer(s):
{{{k^2+6k-27 = 0}}}
This is a quadratic equation which we can solve by factoring or with the quadratic formula. It factors pretty easily:
{{{(k+9)(k-3) = 0}}}
By the Zero Product Property one of these factor must be zero:
{{{k+9 = 0}}} or {{{k-3 = 0}}}
Solving these we get:
{{{k = -9}}} or {{{k = 3}}}
So we have two possible values for k.