SOLUTION: The polynomial f(x)=x^3-x^2-6kx+4k^2 where k is a constant has (x-3)as a factor. Find the possible values of k and for the integral value of k find the remainder when f(x) is divid

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The polynomial f(x)=x^3-x^2-6kx+4k^2 where k is a constant has (x-3)as a factor. Find the possible values of k and for the integral value of k find the remainder when f(x) is divid      Log On


   



Question 1172550: The polynomial f(x)=x^3-x^2-6kx+4k^2 where k is a constant has (x-3)as a factor. Find the possible values of k and for the integral value of k find the remainder when f(x) is divided by x+2.
Answer by ikleyn(52787) About Me  (Show Source):
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The polynomial f(x)=x^3-x^2-6kx+4k^2 where k is a constant has (x-3) as a factor.
(a) Find the possible values of k and
(b) for the highlight%28cross%28integral%29%29 integer value of k find the remainder when f(x) is divided by x+2.
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According to the Remainder theorem, the fact that the polynomial f(x) = x^3 - x^2 - 6kx + 4k^2 has (x-3) as a factor

means that the value of x= 3 is the root of the polynomial.



It gives this equation for k


    f(3) = 0 = 3^3 - 3^2 - 6*3*k + 4k^2,   or

           4k^2 - 18k + 18 = 0,            which is equivalent to

           2k^2 -  9k +  9 = 0.


The roots of the equation are (use the quadratic formula)  k= 4  and  k= 3%2F2.


Of these two roots, the integer value for k is 4 (four).


At k = 4, the polynomial takes the form  f(x) = x^3 - x^2 - 6*4x + 4*4^2 = x^3 - x^2 - 24x + 64.


The reminder of this polynomial, when divided by (x+2),  it its value at x= -2  (here I apply the Remainder theorem again)


    f(-2) = (-2)^3 - (-2)^2 - 24*(-2) + 64 = 100.


ANSWER.  (a)  the possible values of k are  k= 4  and  k= 3%2F2.

         (b)  for the integer value of k, the remainder when f(x) is divided by x+2 is equal to 100.

Solved.     //     All questions are answered.


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   Theorem   (the remainder theorem)
   1. The remainder of division the polynomial  f%28x%29  by the binomial  x-a  is equal to the value  f%28a%29  of the polynomial.
   2. The binomial  x-a  divides the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.
   3. The binomial  x-a  factors the polynomial  f%28x%29  if and only if the value of  a  is the root of the polynomial  f%28x%29,  i.e.  f%28a%29+=+0.


See the lessons
    - Divisibility of polynomial f(x) by binomial x-a and the Remainder theorem
    - Solved problems on the Remainder thoerem
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.