Simplifying Rational Polynomial Functions

Hi, and welcome to this video on simplifying rational polynomial functions!

Remember, a rational expression is a ratio of polynomial expressions, and dividing by zero is “undefined,” so it is really important to note that the denominator of the ratio of polynomials must never equal zero. The values of \(x\) that will result in a zero in the denominator are called “excluded values” or “domain restrictions.” These values must never be used in the expression.

The notation used to represent a rational function, \(f(x)\), is: \(f(x)=\frac{p(x)}{g(x)}\), where \(p(x)\) and \(g(x)\) are polynomials, and \(g(x)\neq 0\).

In order to simplify a rational expression, the polynomials of the numerator and the denominator must be factored, if possible, and the domain restrictions are then determined. The final step is to cancel out like factors from the numerator and denominator, because they divide to one.

Let’s look at a few examples to put this process into action.

Example 1

We’re going to look at:

\(\frac{x^2-8x+15}{x^2-9}\)

This example shows a rational expression that has a trinomial in the numerator and a binomial in the denominator.

Now let’s walk through simplifying this expression.

First, let’s factor the polynomials. So we get: \(\frac{(x-3)(x-5)}{(x-3)(x+3)}\).

Second, determine the domain restrictions from the factored denominator. The factor, \((x-3)\), would equal zero if \(x=3\). And the factor, \((x+3)\), would equal zero if \(x=-3\). So if either of these factors equals zero, then the denominator would equal zero. Therefore, the domain restrictions are \(x=3\) and \(x= -3\).

And lastly, the factor, \((x-3)\), is in both the numerator and denominator. Dividing \((x-3)\) by itself results in 1, so these factors can be canceled out of the expression.

Now we have the simplified rational expression, \(\frac{(x-5)}{(x+3)}\), with domain restrictions of 3 and -3. Let’s go through these same steps in another example.

Example 2

Now we’re going to look at:

\(\frac{x^2+10x+25}{x^2+2x-15}\)

First we’re going to factor the polynomials: \(\frac{(x+5)(x+5)}{(x+5)(x-3)}\).

Then, we’re going to determine the domain restrictions from the factored denominator. The factor, \((x+5)\), would equal zero if \(x= -5\). And the factor, \((x-3)\), would equal zero if \(x=3\). If either of these factors equal zero, then the denominator would equal zero. Therefore, the domain restrictions are \(x= -5\) and \(x=3\).

And lastly, the factor, \((x+5)\), is in both the numerator and the denominator, so these factors cancel out of the rational expression.

This leaves us with the simplified expression of, \(\frac{(x+5)}{(x-3)}\), with domain restrictions of -5 and 3.

The more you practice, the easier this process becomes. It is very important to note that identifying the domain restrictions in the denominator must be done immediately after factoring because the factors that are common cancel out in Step 3. (You won’t be able to identify the restrictions if the factors have been canceled out!)

Review

Now, before we go, let’s look at some true or false questions to test your knowledge. Feel free to pause the video to give yourself more time.

1. True or False. The domain restrictions of this rational expression: \(\frac{8x^4-28x^3+16x^2-56x}{x^2-2x-8}\) are \(x=4\) and \(x= -2\).

2. True or False. The rational expression \(\frac{3x^2+9x-12}{x^2-1}\) simplifies to \(\frac{3(x+4)}{(x+1)}\), with domain restrictions of \(x= -4\).

I hope this review was helpful! Thanks for watching, and happy studying!

Simplifying Rational Polynomial Function Practice Questions