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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 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= .
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= .
(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 by the binomial is equal to the value of the polynomial.
2. The binomial divides the polynomial if and only if the value of is the root of the polynomial , i.e. .
3. The binomial factors the polynomial if and only if the value of is the root of the polynomial , i.e. .
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".
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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.