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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?
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 $\rm\:a =\, \ldots$
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.
Have you looked at Polynomial division? If you do it for a general $Q$, $a$ will be a function of $k$
