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

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

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:

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