# MULTIPLICATION OF ALGEBRAIC FRACTION

The process of multiplying algebraic fractions is the same as that of common fractions.

You multiply the numerator by the other numerator and multiply the denominator by the denominator.

Also, you reduce the fraction if the numerator and denominator have common factors

This is illustrated below.

👉This post extends our posts on algebraic fractions👈

Example 1
Simplify $\frac{2y^3+y^2}{y^2-4}\times\frac{y-2}{2y^2-5y-3}$

Solution:
As you will know in this post, $y^2-4$ is difference of two squares. Therefore,
$\frac{2y^3+y^2}{(y+2)(y-2)}\times\frac{y-2}{2y^2-5y-3}$

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

By factorizing
$\frac{y^2(2y+1)(y-2)}{(y+2)(y-2)(2y+1)(y-3)}$
$\frac{y^2\require{cancel}\bcancel{(2y+1)}\require{cancel}\bcancel{(y-2)}}{(y+2)\require{cancel}\bcancel{(y-2)}\require{cancel}\bcancel{(2y+1)}(y-3)}$
$\frac{y^2}{(y+2)(y-3)}$

Example 2
Solve $\frac{4y^2}{y^2-9}\times\frac{y-3}{4y^2-10y-3}$

Solution:
$y^2-9$ is differences of two squares.
$\frac{4y^2}{(y+3)(y-3)}\times\frac{y-3}{4y^2-10y-3}$

Multiplication of numerators and denominators
$\frac{(4y^2)(y-3)}{(y+3)(y-3)(4y^2-10y-3)}$
$\frac{(4y^2)\require{cancel}\bcancel{(y-3)}}{(y+3)\require{cancel}\bcancel{(y-3)}(4y^2-10y-3)}$
$\frac{4y^2}{(y+3)(4y^2-10y-3)}$

Example 3
Simplify $\frac{y^2-y-6}{y^2-9}\times\frac{4}{y-2}$

Solution:
$y^2-9$ is differences of two squares
$\frac{y^2-y-6}{(y+3)(y-3)}\times\frac{4}{y-2}$
$\frac{(y^2-y-6)(4)}{(y+3)(y-3)(y-2)}$
$\frac{(y+3)(y-2)(4)}{(y+3)(y-3)(y-2)}$
$\frac{\require{cancel}\bcancel{(y+3)(y-2)}(4)}{\require{cancel}\bcancel{(y+3)(y-2)}(y-3)}$
$\frac{4}{y-3}$

Example 4
Solve $\frac{2x^2}{4}\times\frac{3x+4}{x^2}$

Solution:
$\frac{(2x^2)(3x+4)}{4(x^2)}$
$\frac{(2x^2)(3x+4)}{4x^2)}$

If $4x^2$ can be rewritten as $2(2x^2)$, then
$\frac{(2x^2)(3x+4)}{2(2x^2)}$
$\frac{\require{cancel}\bcancel{(2x^2)}(3x+4)}{2\require{cancel}\bcancel{(2x^2)}}$
$\frac{3x+4}{2}$

Example 5
Simplify $\frac{x^2-4x+4}{x^3-6x^2+12x-8}\times\frac{x-2}{x^2}$

Solution:
By multiplication
$\frac{(x^2-4x+4)(x-2)}{(x^3-6x^2+12x-8)(x^2)}$

By factorization
$\frac{(x-2)(x-2)(x-2)}{(x-2)(x-2)(x-2)(x^2)}$
$\frac{\require{cancel}\bcancel{(x-2)(x-2)(x-2)}}{\require{cancel}\bcancel{(x-2)(x-2)(x-2)}(x^2)}$
$\frac{1}{x^2}$

Example 6
Write $\frac{(x^2-4)(x+2)^2}{x^3+6x^2+12x+8}\times\frac{x^2-5x+6}{x-3}$ in the simplest form possible.

Solution:
$x^2-4$ is difference of two squares.
$\frac{(x+2)(x-2)(x+2)^2}{x^3+6x^2+12x+8}\times\frac{x^2-5x+6}{x-3}$

Multiplying numerators and denominators
$\frac{(x+2)(x-2)(x+2)^2(x-2)(x-3)}{(x^3+6x^2+12x+8)(x-3)}$

By factorization
$\frac{(x+2)(x-2)(x+2)(x+2)(x-2)(x-3)}{(x+2)(x+2)(x+2)(x-3)}$
$\frac{\require{cancel}\bcancel{(x+2)(x+2)(x+2)(x-3)}(x-2)(x-2)}{\require{cancel}\bcancel{(x+2)(x+2)(x+2)(x-3)}}$
$\frac{(x-2)(x-2)}{1}$
$(x-2)(x-2)$

Related posts
Example 7
Write $\frac{x^2+2x-8}{x^2+x-12}\times\frac{x^2}{x-2}$

Solution:
$\frac{(x^3+2x-8)(x^2)}{(x^2+x-12)(x-2)}$
Factorizing
$\frac{(x-2)(x+4)(x^2)}{(x+4)(x-3)(x-2)}$
$\frac{\require{cancel}\bcancel{(x+4)(x-2}(x^2)}{\require{cancel}\bcancel{(x+4)(x-2)}(x-3)}$
$\frac{x^2}{x-3}$

Voilà, we just learned multiplication of algebraic. Next up is a division of algebraic fractions, we will be dedicating this post to learning just that.

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