Multiplying and Dividing Rational Expressions

Lesson Features »

Lesson Priority: High

  • Learn how to cancel common factors before multiplying rational polynomial expressions
  • Understand why failing to cancel factors before multiplying actually makes the problem impossible to simplify
  • Quickly turn the division of a rational polynomial expression into a multiplication problem
Lesson Description

When we multiply fractions that contain variable expressions, we might be wasting a LOT of time if we don't first factor them. When we factor first, the result might contain factors that cancel out and thus make the final answer much more simple. We also can handle the division of two rational expressions in a similar way, by first turning it into multiplication, and then proceeding with the same process.

Practice Problems

Practice problems and worksheet coming soon!


Success Through Reduction

Recently, we saw how general rational expressions with linear and polynomial numerators and denominators were often simplifiable by cancelling common factors. This idea is monumentally important to multiplying those types of expressions - it's not only less work for us overall, but sometimes it is just plain impossible otherwise.The idea of reducing fractions before we multiply is not entirely new. Unlike addition and subtraction of fractions », we do not need to have equal denominators or any other kind of pre-requisite to multiply. We simply multiply across each the numerators and denominators:$$\frac{1}{2} \cdot \frac{5}{7} \cdot \frac{3}{8} = \frac{15}{112}$$However, if we have bigger numbers with common factors, we are often taught to do some cancelling before we multiply - not because we have to, but because it's efficient (which means less work for the lazy mathematician!).For example, what is your instinct when you see the following problem?$$\frac{16}{120} \cdot \frac{14}{28} \cdot \frac{42}{49}$$Was it to multiply $16$, $14$, and $42$? And then multiply $120$, $28$, and $49$ to ultimately get $9408/164640$? More likely than not at least one of your teachers has ingrained the will to simplify first before you multiply, and whether or not you ever considered why, hopefully this example makes it clear that we will do less work if we do. Unless simplifying $9408/164640$ without a calculator sounds like a good time to you.Instead,$$\frac{2}{15} \cdot \frac{1}{2} \cdot \frac{6}{7}$$$$\frac{\cancel{2}^1}{\cancel{15}^5} \cdot \frac{1}{\cancel{2}^1} \cdot \frac{\cancel{6}^2}{7}$$$$=\frac{2}{35}$$

Factor Everything

When we're asked to multiply polynomial rational expressions, we will encounter this same phenomenon. If we blindly multiply across, we'll end up with a mess. Instead, we will factor every numerator and denominator of every fraction in every problem every time. When we do, we'll see what common factors exist, cancel those factors, and knock down homework problems in mere minutes.
Pro Tip
You might not have guessed it from the title of this skill, but the only real computational challenge here is factoring. Like we've seen in the past, strong factoring skills payoff big time in many Algebra chapters. If you're still not a master of all things factoring, this is the right time to go back and become one.GCF Factoring »Factoring Trinomials »Factoring Advanced Trinomials »Factor by Grouping »Differences of Squares »Advanced Differences of Squares »Sums and Differences of Cubes »You'll see any and all of them in this lesson. Sometimes, teachers will get mean and give you cubics and quartics where you need to wrestle out its factors by guess-and-check or with synthetic division. This is uncommon but requires a lot of the advanced polynomial » work we recently covered. We'll look at this in the last section » of this lesson.
Let's try an example.
Example 1Multiply and simplify, leaving your answer in factored form.$$\frac{x^2 - 3x - 18}{x^2 - 25} \cdot \frac{x^2 - 10x + 25}{x^2 - 4x - 21} $$$\Rightarrow$ Factor each numerator and denominator.$$\frac{(x+3)(x-6)}{(x+5)(x-5)} \cdot \frac{(x-5)(x-5)}{(x+3)(x-7)}$$Just like when we learned how to "cancel as we go" when multiplying fractions, any factor we see in a numerator can cancel with any numerator we see in a denominator.$$\frac{\cancel{(x+3)}(x-6)}{(x+5)\cancel{(x-5)}} \cdot \frac{\cancel{(x-5)}(x-5)}{\cancel{(x+3)}(x-7)}$$$$=\frac{(x-6)}{(x+5)} \cdot \frac{(x-5)}{(x-7)}$$$$=\frac{(x-6)(x-5)}{(x+5)(x-7)}$$Typically, we will also leave our answers in factored form, which is wonderful if you're a lazy efficient mathematician like me.It was monumentally important that the first step was to factor. If we multiplied first instead of factored first, we would have been looking at$$\frac{x^2 - 3x - 18}{x^2 - 25} \cdot \frac{x^2 - 10x + 25}{x^2 - 4x - 21}$$$$=\frac{x^4-13x^3+37x^2+105x-450}{x^4-4x^3-46x^2+100x+525}$$I don't know about you, but I'm not very interested in factoring that mess! And without computer help, it would take you a very long time indeed to do it.Now that we've seen how this usually works, try the next one on your own.
Example 2Multiply and simplify, leaving your answer in factored form.$$\frac{x^2-10x+24}{3x^2-34x -11} \cdot \frac{2x^2+x-6}{x^2-4x-8}$$
Show solution
$\blacktriangleright$ Again, factor right away.$$\frac{(x-4)(x-6)}{(3x+1)(x-11)} \cdot \frac{(2x-3)(x+2)}{(x-6)(x+2)}$$Cancel what you can:$$\frac{(x-4)\cancel{(x-6)}}{(3x+1)(x-11)} \cdot \frac{(2x-3)\cancel{(x+2)}}{\cancel{(x-6)}\cancel{(x+2)}}$$$$=\frac{x-4}{(3x+1)(x-11)} \cdot \frac{2x-3}{1}$$$$=\frac{(x-4)(2x-3)}{(3x+1)(x-11)}$$

Be Ready to Factor Anything

Teachers, depending on their appetite for insanity, will ask you not only factor quadratics, but also special cubics and maybe even quartics. First make sure you're up to speed on all the aforementioned factoring forms, and then try the next two examples on your own.
Example 3Multiply and simplify.$$\frac{6x^2 + x - 15}{x^2-5x-36} \cdot \frac{4x^2+22x+24}{27x^3 + 125}$$
Show solution
$\blacktriangleright$ Each piece is quadratic, except for the sum of cubes $27x^3 + 125$. Recall that a sum of cubes factors as$$a^3 + b^3 = (a + b)\big(a^2 - ab + b^2\big)$$Recall also that a sum or difference of cubes will leave you with an irreducible quadratic factor, so don't waste your time trying to factor the quadratic piece further. All in all, we have$$\frac{(3x + 5)(2x-3)}{(x+4)(x-9)} \cdot \frac{2(x+4)(2x+3)}{(3x+5)\big( 9x^2 - 15x + 25 \big)}$$$$\frac{\cancel{(3x + 5)}(2x-3)}{\cancel{(x+4)}(x-9)} \cdot \frac{2\cancel{(x+4)}(2x+3)}{\cancel{(3x+5)}\big( 9x^2 - 15x + 25 \big)}$$$$\frac{2(2x-3)(2x+3)}{(x-9)\big( 9x^2 - 15x + 25 \big)}$$
Example 4Multiply and simplify, leaving your answer in factored form.$$\frac{x^3 - 2x^2 + 4x - 8}{x^3 - 8} \cdot \frac{x^2 + 4x - 32}{x^4 - 12x^2 - 64}$$
Show solution
$\blacktriangleright$ First and foremost, for the love of all that is holy, please remember that you cannot cancel the $x^3$ and $8$ in the left-hand fraction with one another. One of the many Mr. Math mantras is: YOU CAN ONLY CANCEL FACTORS!You will, however, have great luck on the first fraction's numerator by trying some factoring by grouping:$$x^3 - 2x^2 + 4x - 8 = x^2(x-2) + 4(x-2)$$$$=\left(x^2 + 4\right)(x-2)$$The denominator of the first fraction is a difference of cubes. It is also worth noting is that the denominator of the second fraction is a "special quadratic", which we could describe in this particular case as "quadratic in $x^2$". Putting all of these thoughts together:$$\frac{x^3 - 2x^2 + 4x - 8}{x^3 - 8} \cdot \frac{x^2 + 4x - 32}{x^4 - 12x^2 - 64}$$$$\Rightarrow \frac{(x^2+4)(x-2)}{(x-2)(x^2+2x+4)} \cdot \frac{(x-4)(x+8)}{(x^2+4)(x^2-16)}$$$$\frac{\cancel{(x^2+4)}\cancel{(x-2)}}{\cancel{(x-2)}(x^2+2x+4)} \cdot \frac{(x-4)(x+8)}{\cancel{(x^2+4)}(x^2-16)}$$$$\frac{1}{x^2+2x+4} \cdot \frac{(x-4)(x+8)}{x^2-16}$$A few things to note here: first, as with Example 3, don't bother trying to further factor the quadratic from the difference of cubes. Second, the quadratic factor we are left with $x^2 - 16$ is further factorable, as it is itself a difference of squares. This is something we need to be aware of because it's possible more factors will cancel because of it.$$=\frac{1}{x^2+2x+4} \cdot \frac{\cancel{(x-4)}(x+8)}{(x+4)\cancel{(x-4)}}$$$$=\frac{x+8}{\left( x^2+2x+4 \right) (x+4)}$$
Pro Tip
Always double check your factored results for further factoring due to differences of squares or GCF factoring. It's easy to miss and can cause you to violate your instructions to "simplify as much as possible", resulting in deflated grades.

Rational Expression Division

Fortunately, as we've seen in the past with fractions, division is really multiplication in disguise. We need only make a quick adjustment to make it so.
Define: Division of Fractions(revisited from Pre-Algebra)$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$$
This mechanic does not change when we look at algebraic expressions instead of integers. For example,$$\frac{x+3}{x-8} \div \frac{x-4}{x+7} = \frac{x+3}{x-8} \cdot \frac{x+7}{x-4}$$
You Should Know
Because fraction multiplication is familiar and somewhat intuitive, and because fraction division can be instantly and painlessly turned into fraction multiplication, you will always turn fraction division into fraction multiplication in every scenario you will ever encounter in any math course. That's just the best way to manage fraction division, plain and simple.
Example 5Divide. Simplify as much as possible but leave your answer in factored form.$$\frac{3x^2 + 6x - 24}{x^2 + 11x + 30} \div \frac{2x^2 + 7x - 4}{x^3 - 25x}$$$\blacktriangleright$ First we will flip the second fraction, and then we'll be looking at the same kind of problem we've been looking throughout this lesson.$$\frac{3x^2 + 6x - 24}{x^2 + 11x + 30} \cdot \frac{x^3 - 25x}{2x^2 + 7x - 4}$$The name of the game from here is factoring, just like before. Don't forget to use GCF factoring when you can - even when it's not 100% essential, it makes your life easier and this problem is no exception.$$\frac{3(x+4)(x-2)}{(x+5)(x+6)} \cdot \frac{x(x+5)(x-5)}{(2x-1)(x+4)}$$$$=\frac{3x\cancel{(x+4)}(x-2)\cancel{(x+5)}(x-5)}{\cancel{(x+5)}(x+6)(2x-1)\cancel{(x+4)}}$$$$=\frac{3x(x-2)(x-5)}{(x+6)(2x-1)}$$

Amass Mastery

One way that teachers enjoy marking down your grade is by using factors that are negatives of one another instead of identical to one another. These factors cancel to $-1$ instead of $1$, but the folly of many students is that they don't cancel them at all, thinking that they cannot cancel.
Example 6Multiply. Simplify as much as possible but leave your answer in factored form.$$\frac{81-x^2}{3x^2 -5x -12} \cdot \frac{3x^3 + 4x^2 + 9x + 12}{x^2-4x -45} $$$\blacktriangleright$ Upon inspection, we're looking at a difference of squares, a factor by grouping cubic, and two factorable quadratics. Factoring all of these yields$$\frac{(9+x)(9-x)}{(3x+4)(x-3)} \cdot \frac{\big( x^2 + 3 \big)(3x+4)}{(x-9)(x+5)}$$The factor of interest for this example is the $9-x$ and $x-9$. These factors are not identical, but they nearly are. $9-x$ is really $(-1)(x-9)$, so $9-x$ and $x-$ cancel to $-1$.$$\frac{(9+x)\cancel{(9-x)}(-1)}{\cancel{(3x+4)}(x-3)} \cdot \frac{\big( x^2 + 3 \big)\cancel{(3x+4)}}{\cancel{(x-9)}(x+5)}$$$$=-\frac{(x+9)\big( x^2 + 3 \big)}{(x-3)(x+5)}$$
The factors $(a-b)$ and $(b-a)$ are not the same factor and do not cancel to $1$ when one is each present in the numerator and denominator. However, if you factor a $-1$ out of either factor, they will cancel:$$\frac{a-b}{b-a} = \frac{\cancel{a-b}}{-1\cancel{(a-b)}} = -1$$This is a common trick employed by teachers to separate the A's from the B's! Look for it on quizzes and tests.
The other common way to be thrown a curveball on a quiz is to be asked to divide a rational expression by a polynomial. Even though you aren't dividing by a fraction, you should still act like you are. It's not so much about the fact that you are or aren't dividing by a fraction, it's that you are performing division with fractions involved.
Example 7Divide. Simplify as much as possible but leave your answer in factored form.$$\frac{4x^2 - 23x -6}{x^2 + 4x + 3} \div x^2 - 3x - 18$$$\blacktriangleright$ When division involving fractions is involved, get things to look like multiplication, if possible. We can do that for this problem:$$\frac{4x^2 - 23x -6}{x^2 + 4x + 3} \div x^2 - 3x - 18$$$$\longrightarrow \frac{4x^2 - 23x -6}{x^2 + 4x + 3} \cdot \frac{1}{x^2 - 3x - 18}$$Now factor and proceed as we have in prior examples.$$\frac{(4x+1)(x-6)}{(x+3)(x+1)} \cdot \frac{1}{(x-6)(x+3)}$$$$\frac{(4x+1)\cancel{(x-6)}}{(x+3)(x+1)} \cdot \frac{1}{\cancel{(x-6)}(x+3)}$$$$\frac{(4x+1)}{(x+3)^2 (x+1)}$$
When you are performing a division that involves a fraction, you should proceed with the "reciprocal" process of turning the problem into a multiplication problem, even if you aren't dividing by a fraction. Just having fractions involved makes this necessary. You certainly wouldn't be pleased if you tried to write Example 7 as$$\frac{\frac{4x^2 - 23x -6}{x^2 + 4x + 3}}{x^2 - 3x - 18}$$because complex fractions » are not your friend (in fact the whole point of the lesson on them is to somehow make them into normal fractions).

Make It Mean

If you are presented with a rational expression multiplication or division problem that has a cubic or fourth degree polynomial, and you've already tried GCF Factoring » and Factoring by Grouping », then you need to use some rational roots » and polynomial long division » knowledge to figure out how that polynomial breaks down into factors.That's the bad news. The good news is, you have a lot of hidden help right in front of you, in terms of what numbers to try. If you're asked to break down a polynomial beast like that, you know that it will break into pieces that will cancel with something else somewhere in the problem. This limits the possibilities that you have to cycle through with the "guess-and-check" part of your work.Let's look at an example of this idea in action.
Example 8Multiply.$$\frac{x^3+3x^2-10x-24}{3x^3 - 7x^2 - 6x} \cdot \frac{x^4 - 81x^2}{x^2 + 4x + 4}$$In the first fraction, the denominator can be turned quadratic by removing the GCF of $x$, and factored from there. Ditto with the fourth degree in the second fraction's numerator. The numerator of the first fraction however, is not easily factorable, and requires us to crack it open by force, so to speak.The rational roots theorem » tells us that this cubic polynomial will have a real, whole number root that is a factor of $24$, positive or negative. If we are shooting blind, that leaves us with $1$, $-1$, $2$, $-2$, $3$, $-3$, $4$, $-4$, $6$, $-6$, $12$, and $-12$. That's a lot of potential guesswork! Instead of resigning to a blind guess-and-check, let's factor everything else.$$\frac{x^3+3x^2-10x-24}{x(x-3)(3x+2)} \cdot \frac{x^2 (x+9)(x-9)}{(x+2)(x+2)}$$If the factored form of that cubic polynomial will eventually cancel with something, it will be factors in either denominator. Since the cubic has a leading coefficient of $1$, we know the $(3x+2)$ factor in the denominator won't be part of it. But that only leaves $(x+2)$ and $(x-3)$! Truly wonderful news to turn twelve suspects into two! Let's use polynomial long division to see what happens when we try to divide $(x-3)$ into it.$$x^3+3x^2-10x-24 \div x-3$$$$\longrightarrow x^2+6x+8$$Of course, if you're working by hand, synthetic division is faster and requires fewer pen strokes. In any case, we have a result we can run with, since the remaining quadratic will factor further.$$\frac{(x-3)\big(x^2+6x+8\big)}{x(x-3)(3x+2)} \cdot \frac{x^2 (x+9)(x-9)}{(x+2)(x+2)}$$$$\frac{\cancel{(x-3)}\cancel{(x+2)}(x+4)}{\cancel{x}\cancel{(x-3)}(3x+2)} \cdot \frac{x^\cancel{2} (x+9)(x-9)}{\cancel{(x+2)}(x+2)}$$$$\frac{x+4}{(3x+2)} \cdot \frac{x (x+9)(x-9)}{(x+2)}$$$$=\frac{x(x+4)(x+9)(x-9)}{(3x+2)(x+2)}$$

Put It To The Test

For each question, multiply or divide, and as always, simplify to the fullest extent but leave your answers in factored form.
Example 9$$\frac{x^2-3x-10}{x^2+5x+4} \cdot \frac{x^2-x-12}{x^2-2x-15}$$
Show solution
$\blacktriangleright$ Factor the quadratics.$$\frac{(x-5)(x+2)}{(x+4)(x+1)} \cdot \frac{(x-4)(x+3)}{(x-5)(x+3)}$$$$=\frac{\cancel{(x-5)}(x+2)}{(x+4)(x+1)} \cdot \frac{(x-4)\cancel{(x+3)}}{\cancel{(x-5)}\cancel{(x+3)}}$$$$=\frac{x-2}{(x+4)(x+1)} \cdot \frac{x-4}{1}$$$$=\frac{(x-2)(x-4)}{(x+4)(x+1)}$$
Example 10$$\frac{x^3 + 4x^2 + 4x}{x^2 - 4x + 3} \cdot \frac{2x^2 - 3x + 1}{x^3 + 2x^2 - 4x - 8}$$
Show solution
$\blacktriangleright$ Look toward the possibilities of GCF Factoring and Factor By Grouping, when you see cubics. We'll use both in this problem.$$\frac{x[x^2 + 4x + 4]}{(x-1)(x-3)} \cdot \frac{(x-1)(2x-1)}{(x^2-4)(x+2)}$$$$\frac{x(x+2)(x+2)}{(x-1)(x-3)} \cdot \frac{(x-1)(2x-1)}{(x+2)(x-2)(x+2)}$$$$\frac{x\cancel{(x+2)}\cancel{(x+2)}}{\cancel{(x-1)}(x-3)} \cdot \frac{\cancel{(x-1)}(2x-1)}{\cancel{(x+2)}(x-2)\cancel{(x+2)}}$$$$\frac{x(2x-1)}{(x-3)(x-2)}$$
Example 11$$\frac{16-x^2}{3x^2+2x-15} \cdot \frac{36x^2-100}{2x^2 + 8x - 64}$$
Show solution
$\blacktriangleright$ There are two subtle tricks to this problem. First, we will see that we have opposite factors that cancel to $-1$ instead of the usual identical factors that cancel "cleanly" to $1$. Second, we have may be tempted to see the difference of squares in the second fraction and factor it as is, doing nothing more with the factors. If we do this, we are at risk of not completely simplifying. Here's why:$$36x^2-100 = (6x+10)(6x-10)$$But we can factor a $2$ out of each of those. This is important since we'll see a $3x-5$ factor later to cancel with. The way we handle everything correctly from the start is to keep to our usual advice - factor out a GCF first if possible (in this case, $4$).When we put everything together correctly, here's how things break down.$$\frac{(4-x)(4+x)}{(3x+5)(x-1)} \cdot \frac{4(3x+5)(3x-5)}{2(x-4)(x+8)}$$The $(4-x)$ factor cancels with the $(x-4)$ factor, leaving a negative one.All in all, when we cancel what we can, we will be left with$$\frac{2(4+x)(3x-5)}{(x-1)(x+8)}$$
Example 12$$\frac{6x^2 - 7x + 6}{x^2 + 10x + 25} \cdot \frac{x^2 + 7x + 10}{x^4 + 2x^3 - 216x - 432}$$
Show solution
$\blacktriangleright$ The three quadratics look like business as usual, but that fourth order may initially seem alarming. Fortunately, one of the first things you should check for in that case (factoring by grouping) will work.$$\frac{(6x-1)(x-6)}{(x+5)(x+5)} \cdot \frac{(x+2)(x+5)}{(x^3-216)(x+2)}$$Now, before we start canceling factors and are tempted to wrap up this exercise, let's go further with the difference of cubes. We should always try to fully factor everything we can before we get caught up with canceling stuff, lest we risk turning in a problem that is only partially simplified.Recall that the difference of cubes $x^3-216$ factors as $(x-6)\big(x^2 + 6x + 36\big)$.$$\frac{(6x-1)(x-6)}{(x+5)(x+5)} \cdot \frac{(x+2)(x+5)}{(x-6)\big(x^2 + 6x + 36\big)(x+2)}$$$$\frac{(6x-1)}{(x+5)} \cdot \frac{1}{x^2 + 6x + 36}$$$$\frac{6x-1}{(x+5)\big(x^2 + 6x + 36\big)}$$
Example 13$$\frac{2x^3+14x^2-60x}{2x^2 - 3x - 2} \div \frac{x^3 + 1000}{x^2-5x+6}$$
Show solution
$\blacktriangleright$ First and foremost, let's recognize the division symbol and instantly change this problem to a more familiar multiplication problem.$$\frac{2x^3+14x^2-60x}{2x^2 - 3x - 2} \cdot \frac{x^2-5x+6}{x^3 + 1000}$$Then it's time to factor what we can. As usual start by taking a glance at the kind of stuff you've been given. Notably, we see a sum of cubes, a GCF cubic, and two quadratics that should factor.$$\frac{2x(x-3)(x+10)}{(x-2)(2x+1)} \cdot \frac{(x-2)(x-3)}{(x+10)\big(x^2 - 10x + 100\big)}$$$$\frac{2x(x-3)\cancel{(x+10)}}{\cancel{(x-2)}(2x+1)} \cdot \frac{\cancel{(x-2)}(x-3)}{\cancel{(x+10)}\big(x^2 - 10x + 100\big)}$$$$\frac{2x(x-3)^2}{(2x+1)\big(x^2 - 10x + 100\big)}$$
Example 14$$\frac{x^3 + x^2 + x + 1}{x^2 - 6x - 16} \div x^2 - 1$$
Show solution
$\blacktriangleright$ First thing's first - we need to make it into a multiplication problem so we can process it like any other problem we've done. While it might throw you off to look at division by a single object, know that we can quickly write that object as a fraction if that makes it easier on the eyes.$$\frac{x^3 + x^2 + x + 1}{x^2 - 6x - 16} \div \frac{x^2 - 1}{1}$$Now we can flip it.$$\frac{x^3 + x^2 + x + 1}{x^2 - 6x - 16} \cdot \frac{1}{x^2 - 1}$$Business as usual now - time to factor!$$\frac{(x+1)\big(x^2+1\big)}{(x-8)(x+2)} \cdot \frac{1}{(x+1)(x-1)}$$$$\frac{x^2 + 1}{(x-8)(x+2)(x-1)}$$Note that the numerator is a sum of squares and cannot factor.
Example 15$$\frac{3x^2-3x-18}{3x^5+32x^4+19x^3-226x^2+72x} \cdot \frac{x^4 - 16x^2}{x^2 + 6x - 27}$$
Show solution
$\blacktriangleright$ As always, consider at each piece in turn and don't panic - even though a dreaded fifth degree is waiting for us this time! In the first fraction's numerator, recognize and remove the GCF of $3$ to leave a manageable quadratic, and the GCF of $x$ in the denominator to leave a fourth degree that does not have any factor by grouping pattern and will require some long division. In the second fraction, the numerator is a typical difference of squares once we remove the GCF of $x^2$, and the denominator is a business-as-usual quadratic.$$\frac{3(x-3)(x+2)}{x\big[3x^4+32x^3+19x^2-226x+72\big]} \cdot \frac{x^2(x+4)(x-4)}{(x+9)(x-3)}$$To handle that fourth degree polynomial, remember what we saw earlier for tips - namely that even though we are resorting to rational roots and "guess and check" long division processes, that we should not be wasting our time randomly seeking roots of this polynomial. Instead, look to the other factors in the problem that are candidates for cancelling with it: namely $(x-3)$, $(x+2)$, $(x+4)$, and $(x-4)$.A few guess and check iterations should tell you that $(x+4)$ should do nicely for us.$$3x^4+32x^3+19x^2-226x+72 \div x+4$$$$\longrightarrow 3x^3 + 20x^2 -61x + 18$$Regrettably, we still need to proceed with some rational roots guess and check to further break down this remaining third order polynomial result. $2$ and $3$ are really good candidates for roots of this cubic polynomial, based on the constant term $18$. The factor $(x-2)$ will work.$$3x^3 + 20x^2 -61x + 18 \div x-2$$$$\longrightarrow 3x^2 +26x -9$$Finally, we are left with a manageable quadratic, since $3x^2 +26x -9$ factors into $(3x-1)(x+9)$.Putting it all together:$$\frac{3(x-3)(x+2)}{x(3x-1)(x+4)(x-2)(x+9)} \cdot \frac{x^2(x+4)(x-4)}{(x+9)(x-3)}$$$$\frac{3\cancel{(x-3)}\cancel{(x+2)}}{\cancel{x}(3x-1)\cancel{(x+4)}\cancel{(x-2)}(x+9)} \cdot \frac{x^\cancel{2}\cancel{(x+4)}(x-4)}{(x+9)\cancel{(x-3)}}$$$$\frac{3}{(3x-1)(x+9)} \cdot \frac{x(x-4)}{(x+9)}$$$$\frac{3x(x-4)}{(3x-1)(x+9)^2}$$
Lesson Takeaways
  • Understand why it's bad for everyone if you just blindly start multiplying variable rational expressions without trying to factor and cancel
  • Be prepared to factor in all kinds of ways when you're asked to multiply rational expressions
  • Know how to quickly turn division into multiplication so you can handle both problem types the same way
  • Be privy to the special factoring forms and common tricks that frequently appear on tests
  • Although not common, be ready to break down third and fourth degree polynomials "organically" using polynomial root knowledge we studied in recent lessons

Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.

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