Question 1012505
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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 <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-polynomial-f%28x%29-by-binomial-x-a.lesson>Divisibility of polynomial f(x) by binomial x-a</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:

{{{(k^2)*1^3 - 6k*1 + 9}}} = {{{0}}}.

It is your equation to determine k:

{{{k^2 - 6k + 9}}} = {{{0}}}.

To solve it, notice that the left part is nothing else as {{{(k-3)^2}}}.

So, your equation is 

{{{(k-3)^2}}} = {{{0}}},

and its solution is k = 3.

There is no other solution.

<U>Answer</U>. k = 3. The answer is C).
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