# Reducing Rational Expressions – Polynomials EVERYWHERE!

Hey guys! Welcome to this video on Simplifying Rational Expressions.

A rational expression just refers to a fraction with a polynomial in the numerator, and a polynomial in the denominator.

Here are a few examples:

1. $$\frac{x^{2}-16}{x+4}$$

2. $$\frac{x^{2}-2x-8}{x^{2}-9x+20}$$

3. $$\frac{4x+4}{x^{4}-x^{2}}$$

One thing that we need to keep in mind when working with rational expression is that divisibility by 0 is not allowed. Just like when dealing with regular numbers, you cannot divide by 0. So, when dealing with a rational expression, we always assume that whatever x is, it will not give us division by 0.

Alright let’s take a look at how to reduce a rational expression. We’re actually doing the same thing we would do when reducing a regular fraction.

So, let’s say we have $$\frac{18}{8}$$. When we reduce this, we can cancel our like terms. So we can rewrite this as:

$$\frac{9(2)}{4(2)}$$
.

We can cancel our 2s here giving us:

$$\frac{9}{4}$$
.

So now we have a fraction reduced down to its simplest form. There is not another number that both our numerator and denominator are divisible by.

It works the same way with a rational expression.

Let’s try reducing our first example.

1. $$\frac{x^{2}-16}{x+4}$$

We can rewrite our numerator, once we factor this out, as:

$$\frac{(x+4)(x-4)}{(x+4)}$$
.

And once we do this, we can see that our $$(x+4)$$s will cancel out. So we cancel that out, leaving us with:

$$x-4$$
.

Now, we need to be careful when canceling terms. The only reason we were able to cancel out our $$(x+4)$$s here was because they are both being multiplied in the numerator and the denominator. This would not work if our top was: $$\frac{(x+4)+(x-4)}{(x+4)}$$.

Let’s now move on to our second example, which is a bit trickier.

2. $$\frac{x^{2}-2x-8}{x^{2}-9x+20}$$

We can do the same thing that we did in our first example by rewriting our numerator and denominator. So that would give us:

$$=\frac{(x-4)(x+2)}{(x-5)(x-4)}$$
.

So, we can go ahead here and cancel our $$(x-4)$$s, which would leave us with:

$$=\frac{x+2}{x-5}$$
.

For our last example we have:

3. $$\frac{4x+4}{x^{4}-x^{2}}$$

To reduce it, we can rewrite our numerator by factoring out a 4. Which would give us $$4(x+1)$$. In the denominator we can factor out an $$x^{2}$$, which would give us $$x^{2}(x^{2}-1)$$.

$$=\frac{4(x+1)}{x^{2}(x^{2}-1)}$$
.

But, notice, we can factor this out even further so we can get something to cancel out with our numerator here.

$$=\frac{4(x+1)}{x^{2}(x+1)(x-1)}$$
.

At this point, we can cancel out our $$(x+1)$$s here, leaving us with:

$$=\frac{4}{x^{2}(x-1)}$$
.

There is no further reduction we can do, so we now have it in our simplest form.

I hope that this video has been helpful for you. For further help, be sure to check out more of our videos by subscribing to our channel below.

See you guys next time!

## Practice Questions

Question #1:

True or false: we can reduce $$\frac{x+8}{x-8}$$.

True

False

You cannot cancel out anything between the numerator and the denominator (because we’re adding 8 in the numerator, and subtracting by 8 in the denominator), so $$\frac{x+8}{x-8}$$ is as simplified as it’s going to get.

Question #2:

True or false: $$\frac{(x+5)-(2x+1)}{10(2x+1)}$$ reduces to $$\frac{x+5}{10}$$.

True

False

Because we are subtracting $$2x+1$$ from $$x+5$$, we have to distribute the negative sign and simplify the numerator first:
$$\frac{(x+5)-(2x+1)}{10(2x+1)}=\frac{x+5-2x-1}{20x+10}=\frac{-x+4}{20x+10}=\frac{-x+4}{10(2x+1)}$$
This expression cannot be simplified any further or reduced.

Question #3:

Brandon is solving a problem on his homework. He’s been asked to reduce the expression: $$\frac{7x^2+16x}{4x}$$. Here are the steps that he took and his final answer:
Step 1: $$\frac{7x^2+16x}{4x}=\frac{x(7x+16)}{4x}$$
Step 2: $$\frac{7x+16}{4}$$
Step 3: $$\frac{7x+4\times4}{4}$$
$$=7x+4$$

Where did Brandon first make a mistake while solving?

Between step 1 and step 2

Between step 2 and step 3

Between step 3 and his final answer

He did not make any mistakes

In hopes to reduce the original expression to the point where the expression no longer looks like a fraction, Brandon made an ‘illegal move.’ Even though it is technically correct that 16 can be written as $$4\times4$$, he cannot cancel out one of these 4s without canceling a 4 from the additive term (a.k.a. “7x”). In other words, if the expression were $$\frac{8x+16}{4}$$ instead, he could have done the following:
$$\frac{8x+16}{4}=\frac{4\times2x+4\times4}{4}=2x+4$$

Question #4:

Simplify the expression: $$\frac{x^3+6x^2-16x}{x-6x-40}$$

$$\frac{x(x-8)(x+2)}{-7x+40}$$
$$\frac{x(x+8)(x-2)}{-5(x-8)}$$
$$\frac{x(x-2)}{-5}$$
$$\frac{x(x+2)}{x-1}$$

The correct answer is C. Let’s walk through it.
First, we’ll simplify the numerator:
$$\frac{x^3+6x^2-16x}{x-6x-40}=\frac{x(x^2+6x-16)}{x-6x-40}=\frac{x(x+8)(x-2)}{x-6x-40}$$

Now, let’s work on the denominator:
$$\frac{x(x+8)(x-2)}{x-6x-40}=\frac{x(x+8)(x-2)}{-5x-40}=\frac{x(x+8)(x-2)}{-5(x+8)}$$

Finally, notice that we can cancel out the $$(x+8)$$ term:
$$\frac{x(x+8)(x-2)}{-5(x+8)}=\frac{x(x-2)}{-5}$$

Question #5:

Simplify the expression: $$\frac{(x+7)+(5x+5)}{x^2+2x}$$

$$\frac{6(6x+1)}{x}$$
$$\frac{x+6}{2}$$
$$\frac{5(x+7)(x+1)}{x(x+2)}$$
$$\frac{6}{x}$$

The correct answer is D. Let’s walk through it.
First, we’ll simplify the numerator:
$$\frac{(x+7)+(5x+5)}{x^2+2x}=\frac{x+7+5x+5}{x^2+2x}=\frac{6x+12}{x^2+2x}=\frac{6(x+2)}{x^2+2x}$$

Now, we’ll simplify the denominator and see if we can reduce the expression:
$$\frac{6(x+2)}{x^2+2x}=\frac{6(x+2)}{x(x+2)}=\frac{6}{x}$$