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.

Guest Jan 30, 2021

#1**0 **

I will use synthetic substitution to speed up the process of these calculations. Unfortunately, showing work for synthetic substitution is quite tricky in computer form, but I will do my best. I know just enough LaTeX so that my work is presentable. I had to be very creative, however, for your benefit.

Since the problem states that \(x-3\) is a factor of \(f(x)=x^3-x^2-6kx+4k^2\), we can also use this fact to reason that \(f(3)=0\). I will substitute in 3 for x using synthetic substitution.

\(3)\overline{\quad 1 \quad -1 \quad\quad -6k \quad\quad 4k^2 }\\ \quad\quad\quad\; +3 \quad\quad\;\; +6 \quad -18k+18\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \quad\quad 1 \quad\quad 2 \quad -6k+6 | \quad 4k^2-18k+18 \)

Let's now solve for the possible values of k knowing that \(4k^2-18k+18=0\):

\(4k^2-18k+18=0\\ 2(2k^2-9k+9)=0\\ 2(2k^2-6x-3x+9)=0\\ 2(2k(k-3)-3(k-3))=0\\ 2(k-3)(2k-3)=0\\ k_1=3\text{ or }k_2=\frac{3}{2}\)

Do you think you can finish the problem from here?

Guest Jan 31, 2021

#2**0 **

By the remainder theorem, for any polynomial p(x), x-a will be a factor of p(x) only if p(a) = 0, where 'a' is a constant. Knowing that, for (x-3) to be a factor of f(x), f(3) must be equal to zero. Therefore, set x = 3 and solve for k:

\(3^3-3^2-6\cdot k \cdot3+4k^2=0\\27-9-18k+4k^2=0\\4k^2-18k+18=0\\2k^2-9k+9=0\\(2k-3)(k-3)=0\\\boxed{k=\frac{3}{2}, k=3}\)

Now, to solve the second question, set k = 3 to get f(x) to equal \(x^3-x^2-18x+36\). This should be easy enough to be solved using the remainder theorem mentioned above.

textot Jan 31, 2021