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When a polynomial
$$P(x)=x^4- 6x^3 +16x^2 -25x + 10$$
is divided by another polynomial
$$Q(x)=x^2 - 2x +k,$$
then the remainder is
$$x+a.$$
I have to find the values of $a$ and $k$.
Can somebody tell me shortest way to get these values? Which theorem should be applied here?
Hint:
Perform a long division between $P(x)$ and $Q(x)$ and you will get a remainder of degree $1$ of the form $Ax + B,$ where both $A$ and $B$ are expressions in terms of $k.$ Equate with the given $x + a.$ First, equate $A = 1$ to get the value of $k.$ Then equate $B = a$ to get the value of $a.$
For example, you will get something like (NOT the actual answer): $(k+1) x + (2k-1).$ This means $1 = k+1$ and $a = 2k-1.$ Two equations in two variables.
Multiplying is easier than dividing, so we let the quotient have undetermined coef's and multiply.
Suppose $\rm\ x^4\!-\!6x^3\!+\!16x^2\!-\!25x\!+\!10\, =\, (x^2\!+cx+b)\,(x^2\!-2x+k)\, +\, x+ a.\, $ Comparing coef's
$\rm\ \ x^3\:$ coef $\rm\: \Rightarrow\: c-2 = -6\:\Rightarrow\: c = -4$
$\rm\ \ x^2\:$ coef $\rm\:\Rightarrow\ b+8+k\, =\, 16\ \ \Rightarrow\ \ \ \ b\ +\ k\, =\, 8$
$\rm\ \ x^1\:$ coef $\rm\:\Rightarrow\: -2b\!-\!4k = -26\:\Rightarrow\:-b-2k\, =-13$
Adding the prior two equations yields $\rm\,\ -k = -5,\:$ so $\rm\:k = 5,\:$ so $\rm\:b = 8-k = 3,\:$ so $\rm\:a =\, \ldots$
Have you looked at Polynomial division? If you do it for a general $Q$, $a$ will be a function of $k$
