# Thread: impossible question for me!!!!!

1. ## impossible question for me!!!!!

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

2. Originally Posted by jashansinghal
When $\displaystyle f=x^4 - 6.x^3 + 16.x^2 - 25. x + 10$ is divided by $\displaystyle g= x^2 - 2.x + k$ the remainder is $\displaystyle r = x+a$ 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.

3. Originally Posted by jashansinghal
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
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 $\displaystyle k = 5$.