# impossible question for me!!!!!

• Jul 3rd 2009, 02:03 AM
jashansinghal
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
• Jul 3rd 2009, 02:26 AM
Swlabr
Quote:

Originally Posted by jashansinghal
When $f=x^4 - 6.x^3 + 16.x^2 - 25. x + 10$ is divided by $g= x^2 - 2.x + k$ the remainder is $r = x+a$ 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 $f=gh+r$ where $h$ is some unknown polynomial and $f$, $g$, and $r$ are as above, because that's what division with a remainder means. So, simply substitute unknowns into the polynomial $h$ and then multiply out and pair off the coefficients, and you'll get your result.
• Jul 3rd 2009, 02:29 AM
Prove It
Quote:

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

$\frac{x^4 - 6x^3 + 16x^2 - 25x + 10}{x^2 - 2x + k} = x^2 - 4x + 8 - k + \frac{(2k - 9)x + k^2 - 8k + 10}{x^2 - 2x + k}$.

So the remainder is:

$x + a = (2k - 9)x + k^2 - 8k + 10$.

Equating like co-efficients of $x$ gives

$2k - 9 = 1$ and $k^2 - 8k + 10 = a$.

Solving for $k$ in the first equation gives:

$2k = 10$

$k = 5$.

Substituting into the second equation gives:

$5^2 - 8\cdot 5 + 10 = a$

$25 - 40 + 10 = a$

$a = -5$.

So $a = -5$ and $k = 5$.