SOLUTION: please help me out with this question find the value(s) of k that make the 2nd polynomial a factor of the first one a)x^3-kx^2+5x+k;x-2 b)x^3+4x^2-11x+k;x+2

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Question 666071: please help me out with this question
find the value(s) of k that make the 2nd polynomial a factor of the first one
a)x^3-kx^2+5x+k;x-2
b)x^3+4x^2-11x+k;x+2

Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website!
Since the the binomial is a factor of the trinomial we are essentially given a root of the trinomial. This means that the trinomial evaluated at that root is equal to zero. We will use this fact to find k.
Given the trinomial
(1) x^3 - kx^2 + 5x +k
which we know has a factor
(2) (x-2) which in turn means that we have a root at
(3) (x-2) = 0 or
(4) x = 2, OK so far?
This means that the trinomial of (1) is equal to zero when x = 2. Let's do that and (1) evaluates to
(5) 2^3 - k*2^2 + 5*2 + k = 0
Solve (5) for k by simplifiction
(6) 8 - 4k + 10 + k =0 or
(7) 18 - 3k = 0 or
(8) k = 6
Let's check this value of k by substituting k = 6, into (1) and get the trinomial
(9) x^3 -6x^2 + 5x +6.
Is this equal to zero at x = 2?
Is (2^3 -6*(2^2) + 5*2 +6 = 0)?
Is (8-24+10+6 = 0)?
Is (24-24 = 0)?
Is (0=0)? Yes
Answer to a): The value of k that makes trinomial have a factor (x-2) is 6.
Now you can solve the second problem the same way. It's a little easier, I'll let you do it. The answer is k = -30