# Theorem for Dividing Polynomials

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?

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Incidentally, there is a post on the front page about polynomial long division. Please take a look, you might find some of the answers over there, helpful. – user2468 Jul 14 '12 at 3:44

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.

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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 $\rm\:a =\, \ldots$

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+1 , as it is much simpler. – gpuguy Jul 14 '12 at 5:22

Have you looked at Polynomial division? If you do it for a general $Q$, $a$ will be a function of $k$

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Yes, but the issue is the Remainder will have only one unknown i.e k, and I have two unknowns a and k, where to drive 2 equations to solve for a and k? – gpuguy Jul 14 '12 at 3:30
@gpuguy: have you tried doing the division? (See this for one such algorithm). If you did it properly, you should notice that the coefficient of $x$ in the remainder is not (yet) equal to $1$... – J. M. Jul 14 '12 at 3:36