# 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!

## Practice Questions

Question #1:

What is $$\frac{x^2+2x}{x^2+x-2}$$ in its most simplified form?

$$\frac{x}{x-1}$$, with domain restrictions $$x=1$$ and $$x=-2$$.

$$\frac{x}{x+2}$$, with domain restrictions $$x=1$$ and $$x=2$$.

$$\frac{x+2}{x-1}$$, with domain restrictions $$x=-1$$ and $$x=-2$$.

$$\frac{x-1}{x+2}$$, with domain restrictions $$x=-1$$ and $$x=2$$.

When simplifying rational polynomial functions, we start by factoring the numerator and denominator. Since a fraction would be undefined if there is a zero in the denominator, we must identify the domain restrictions, which is any x-value that would make the denominator equal to zero. Then we cancel anything that is in both the numerator and the denominator, since it equals to 1. Factoring the numerator and denominator results in:
$$\frac{x(x+2)}{(x-1)(x+2)}$$

We will look at the denominator to identify the domain restrictions, which are $$x=1$$ and $$x=-2$$, then the $$(x + 2)$$ from the numerator and denominator gets canceled and we are left with $$\frac{x}{x-1}$$.

Question #2:

Simplify the expression: $$\frac{x^2-5x}{x^2-8x}$$.

$$\frac{x}{x-8}$$, with domain restrictions $$x=0$$ and $$x=8$$.

$$\frac{x}{x-5}$$, with domain restrictions $$x=0$$ and $$x=5$$.

$$\frac{x-5}{x-8}$$, with domain restrictions $$x=0$$ and $$x=8$$.

$$\frac{x-8}{x-5}$$, with domain restrictions $$x=0$$ and $$x=5$$.

To simplify an expression with polynomials in the numerator and denominator we will start by factoring:
$$\frac{x(x-5)}{x(x-8)}$$

The domain restriction, based on the factored denominator, is $$x=0$$ and $$x=8$$. The x in the numerator and denominator are canceled and we are left with $$\frac{x-5}{x-8}$$.

Question #3:

Which expression shows the most simplified form of the expression $$\frac{x^2-9}{x^2-2x-3}$$?

$$\frac{x+1}{x+3}$$, with domain restrictions $$x=-1$$ and $$x=-3$$.

$$\frac{x+3}{x+1}$$, with domain restrictions $$x=-1$$ and $$x=3$$.

$$\frac{x-3}{x+1}$$, with domain restrictions $$x=1$$ and $$x=-3$$.

$$\frac{x+3}{x-3}$$, with domain restrictions $$x=-3$$ and $$x=3$$.

The expression, $$\frac{x^2-9}{x^2-2x-3}$$, can be simplified by first factoring the numerator and denominator:
$$\frac{(x+3)(x-3)}{(x+1)(x-3)}$$

The domain restrictions are any values for x that would make the denominator of the rational polynomial a zero, which in this case are $$x=-1$$ and $$x=3$$. The $$(x–3)$$ factors can be canceled, since it is equal to 1, and we are left with $$\frac{x+3}{x+1}$$, the most simplified form of the expression.

Question #4:

Which shows the expression, $$\frac{2x^2-3x+1}{x^2-1}$$, in its most simplified form?

$$\frac{x-1}{x+1}$$, with domain restrictions $$x=1$$ and $$x=-1$$.

$$\frac{x+1}{x-1}$$, with domain restrictions $$x=1$$ and $$x=-1$$.

$$\frac{x-1}{2x-1}$$, with domain restrictions $$x=\frac{1}{2}$$ and $$x=1$$.

$$\frac{2x-1}{x+1}$$, with domain restrictions $$x=1$$ and $$x=-1$$.

We will start by factoring the numerator and denominator of the rational polynomial expression:
$$\frac{(2x-1)(x-1)}{(x+1)(x-1)}$$

The domain restrictions are $$x=1$$ and $$x=-1$$, because either of these two values would make the denominator have a factor of zero, which would make it undefined. The factors $$(x–1)$$ from both the numerator and denominator cancel out, because it is equal to 1 and anything times 1 is itself. The expression in its most simplified form is $$\frac{2x-1}{x+1}$$.

Question #5:

Simplify the expression: $$\frac{3x^2-16x-12}{x^2-4x-12}$$.

$$\frac{x-6}{x+2}$$, with domain restrictions $$x=2$$ and $$x=-6$$.

$$\frac{x+2}{x-6}$$, with domain restrictions $$x=-2$$ and $$x=6$$.

$$\frac{3x+2}{x+2}$$, with domain restrictions $$x=-2$$ and $$x=6$$.

$$\frac{x+2}{3x+2}$$, with domain restrictions $$x=-\frac{2}{3}$$ and $$x=-2$$.

To simplify the expression, $$\frac{3x^2-16x-12}{x^2-4x-12}$$, we will start by factoring both the numerator and denominator:
$$\frac{(3x+2)(x-6)}{(x+2)(x-6)}$$

The domain restrictions are any value for x that would make the denominator have a factor of zero. In this case the values $$x=-2$$ and $$x=6$$ would make the denominator have a factor of zero. The factor $$(x–6)$$ in the numerator and denominator equals 1, so we can cancel those two factors and what remains is $$\frac{3x+2}{x+2}$$.