when (x^4 - 6.x^3 + 16.x^2 - 25. x + 10) is divided by( x^2 - 2.x + k)
the remainder is = x+a
find the values of k and a
There are, I am sure, many subtle ways of solving this problem, but the most obvious one which springs to mind is a simple cheat:
We know that $\displaystyle f=gh+r$ where $\displaystyle h$ is some unknown polynomial and $\displaystyle f$, $\displaystyle g$, and $\displaystyle r$ are as above, because that's what division with a remainder means. So, simply substitute unknowns into the polynomial $\displaystyle h$ and then multiply out and pair off the coefficients, and you'll get your result.
After using long division, you should find
$\displaystyle \frac{x^4 - 6x^3 + 16x^2 - 25x + 10}{x^2 - 2x + k} = x^2 - 4x + 8 - k + \frac{(2k - 9)x + k^2 - 8k + 10}{x^2 - 2x + k}$.
So the remainder is:
$\displaystyle x + a = (2k - 9)x + k^2 - 8k + 10$.
Equating like co-efficients of $\displaystyle x$ gives
$\displaystyle 2k - 9 = 1$ and $\displaystyle k^2 - 8k + 10 = a$.
Solving for $\displaystyle k$ in the first equation gives:
$\displaystyle 2k = 10$
$\displaystyle k = 5$.
Substituting into the second equation gives:
$\displaystyle 5^2 - 8\cdot 5 + 10 = a$
$\displaystyle 25 - 40 + 10 = a$
$\displaystyle a = -5$.
So $\displaystyle a = -5$ and $\displaystyle k = 5$.