SOLUTION: If K is a constant, what is the value of K such that the polynomial K^2X^3-6kx+9 is divisible by X-1? A. K=1 B. K=-1 C. K=3 D. K=-3

Algebra ->  Finance -> SOLUTION: If K is a constant, what is the value of K such that the polynomial K^2X^3-6kx+9 is divisible by X-1? A. K=1 B. K=-1 C. K=3 D. K=-3      Log On


   



Question 1012505: If K is a constant, what is the value of K such that the polynomial K^2X^3-6kx+9 is divisible by X-1?
A. K=1
B. K=-1
C. K=3
D. K=-3

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
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If K is a constant, what is the value of K such that the polynomial K^2X^3-6kx+9 is divisible by X-1?
A. K=1
B. K=-1
C. K=3
D. K=-3
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Apply the Remainder Theorem (see the lesson Divisibility of polynomial f(x) by binomial x-a in this site).


The Remainder Theorem says:

   If the polynomial f(x) is divisible by a bynomial x-a then the number "a" is a root of the polynomial f(x), i.e. f(a) = 0.

In our case, since the given polynomial is divisible by X-1, the value of 1 is the root of the polynomial:

%28k%5E2%29%2A1%5E3+-+6k%2A1+%2B+9 = 0.

It is your equation to determine k:

k%5E2+-+6k+%2B+9 = 0.

To solve it, notice that the left part is nothing else as %28k-3%29%5E2.

So, your equation is 

%28k-3%29%5E2 = 0,

and its solution is k = 3.

There is no other solution.

Answer. k = 3. The answer is C).