Algebraic division

In a nutshell

An algebraic fraction can be written in the form $\cfrac{P(x)}{Q(x)}$ , where $P(x)$ and $Q(x)$ are polynomials. If the order of $P(x)$ is larger than or equal to the order of $Q(x)$ , this is an improper algebraic fraction. These types of fractions must be turned into mixed fractions in order to be expressed in terms of partial fractions.

Note: The order of a polynomial is given by the highest power of $x$ .

Algebraic division

There are two different methods used to convert an improper fraction into a mixed one:

Algebraic division
$F(x) = Q(x) \times \text{divisor} \ + \ \rm remainder$ relation

Algebraic division

procedure

1.	Identify $P(x), \ Q(x)$ in your division.
2.	Write down the elements of the algebraic division.
3.	Compare the first element of $P(x)$ and $Q(x)$ to find out the multiplier (the term you multiply $Q(x)$ by to get $P(x)$ .
4.	Multiply $Q(x)$ by the multiplier and subtract that result from $P(x)$ .
5.	Repeat steps from 1-4 until you reach the final answer.

Example 1

Simplify $\cfrac{6x^3 + 13 x^2 -4}{3x+2}$ 

First, write down the elements:

$\begin{array}{r}3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\end{array}$

Compare the first element of $P(x)$ and $Q(x)$ , and write down the multiplier:

$\begin{array}{r}2x^2 \hspace{19mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\end{array}$

Multiply $Q(x)$ by the quantity found in the previous step, and subtract the result from $P(x)$ :

$\begin{array}{r}2x^2 \hspace{19mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\\- \hspace{25mm}\\ 6x^3 + 4x^2\hspace{9.5mm}\\\overline{\smash{}\,0x^3 +9x^2 \ \ -4\hspace{1.5mm}}\end{array}$

Repeat these steps until you reach the final answer.

$\begin{array}{r}2x^2+3x -2 \hspace{5mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\\- \hspace{25mm}\\ 6x^3 + 4x^2\hspace{9.5mm}\\\overline{\smash{}\,\hspace{5mm} +9x^2 \hspace{5mm} -4\hspace{1.5mm}}\\- \hspace{22mm}\\9x^2+ 6x \hspace{4.8mm}\\\overline{\smash{}\,\hspace{10mm} \ -6x-4 }\\- \hspace{11mm}\\-6x - 4 \\\overline{\smash{}\,\hspace{10mm} \ 0 }\end{array}$

Therefore $\underline{\cfrac{6x^3+13x^2-4}{3x+2} =2x^2+3x-2 + 0}$ 

Note: You can factorise your result, but in this case it is not necessary.

$\bf F(x) = Q(x) \times \textbf{divisor} \ + \ \rm\bf remainder$ relation

This method is really useful when dealing with expressions with the form:

$\ F(x) = Q(x) \times D(x) \ + \ R(x)$

Where $D(x)$ is the divisor and $R(x)$ the remainder.

Procedure

1.	Multiply out the brackets if necessary.
2.	Group together the cubic, quadratic, linear and constant terms.
3.	Equate these terms on both sides of the expression.
4.	Solve for the values, using substitution.

Example 2

Given that ${6x^3+13x^2-4}=({3x+2})\times (Ax^2+Bx+C) + R$ , find the values $A, \ B, \ C$ and $R$ 

Multiply out the brackets.

${6x^3+13x^2-4}=3Ax^3 + 3Bx^2 + 3Cx + 2Ax^2 + 2Bx + 2C + R$

Group the cubic, quadratic, linear and constant terms together:

${6x^3+13x^2-4}=3Ax^3 + x^2 (3B + 2A) + x (3C + 2B) + (2C + R)$

Equate the cubic, quadratic, linear and constant terms:

$\begin{cases}\begin{aligned}6x^3 &= 3Ax^3\\13x^2 &= x^2 (3B+2A) \\0x & = x(3C + 2 B)\\-4 &= 2C + R\end{aligned}\end{cases}$

From the first equation you have that $\underline{A = 2}$ . Now onto the other equations:

$\begin{cases}\begin{aligned}13x^2 &= x^2 (3B+4) \implies \underline{B = 3}\\0x & = x(3C + 2 B)\implies \underline{C=-2}\\-4 &= 2C + R \implies \underline{R=0}\end{aligned}\end{cases}$

In conclusion:

$\underline{6x^3 +13x^2 -4 =(3x+2)\times (2x^2 +3x -2) }$