# DIVISION OF ALGEBRAIC FRACTION

Like common fractions, algebraic fractions are also divisible. The process is the same as common fractions.

First, you will Invert the fraction. Next, you Change $÷$ to $\times$. Thirdly, You multiply the fraction. And lastly, you reduce the fraction, if there are any common factors.

The process is illustrated below

๐This post extends our posts on algebraic fractions ๐

Example 1
Write $\frac{2y}{2y^2+3y-5}÷\frac{y^3}{y^2-y}$ in the simplest form possible.

Solution:
First, we invert the fraction and change the division sign$(÷)$ to multiplication.

$\frac{2y}{2y^2+3y-5}\times\frac{y^2-y}{y^3}$

Multiply numerators and denominators
$\frac{(2y)(y^2-y)}{(2y^2-3y-5)(y^3)}$

By factorization
$\frac{(2)(y)(y)(y-1)}{(2y+5)(y-1)(y)(y)(y)}$
$\frac{(2)\require{cancel}\bcancel{(y)(y)(y-1)}}{(2y+5)\require{cancel}\bcancel{(y)(y)(y-1)}(y)}$
$\frac{2}{y(2y+5)}$

Example 2
Simplify $\frac{2x^2+9x+4}{x-3}÷\frac{8x+4}{x^2-9}$

Solution:
Invert the fraction
$\frac{2x^2+9x+4}{x-3}\times\frac{x^2-9}{8x+4}$

$x^2-9$ is difference of two squares.
$\frac{2x^2+9x+4}{x-3}÷\frac{(x+3)(x-3)}{8x+4}$
$\frac{(2x^2+9x+4)(x+3)(x-3)}{(x-3)(8x+4)}$
$\frac{(2x+1)(x+4)(x+3)(x-3)}{(x-3)(4)(2x+1)}$
$\frac{\require{cancel}\bcancel{(2x+1)(x-3)}(x+4)(x+3)}{\require{cancel}\bcancel{(2x+1)(x-3)}(4)}$
$\frac{(x+4)(x+3)}{4}$

Example 3
Write $\frac{x^2-7x+12}{x-3}÷\frac{x-4}{x^2}$ in the simplest form possible.

Solution:
Overturn the fraction
$\frac{x^2-7x+12}{x-3}\times\frac{x^2}{x-4}$

$\frac{x^2-7x+12)(x^2)}{(x-3)(x-4)}$
$\frac{(x-3)(x-4)(x^2)}{(x-3)(x-4)}$
$\frac{\require{cancel}\bcancel{(x-3)(x-4)}(x^2)}{\require{cancel}\bcancel{(x-3)(x-4)}}$
$\frac{x^2}{1}$
$x^2$

Example 4
Solve $\frac{x^3-2x^2-13x-10}{x^2}÷\frac{x^2+3x+2}{x^4}$

Solution:
First, we overturn the fraction
$\frac{x^3-2x^2-13x-10}{x^2}\times\frac{x^4}{x^2+3x+2}$

Multiplications of numerators and denominators
$\frac{(x^3-2x^2-13x-10)(x^4)}{(x^2)(x^2+3x+2)}$

By factorizing
$\frac{(x-5)(x+1)(x+2)(x^2)(x^2)}{(x^2)(x+1)(x+2)}$
$\frac{\require{cancel}\bcancel{(x+1)(x+2)(x^2)}(x-5)(x^2)}{\require{cancel}\bcancel{(x+1)(x+2)(x^2)}}$.
$\frac{(x-5)(x^2)}{1}$
$(x-5)(x^2)$

Related post

Example 5
Write $\frac{2x+1}{8x^2}÷\frac{2x+1}{x^2}$ in the simplest form

Solution:
Invert the fractions
$\frac{2x+1}{8x^2}\times\frac{x^2}{2x+1}$

$\frac{(2x+1)(x^2)}{(8)(x^2)(2x+1)}$
$\frac{\require{cancel}\bcancel{(2x+1)(x^2)}}{\require{cancel}\bcancel{(2x+1)(x^2)}(8)}$
$\frac{1}{8}$
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