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Algebraic long division

Algebraic Long Division 1

Given a polynomials, $P(x)$ and $S(x)$ where $\mbox{deg}(P) \geq \mbox{deg}(S)$, it is possible to express $P(x)$ as $$ P(x) = S(x)Q(x) + R(x) $$ Where $Q(x)$ and $R(x)$ are polynomials. We call $R(x)$ the remainder and $Q(x)$ the quotient

Algebraic Long Division 1

Determine the quotient and remainder when $x^3 + 3x^2 - 14x - 12$ is divided by $x^2 + 4x + 1$
solution - press button to display
$$ \begin{array}{c|cccccc} & &&&&x & + & 4 \\ \hline x^2 + 4x + 1 & x^3 &+ & 8x^2 &+ & 11x & - & 4 \\ & x^3 &+& 4x^2 &+& x & & \\ \hline & & &4x^2 & + & 10x & - & 4 \\ & & &4x^2 & + & 16x & + & 4 \\ \hline & & & & - & 6x & - & 8 \end{array} $$ The quotient is $x+4$ and the remainder $-6x - 8$

Algebraic Long Division 2

Determine the remainder when $x^4+3x^3 + 6x^2 + 2$ is divided by $x^2 + 4x + 2$
solution - press button to display

We shall separate out each iteration of the process:

In the first instance, $x^4/ x^2 = x^2$ so we have

$$x^4 +3x^3 +6x^2+0x+2 - (x^2)(x^2 + 4x + 2) = -x^3 + 4x^2 +0x +2$$

In the second instance, $-x^3 / x^2 = - x$ so we have

$$-x^3+4x^2+0x+2 - (-x)(x^2+4x+2) = 8x^2 +2x +2$$

In the third instance, $8x^2 /x^2 = 8$,so we have

$$8x^2 +2x +2 - (8)(x^2+4x+2) = -30x-14$$

The process now halts as the remainder has a degree strictly lower than the divisor.

The Quotient is $x^2 -x + 8$ and the Remainder is $-30x-14$

This process is normally tabulated, but the expanded view is sometimes helpful to elucidate the process.

Algebraic Long Division 3

Determine the value of $k$ such that the remainder is a constant when $x^3 + 3x^2 + kx + 2$ is divided by $x^2 + 2x + 1$.
solution - press button to display
This question is largely interesting because of the presence of an additional parameter $k$. As before, we proceed with the tabulated method; $$ \begin{array}{c|cccc} & & & x & +1 \\ x^2 + 2x + 1 & x^3 & + 3x^2 & +kx & +2 \\ \hline & x^3 & 2x^2 & + x \\ \hline & & x^2 & +(k-1)x & + 2 \\ & & x^2 & + 2x & 1 \\ \hline & & & (k-3)x & 1 \end{array} $$ Immediately we can see that we have a constant remainder of $1$ if and only if $k=3$