Hey guys! Welcome to this video on simplifying rational expressions.
What is a Rational Expression?
A rational expression just refers to a fraction with a polynomial in the numerator, and a polynomial in the denominator.
Rational Expression Examples
Here are a few examples:
- \(\frac{x^{2}-16}{x+4}\)
- \(\frac{x^{2}-2x-8}{x^{2}-9x+20}\)
- \(\frac{4x+4}{x^{4}-x^{2}}\)
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.
How to Reduce Rational Expressions
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.
Example #1
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:
We can cancel our 2s here giving us:
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.
We can rewrite our numerator, once we factor this out, as:
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: \(\frac{(x+4)+(x-4)}{(x+4)}\).
Example #2
Let’s now move on to our second example, which is a bit trickier.
We can do the same thing that we did in our first example by rewriting our numerator and denominator. So that would give us:
So, we can go ahead here and cancel our \((x-4)\)s, which would leave us with:
Example #3
For our last example we have:
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 \(x^{2}(x^{2}-1)\).
But, notice, we can factor this out even further so we can get something to cancel out with our numerator here.
At this point, we can cancel out our \((x+1)\)s here, leaving us with:
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!
See you guys next time!
Rational Expression Practice Questions
True or false: we can reduce \(\frac{x+8}{x-8}\).
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.
True or false: \(\frac{(x+5)-(2x+1)}{10(2x+1)}\) reduces to \(\frac{x+5}{10}\).
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.
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?
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:
\(\frac{8x+16}{4}=\frac{4\times2x+4\times4}{4}=2x+4\)
Simplify the expression: \(\frac{x^3+6x^2-16x}{x-6x-40}\)
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:
\(\frac{x(x+8)(x-2)}{-5(x+8)}=\frac{x(x-2)}{-5}\)
Simplify the expression: \(\frac{(x+7)+(5x+5)}{x^2+2x}\)
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:
\(\frac{6(x+2)}{x^2+2x}=\frac{6(x+2)}{x(x+2)}=\frac{6}{x}\)
