# Simplifying Polynomial Rational Expressions

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Objectives
• Learn methods for simplifying rational expressions that contain binomials or polynomials
• Use factoring methods to either simplify rational expressions or classify them as irreducible
Lesson Description

In this very important lesson, we'll see how to simplify rational expressions with objects like quadratics that can first be factored, which can create canceling linear factors. In short, we'll see how factoring allows us to reduce a fraction with variables that otherwise could not be simplified.

Practice Problems

Practice problems and worksheet coming soon!

## Cleaning Up a Mess

Rational expressions are similar to numeric fractions in that they both have numerators and denominators, but we often think of them very differently. With numeric fractions, there is much less mystery - a fixed number is a fixed number, and while we can re-write something like $1/2$ as $5/10$ when there is reason to do so, or talk about adding or multiplying fractions together, the fraction on its own doesn't leave much to analyze. General rational expressions, however, provide us with much more entertainment.Before we work on things like adding, subtracting, multiplying, or dividing rational expressions in the next few lessons, we need to be experts on simplifying a given rational expression on its own. Various operations and processes in advanced math with yield some sort of rational expression, and it's our job to take that result and simplify it - especially if we will then proceed to do something else with it.Just like numeric fractions, the key to simplifying a rational expression will be to find identical factors in both the numerator and denominator, and cancel them one-for-one with one another.

## First Examples

Recall the type of simplification we learned in the Algebra One lesson on simplifying basic rational expressions » . Here's an example.$$\frac{2x^2y}{wx}$$The $x$ factor in the denominator can cancel with a factor of $x$ in the numerator.$$\frac{2xxy}{wx}$$$$\frac{2x\cancel{x}y}{w\cancel{x}}$$$$\frac{2xy}{w}$$When simplifying general rational expressions, we will be proceeding similarly, except the factors that we cancel will often be binomial and polynomial terms, not mere monomials.

Example 1Simplify.$$\frac{(x+2)(x+3)}{(x+3)(x-4)}$$$\blacktriangleright$ The $(x+3)$ factor occurs in both the numerator and denominator. These cancel with one another.$$\frac{(x+2)\cancel{(x+3)}}{\cancel{(x+3)}(x-4)}$$$$\frac{(x+2)}{(x-4)}$$We will also often be asked to do some factoring ourselves before we can cancel these factors.

Example 2Simplify.$$\frac{x^2+5x+6}{x^2-x-12}$$$\blacktriangleright$ First we will factor, and then cancel out any matching factors that appear in both the numerator and denominator.$$\frac{(x+2)(x+3)}{(x+3)(x-4)}$$$$\frac{(x+2)\cancel{(x+3)}}{\cancel{(x+3)}(x-4)}$$$$\frac{(x+2)}{(x-4)}$$
Remember!
We can only cancel factors. Students working on rational expressions containing polynomials of some kind are very often tempted to cancel terms that are not appropriate to cancel. If your denominator is $2x+1$, then $2x$ and $1$ are each NOT factors. The entire $(2x+1)$ term is a single factor and can only cancel with a $(2x+1)$ term in the numerator.
We will use the skill of simplifying rational algebraic expressions here and there for the foreseeable future, and even into parts of intermediate level Calculus. We typically end up with either pre-factored expressions, like we saw in Example 1 above, or with expressions that we need to factor, like Example 2 above.
You Should Know
Usually, quiz and test questions that require factoring will focus on quadratic expressions, while more complicated expressions will be given to you pre-factored.

## Pre-Factored Simplifying

As we saw in Example 1 above, simplifying a rational expression that is pre-factored consists only of identifying the matching factors in the numerator and the denominator. This is relatively less challenging than having to factor things ourselves, though there are often more factors involved. Pre-factored rational expressions will appear less often on quizzes and tests, and will also usually only come across our desks as a result of other operations (like multiplying and dividing rational expressions »), not as a task given to us verbatim.

Example 3$$\frac{(x-2)(2x+8)(x-9)}{(x-9)(x+2)(x+4)}$$$\blacktriangleright$ To simplify this expression, cancel each factor that appears in both the numerator and denominator. We still need to see the common $2$ to factor out in the middle factor of the numerator.$$\frac{(x-2)\big[(2)(x+4)\big](x-9)}{(x-9)(x+2)(x+4)}$$$$\frac{(x-2)\big[(2)\cancel{(x+4)}\big]\cancel{(x-9)}}{\cancel{(x-9)}(x+2)\cancel{(x+4)}}$$$$\frac{2(x-2)}{x+2}$$Typically on a test, a simplify question on a pre-factored rational expression will have more manipulation steps than this example has, but it helps illustrate the idea of canceling the linear factors as units, and reminds us that we cannot cancel individual pieces of factors, nor factors that are "similar" (e.g. $(x-2)$ and $(x+2)$ cannot cancel).

Example 4$\frac{18xz^2(x+2y)(y-x)(xz-z)}{x^3 y^2 (3x+3y)(xy-z)(x-1)(x-y)}$$\blacktriangleright Similar to the last problem, this expression is pre-factored, so we just need to take care of some minor GCF factoring if needed, then keep a keen eye for what does and doesn't match up, on top and bottom. The terms that require a little GCF work are (xz-z) in the numerator and (3x+3y) in the denominator.\frac{18xz^2 (x+2y)(y-x)\big[z(x-1)\big] }{x^3 y^2 \big[3(x+y)\big] (xy-z)(x-1)(x-y)}$$\frac{\cancel{18}\cancel{x}z^3 (x+2y)(y-x)\cancel{(x-1)}}{\cancel{3}\cancel{x^3}y^2 (x+y)(xy-z)\cancel{(x-1)}(x-y)}$$\frac{6z^3 (x+2y)(y-x)}{x^2 y^2 (x+y) (xy-z)(x-y)}We're nearly done, but there is one more factor to cancel. If we factor a (-1) out of the factor (y-x), we get (-1)(x-y).$$\frac{6z^3 (x+2y)\big[(-1)\cancel{(x-y)}\big]}{x^2 y^2 (x+y) (xy-z)\cancel{(x-y)}}\frac{-6z^3 (x+2y)}{x^2 y^2 (x+y) (xy-z)}$$Pro Tip Factoring out a GCF is more common and clear to us when the common piece is numeric, such as factoring a 3 out of (6x-9). However, it is infrequently required of us to recognize that something like (x-y) and (y-x) actually cancel with one another to leave behind a (-1).$$\frac{y-x}{x-y} = -1$$This doesn't violate our general cancellation rule: this only happens because we can factor out a (-1) out of either term and leave ourselves with identical factors that can cancel. We still cannot do anything to make a factor like (x+2) cancel with (x-2), because there is no way to manipulate these factors to be the same objects (but again we could make (x-2) cancel with (2-x) to leave behind a (-1)). Example 5$$\frac{(x-2)^2 (x+6)^3 (x-3)^4}{(x+8)^2 (x+6)^5 (2x-6)^2}$$\blacktriangleright This time, several of the factors we are dealing with have multiplicity. Additionally, there is a GCF inside an object that is raised to a power. We must proceed using the fact that$$(2x-6) = 2(x-3)\therefore (2x-6)^2 = \big[2(x-3)\big]^2=2^2 \cdot (x-3)^2 = 4(x-3)^2$$Putting that result into the expression, and cancelling out multiplicities with one another:$$\longrightarrow \frac{(x-2)^2 \cancel{(x+6)^3} \cancel{(x-3)^{4}}}{(x+8)^2 \cancel{(x+6)^{5}} 4\cancel{(x-3)^2}}$$=$$\frac{(x-2)^2 (x-3)^2}{4 (x+8)^2 (x+6)^2}$$## The Usual Factoring Suspects We should be fairly well-practiced on factoring by now, particularly on quadratic stuff. Here are the typical factoring techniques you should expect to use when simplifying rational expressions in an Algebra Two level course. In the next several lessons, we will simplify each expression, relying on the stated factoring type.Trinomial FactoringFactoring general quadratics is something we've practiced a lot up to this point, so hopefully both identifying them and factoring them is beginning to become automatic. Example 6$$\frac{x^2-x-30}{x^2+3x-10}$$\blacktriangleright Factor each trinomial and cancel any matching factors.$$\frac{(x-6)\cancel{(x+5)}}{\cancel{(x+5)}(x-2)}$$One small convenience with these problems is that when you factor one of them, the other is easier to factor, since you know with near certainty that one of the factors will be in the other factored form.Difference of SquaresYou will usually encounter simple cases of this, but it is possible to be presented with two-variable difference of squares cases in conjunction with two-variable quadratics (we'll see one below when we Put It to the Test). Example 7$$\longrightarrow \frac{x^2-49}{x^2 + 3x -28}$$\blacktriangleright First, factor both the top and bottom, recognizing the difference of squares in the numerator. Then, cancel factors if possible.$$\frac{\cancel{(x+7)}(x-7)}{(x-4)\cancel{(x+7)}}=\frac{x-7}{x-4}$$Sum or Difference of CubesSome teachers really love to use this formula, and some don't. Generally, you will see sums and differences of cubes here and there along your math journey, and even into parts of Calculus. It's a matter of recognizing the form, and remembering the factoring pattern.$$(a^3 \pm b^3 ) = (a \pm b)\left(a^2 \mp ab + b^2\right)$$Example 8$$\frac{x^2+5x-24}{3x^3-81}$$\blacktriangleright The quadratic numerator factors using reverse FOIL without issue. The denominator, however, requires the difference of cubes formula, once the common 3 is factored out.$$\frac{x^2+5x-24}{3(x^3-27)}\frac{(x-3)(x+8)}{(x-3)(x^2 + 3x + 9)}\frac{x+8}{x^2 + 3x + 9}$$Remember! The quadratic piece that we are left with in the sum or difference of cubes formula is guaranteed to be irreducible. In other words, you will absolutely not be able to factor it. GCF FactoringFor the shrewd student who recognizes that we've referenced this already with pre-factored form examples above, GCF Factoring as a "factoring technique" usually refers to factoring out entire variable monomial expressions from a sum or difference of terms, so it's a little more involved than factoring out something like a common 2. Recall that GCF polynomial factoring » typically involves polynomials with higher degrees than quadratics, and may or may not include more than one variable. We'll know to use this when we see such cases, and we will also know to consider it if we're looking at something unusual (i.e. not a quadratic, difference of squares, or sum / difference of cubes). Example 9$$\frac{6x^4 y^3 + 48x^3 y^3 - 120x^2 y^3}{30x^3y^4 + 60x^2 y^4 - 240 x y^4}$$\blacktriangleright Before you panic at this mess of symbols, always remember that if you're asked to do a task, there's gotta be a way to do it. When we see a bunch of terms that all have the same variables, it's a good sign that GCF Factoring is a good first thought.$$\frac{6x^2 y^3 \big( x^2 + 8x - 20 \big)}{30x y^4 \big( x^2 + 2x - 8 \big)}$$Now, we just have to factor these two quadratics, which is business as usual.$$\frac{6x^2 y^3(x+10)(x-2)}{30x y^4(x+4)(x-2)}$$Cancel the linear matching factors. Also, there are some constants, x, and y terms that can cancel.$$\frac{\cancel{6}x^\cancel{2} \cancel{y^3}(x+10)\cancel{(x-2)}}{\cancel{30} \cancel{x} y^\cancel{4}(x+4)\cancel{(x-2)}}\frac{x(x+10)}{5y(x+4)}$$Factoring by GroupingSimilar to GCF Factoring, "Factoring by Grouping" is a skill that we need to consider looking for if the expression we are given does not look typical. Though you will encounter this only infrequently, Factoring by Grouping does have a tell-tale hallmark - it's a common way to proceed if you are given a cubic polynomial that isn't a sum or difference of cubes. Example 10$$\frac{x^3 - 4x^2 + 6x - 24}{x^2 -9x + 20}$$\blacktriangleright The denominator is a standard issue quadratic. The numerator, however, must be done with Factoring by Grouping.$$\frac{x^2 (x-4) + 6(x-4)}{(x-5)(x-4)}\frac{(x^2+6)\cancel{(x-4)}}{(x-5)\cancel{(x-4)}}$$You Should Know Not all students have studied polynomials extensively when they learn this topic, but if your class has, then some advanced (aka evil) teachers will give you cubic and quartic polynomials to factor here, using such techniques as trial and error on finding factors via the rational roots theorem » and polynomial long division ». This is honestly quite a rare ask when studying rational expressions, but if that's the case for you, work on factoring those monster polynomials with those techniques, and when you're done, you're still looking at the same idea, which is to find and cancel identical factors in the numerator and denominator, if possible. ## Put It To The Test Here are a few more examples for you to try on your own. As always, make sure you look to the practice worksheet for this lesson to really lock in your skills! Example 11$$\frac{8x^2-72}{2x^2-6x-36}$$Show solution$$\blacktriangleright \,\, \frac{8(x+3)(x-3)}{2(x-6)(x+3)}\frac{\cancel{8}\cancel{(x+3)}(x-3)}{\cancel{2}(x-6)\cancel{(x+3)}}\frac{4(x-3)}{x-6}$$Example 12$$\frac{x^3+11x^2+24x}{x^5+512x^2}$$Show solution$$\blacktriangleright \,\, \frac{x(x+3)(x+8)}{x^2(x+8)\big(x^2-8x + 64\big)}\frac{\cancel{x}(x+3)\cancel{(x+8)}}{x^\cancel{2}\cancel{(x+8)}\big(x^2-8x + 64\big)}\frac{x+3}{x\big(x^2-8x + 64\big)}$$Example 13$$\frac{4x^4 y^7 - 81x^2 y}{2x^2 y^6 -5xy^3 -18}$$Show solution$$\blacktriangleright \,\, \frac{x^2 y \big(4x^2 y^6 - 81\big)}{\left(2xy^3 - 9\right)\left(xy^3 + 2\right)}\frac{x^2 y \left(2xy^3 + 9\right)\left(2xy^3 - 9\right)}{\left(2xy^3 - 9\right)\left(xy^3 + 2\right)}\frac{x^2 y \left(2xy^3 + 9\right)\cancel{\left(2xy^3 - 9\right)}}{\cancel{\left(2xy^3 - 9\right)}\left(xy^3 + 2\right)}\frac{x^2 y \left(2xy^3 + 9\right)}{\left(xy^3 + 2\right)}$$Example 14$$\frac{4x^2 (x^2-5x-84)(x^2-4x-12)}{2x^5-18x^4-72x^3}$$Show solution$$\blacktriangleright \,\, \frac{4x^2 (x-12)(x+7)(x+2)(x-6)}{2(x+3)(x-12)}\frac{2x^2 (x+7)(x+2)(x-6)}{x+3}$$Example 15$$\frac{25x^4 - 121y^2}{5x^4 -4x^2 y - 33y^2}$$Show solution$$\blacktriangleright \,\, \frac{\big(5x^2 - 11y\big)\big(5x^2 + 11y\big)}{\big(5x^2 + 11y\big)\big(x^2 - 3y\big)}\frac{\big(5x^2 - 11y\big)}{\big(x^2 - 3y\big)}$\$

Lesson Takeaways
• Be able to identify factors that can cancel, and why pieces of factors cannot cancel
• Even for pre-factored rational expressions, look for common numeric factors to cancel out, and know how to simplify completely by canceling factors match-for-match, with and without factor multiplicity
• Gain familiarity with the different kinds of factoring patterns we typically see for simplifying variable rational expressions
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