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

1. Simplify: Can do so by factoring to see if anything from the numerator and the denominator will cancel out.
2. Multiply across: Just Do It!
3. Simplify: By taking out a common factor if possible
4. 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.
5. 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.

1. Simplify: Can do so by factoring
2. Reciprocate
3. Cancel any pair terms
4. Multiply across: Just Do It!
5. Simplify: By taking out a common factor if possible
6. 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.
7. 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.

Pre-Calc 11: Week 14 Multiplying and Dividing Rational Expressions

Last week in Pre-Calculus 11 we learned about multiplying and dividing rational expressions. A thing to remember about expressions is that it does not have an equal sign and you do not need to solve, you only need to simplify. I will be teaching you what non-permissible values are and how to determine what they are.

Non-permissible value :  Values that cause the fraction to have a denominator with a value of zero. (In math, we cannot divide by zero).

Here is an example of multiplying rational expressions

Steps:

Step 1:  Simplify $\frac{3x}{2(x-3)}\cdot\frac{8(x-3)}{9x^2}$

Non permissible values: $x\neq -3$

• $\frac{3x}{2(x-3)}\cdot\frac{8(x-3)}{9x^2}$
• remove (x-3) from top and bottom because they cancel eachother out
• then it becomes… $\frac {3x}{2}\cdot\frac{8}{9x^2}$

Step 2: Simplify by multiplying across (Just do it)

• $\frac {3x}{2}\cdot\frac{8}{9x^2}$
• $\frac{24x}{18x^2}$

Step 3: Take the highest common factor from both the numerator and the denominator.

In this case 6 is the highest number that goes into both.

• $\frac{24x}{18x^2}$
• $\frac{4x}{18x^2}$

Step 4: Notice that x is on the bottom and the top, if it has a pair it can cancel out. two x’s on the bottom one on the top. when they cancel out each other you will be left with only 1 on the bottom.

• $\frac{4x}{3x^2}$
• $\frac{4}{3x}$

Final Answer: $\frac{4}{3x}$

Dividing Rational Expressions

There are many steps when dividing rational expressions

1. Simplify the fraction: can factor or take out the common denominator.
2. State the non permissible values.
3. Reciprocate the second fraction and it will because a multiplication expression.
4. State the restrictions again because there are new values in the denominator and could be non-permissible.
5. Simplify (cancel out like terms that have a pair on the numerator and denominator.
6. Multiply across (Just do it)
7. Simplify again if possible.

Step 1: Simplify $\frac{x+5}{x-4}\div\frac{x^2 - 25}{3x-12}$

FACTOR:

• $\frac{x+5}{x-4}\div\frac{x^2 - 25}{3x-12}$
• $\frac{x+5}{x-4}\div\frac{(x+5)(x-5)}{3(x-4)}$

Step 2: Non-permissible values

• $x\neq 4$

Step 3: Reciprocate

• $\frac{x+5}{x-4}\div\frac{(x+5)(x-5)}{3(x-4)}$
• $\frac{x+5}{x-4}\cdot\frac{3(x-4)}{(x+5)(x-5)}$

Step 4: Non-permissible values

• $x\neq 4$
• $x\neq 5$
• $x\neq -5$

Step 5: Cross out like terms

• $\frac{x+5}{x-4}\cdot\frac{3(x-4)}{(x+5)(x-5)}$
• $\frac{3}{(x-5)}$

Step 6: Multiply across if possible

• In this example it is not

Step 7: Simplify further if possible

• In this example it is not

Final Answer:  $\frac{3}{(x-5)}$

This is how you Multiply and Divide Rational Expressions