Pre Calculus 11 Week 16: Multiplying and Dividing Rational Expressions

During the Unit of Rational Expressions I have learned a few things like multiplying and dividing rational expressions. Compared to adding and subtracting these expressions it is a tad bit easier. I have learned to Just Do IT!

Steps When Multiplying Rational Expressions

Simplify: Can do so by factoring to see if anything from the numerator and the denominator will cancel out.
Multiply across: Just Do It!
Simplify: By taking out a common factor if possible
Variables: You can simplify variables too! If there is a $x^2$ on the denominator and a single x in the numerator we know that at least one pair can cancel out. The leftover variable will be left on the same side that it is on.
State any Non Permissible values

Example 1: $\frac{10x^2}{32x^3}\cdot\frac{4x}{22x^2}$

Step 1: Simplify: Get the simplest form of each fraction by taking out a common factor.

In this case it is two

$\star$ could be a different number for the two fractions, it’s not always the same.

$\frac{10x^2}{32x^3}\cdot\frac{4x}{22x^2}$
$\frac{5x^2}{16x^3}\cdot\frac{2x}{11x^2}$

You can also simplify the the variables

$\frac{5x^2}{16x^3}\cdot\frac{2x^3}{11x^2}$
$\frac{5}{16x}\cdot\frac{2x}{11}$

Step 2: Multiply across (Just Do It!)

$\frac{5}{16x}\cdot\frac{2x}{11}$
$\frac{10x}{176x}$

Step 3: Simplify by taking out common factors if possible

$\frac{10x}{176x}$
$\frac{5x}{88x}$

Step 4: Variables

$\frac{5x}{88x}$
$\frac{5}{88}$

Step 5: State any Non Permissible Values

$x\neq 0$

This is how you would multiply rational expressions

How to Divide Rational Expressions

Steps When Dividing Rational Expressions

Dividing rational expressions is very similar to that of multiplying. There is just one extra step. It is to reciprocate the fraction that is to right after the divide sign.

Simplify: Can do so by factoring
Reciprocate
Cancel any pair terms
Multiply across: Just Do It!
Simplify: By taking out a common factor if possible
Variables: You can simplify variables too! If there is a $x^2$ on the denominator and a single x in the numerator we know that at least one pair can cancel out. The left over variable will be left on the same side that it is on.
State any Non Permissible values

Example 2: $\frac{3x-27}{x^2 - 1}\cdot\frac{x^2 -7x - 8}{x^2 - 81}\div\frac{5x}{2x - 2}$

Step 1: Simplify (factor)

$\frac{3x-27}{x^2 - 1}\cdot\frac{x^2 -7x - 8}{x^2 - 81}\div\frac{5x}{2x - 2}$
$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\div\frac{5x}{2(x - 1)}$

Step 2: Reciprocate (this gets rid of the dividing sign and makes it a multiplying sign)

$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\div\frac{5x}{2(x-1)}$
$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\cdot\frac{2(x - 1)}{5x}$

Step 3: Cancel out any pair terms

$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\cdot\frac{2(x - 1)}{5x}$
$\frac{3}{1}\cdot\frac{(x-8)}{(x+9)}\cdot\frac{2}{5x}$

Step 4: Multiply Across (Just Do It!)

$\frac{3}{1}\cdot\frac{(x-8)}{(x+9)}\cdot\frac{2}{5x}$
$\frac {2\cdot 3\cdot (x-8)}{5x\cdot (x+9)}$
$\frac {6(x-8)}{5x(x+9)}$

State the Non Permissible Values

$x\neq { -9, -1, 0, 1, 9}$

FINAL ANSWER: $\frac {6(x-8)}{5x(x+9)}$

This is how you would divide rational expressions.

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Pre Calculus 11 Week 16: Multiplying and Dividing Rational Expressions

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