Multiplying and Dividing Rational Expressions, and Cancellation
Parallels Between Fraction Reduction and Cancellation in Rational Expressions
Suppose we want to reduce the following fraction to lowest terms:
$$\frac{14}{35}$$
One way this can be done is by finding the greatest factor (in this case, \(7\)) that evenly divides both \(14\) and \(35\) and then divide both the numerator and denominator by it.
Alternatively, one can factor (or factorize) both the numerator and denominator and then cancel the common factors that one sees on top and bottom with one another. Viz:
$$\frac{14}{35}=\frac{\not 7\cdot 2}{\not 7\cdot 5}=\frac{2}{5}$$
The reason this works is because of how fractions are multiplied, i.e. straight across the top and bottom. So the expression \(\frac{7\cdot 2}{7\cdot 5}\) can be written as
$$\frac{7\cdot 2}{7\cdot 5}=\frac{7}{7}\cdot \frac{2}{5}$$
And the \(\frac{7}{7}\) reduces to \(1\), leaving you with the same final answer as we got above.
Cancellation for radical expressions works in much the same way and for exactly the same reasons.
Simply put, a variable represents an unknown number. As such, so do expressions involving variables. (for instance, if we knew what \(x\) was, we could calculate the numeric value of the expression \((4x+5)\)).
So, if you have an expression like \(\frac{(x+1)(x+3)^2}{(x+1)^2}\), we can write this as a product and cancel common factors in the numerator and denominator, just as with ordinary fractions. Viz:
$$\frac{(x+1)(x+3)^2}{(x+1)^2}=\frac{(x+1)}{(x+1)}\cdot \frac{(x+3)^2}{(x+1)}=\frac{(x+3)^2}{(x+1)}$$
This leads to the following key idea.
After you have cancelled all factors that both the numerator and denominator have in common, the rational expression is said to be completely reduced.
Reduce the following rational expression completely by cancelling all factors that both the numerator and denominator have in common:
$$\frac{x^2(x+1)^3(x-2)}{x^3(x+1)^2}$$
Notice that the numerator and denominator both have a factor of \(x^2\) in common, as well as a factor of \((x+1)^2\). We can reduce the fraction/rational expression by cancelling these common factors out, reducing our exponents accordingly. This gives us
$$\frac{x^2(x+1)^3(x-2)}{x^3(x+1)^2}=\frac{(x+1)(x-2)}{x}$$
where we still have an \((x+1)\) in the numerator because there were three of those there before, and there were two in the denominator (looking at the powers). So, we could remove/cancel two of them, i.e. \((x+1)^2\). Similarly, we had two \(x\)’s being multiplied in the numerator and three in the denominator being multiplied (looking at powers on \(x\) again tells me how many are being multiplied). So, I can cancel off two \(x\)’s from the numerator and denominator because that’s what they have in common. Because we had three \(x\)’s in the denominator, and we cancelled two of them off, we are left with that single \(x\) in the denominator. All this explains how we arrived at the answer above.
Similar to how we cancel and reduce fractions, sometimes it is necessary to factor the numerator and denominator first before any cancellation can take place. This is illustrated in the next quick example (that you should try!).
Reduce the following rational expression completely by cancelling all factors that both the numerator and denominator have in common. Be sure to factor the numerator and denominator first!
$$\frac{x^2+2x+1}{x^2-2x-3}$$
As this rational expression stands, we cannot cancel anything yet, because we have a fraction of sums/differences. That is, we have a bunch of stuff in the numerator being added, and a bunch of stuff in the denominator being added. In order to cancel, we need multiplication in the numerator and denominator first.
To rewrite this expression to get multiplication of factors in the numerator and denominator (as opposed to addition/subtraction of terms) we must factor the numerator and denominators separately. Doing so gives us
$$\frac{x^2+2x+1}{x^2-2x-3}=\frac{(x+1)(x+1)}{(x+1)(x-3)}$$
Notice that after we’ve factored, we have multiplication of factors in both the numerator and denominator, which is precisely what we need to have if we want to cancel stuff (if possible).
In the top and bottom of this expression we have a common factor of \((x+1)\). Thus, we can cancel this off. The final answer is therefore
$$\frac{(x+1)}{(x-3)}$$
For instance, in the rational expression \(\frac{x(x+1)}{x(x-1)}\) you CAN cancel the \(x\)’s out in front of the numerator and denominator because they are factors in a product (i.e. they are being multiplied in the top and bottom).
However, in the rational expression \(\frac{x+2}{x-1}\) you CANNOT cancel \(x\)’s because you have a fraction of sums/differences; i.e. you have addition or subtraction in the numerator and denominator instead of multiplication.
In summary, you can only cancel things being multiplied in the top and bottom, and CANNOT cancel things being added or subtracted.
Multiplying Rational Expressions
Recall that in order to multiply two fractions of numbers, you multiply across the numerator and denominator to get the final answer. For example,
$$\frac{2}{3}\cdot \frac{5}{7}=\frac{2\cdot 5}{3\cdot 7}=\frac{10}{21}$$
Sometimes you’ll need to cancel and reduce the fraction after multiplying, such as in this example:
$$\begin{align}\frac{5}{4}\cdot \frac{2}{3}&=\frac{5\cdot 2}{4\cdot 3}\\&=\frac{10}{12}\\&=\frac{5}{6}\end{align}$$
It turns out that multiplying rational expressions works exactly the same way.
The above key idea is illustrated in the following example.
Multiply the following rational expressions, and reduce (i.e. simplify) as much as possible.
$$\frac{x^2+2x+1}{x^2-2x-3}\cdot \frac{x-3}{x^2+3x+2}$$
Just like ordinary fractions with numbers, we multiply these fractions straight across the tops and straight across the bottom (no need to “cross multiply” or anything weird here). Doing this gives you
$$\frac{x^2+2x+1}{x^2-2x-3}\cdot \frac{x-3}{x^2+3x+2}= \frac{(x^2+2x+1)\cdot (x-3)}{(x^2-2x-3)\cdot (x^2+3x+2)}$$
Now, this is ugly, so we can try to reduce this fraction by factoring everything in both the numerator and denominator respectively. Specifically, we will factor everything contained in parentheses, namely the \((x^2+2x+1)\), \((x^2-2x-3)\), and \((x^2+3x+2)\) right where you see them in the fraction (i.e. replace these quantities with their factorization). Viz:
$$\frac{(x^2+2x+1)\cdot (x-3)}{(x^2-2x-3)\cdot (x^2+3x+2)}=\frac{(x+1)(x+1)(x-3)}{(x-3)(x+1)(x+2)(x+1)}$$
Now, since we have multiplication of factors in the numerator and denominator, we can cancel out terms that both have in common to reduce the fraction as follows:
$$ \frac{(x+1)(x+1)(x-3)}{(x-3)(x+1)(x+2)(x+1)} =\frac{1}{(x-3)(x+2)}$$
Noting that we end up with the \(1\) in the numerator because we cancelled everything that was in the numerator. Above is the final answer.
Dividing Rational Expressions
There are a multitude of different ways fraction division is taught. Perhaps you are familiar with the methods of “cross multiplication” or “keep-change-flip.” We will briefly review the “keep-change-flip” method.
Suppose you want to divide fractions such as in
$$\frac{2}{3}/\frac{4}{5}$$
To divide these, we will keep the first fraction \(\frac{2}{3}\) as it is, change the division sign to multiplication, and flip the second fraction (and then multiply as usual). Hence the “keep-change-flip!” At the end of this, be sure to reduce your fraction if possible.
$$\frac{2}{3}/\frac{4}{5}=\frac{2}{3}\cdot \frac{5}{4}=\frac{10}{12}=\frac{5}{6}$$
We can use this exact same process with rational expressions as well, because variables represent numbers, and expressions involving variables also represent unknown numbers that we could compute if we knew the value of the variable.
This is exhibited in the following example.
Divide the following rational expressions, and reduce (i.e. simplify) as much as possible.
$$\frac{x^2-5x+6}{x^2+5x+6}/ \frac{x^2-6x+9}{x^2+4x+4}$$
We first perform a “keep-change-flip” on this fraction. that is, we will leave the first fraction alone, change the sign from division to multiplication, then flip the second fraction.
$$\frac{x^2-5x+6}{x^2+5x+6}/ \frac{x^2-6x+9}{x^2+4x+4}= \frac{x^2-5x+6}{x^2+5x+6}\cdot \frac{x^2+4x+4}{x^2-6x+9}$$
Now the rest is just multiplication as usual, then factoring and cancelling as much as possible. Viz:
$$\begin{align}\frac{x^2-5x+6}{x^2+5x+6}\cdot \frac{x^2+4x+4}{x^2-6x+9}&=\frac{(x^2-5x+6)(x^2+4x+4)}{(x^2+5x+6)(x^2-6x+9)}\\&=\frac{(x-2)(x-3)(x+2)(x+2)}{(x+2)(x+3)(x-3)(x-3)}\\&=\frac{(x-2)(x+2)}{(x+3)(x-3)}\end{align}$$
Note that nothing else really can be done in the final answer that you see above, unless you want to re-distribute in both the numerator and denominator. The above answer is likely to make most professors and teachers happy.
\(\frac{x+2}{x}\)
\(\frac{1}{3y^2-1}\)
\(-\frac{2}{x+1}=\frac{-2}{x+1}=\frac{2}{(-1)(x+1)}\)
\(\frac{1}{2}\)
\(\frac{x+1}{x+2}\)
\(\frac{2x+1}{x+1}\)
\(5x\)
\(\frac{x+2}{x+4}\)
\(\frac{x-5}{x-4}\)
\(\frac{25x^2}{3}\)
\(\frac{x+2}{2(x+5)}\)
\(\frac{(x-5)(6x^2-4x+2)}{(x-1)(3x+1)(x+3)}\)