Before moving on, we need to ensure that our algebraic simplification skills are "up to snuff." This section briefly covers some standard factorization techniques necessary for Calculus. These techniques are not just theoretical, but they are the backbone of many Calculus problems. We also review simplifying compound rational expressions and rationalizing (both numerators and denominators). As is the case for most review topics, the level here might seem higher than what you learned in your Intermediate Algebra class, but I assure you that the base content is the same. The material in this section was included from years of experience teaching Calculus and seeing students, semester after semester, struggle to succeed in the subject due to holes in their prerequisite Algebra skills.
Nontrivial Factoring Techniques
Up to this point in Algebra, it is assumed the reader has proficiency with the following factoring techniques:
- factoring out the GCF (greatest common factor) from an algebraic expression
- factoring a polynomial by grouping
- factoring a binomial using the Difference of Squares formula, if possible
- factoring a trinomial having a lead coefficient of 1
- factoring a trinomial having a lead coefficient other than 1 (or 0)
Factoring Using the Sum and Difference of Cubes
We begin this subsection by augmenting our factoring knowledge with the Sum and Difference of Cubes formula.
Theorem: Sum and Difference of Cubes
\[ F^3 \pm L^3 = (F \pm L)(F^2 \mp FL + L^2) \nonumber \]
As with all the factoring techniques you learned in Algebra, this new formula will arise as a required peripheral skill while working on something in Calculus. You could be twenty minutes into a robust problem when you suddenly need to use the Difference of Cubes formula to move forward. Calculus is so dependent on your mastery of these prerequisite topics that it is commonly the case that students fail Calculus not due to their lack of ability to understand concepts in Calculus, but as a result of not knowing the fundamental Algebra that Calculus requires.
Example \(\PageIndex{1}\)
You are working on a complex Calculus problem during an exam and arrive at the following expression.\[ \dfrac{8x^3 - 27 }{3 - 2x} \nonumber \]You are told that \( x \neq \frac{3}{2} \).
You cannot move forward on the problem without somehow removing that denominator (this is something that will be necessary, for reasons to be revealed in Calculus). Simplify the expression to an equivalent expression not involving the denominator \( 3 - 2x \).
- Solution
-
Using the difference of cubes formula, we get\[ \begin{array}{rcl}
\dfrac{8x^3 - 27}{3 - 2x} & = & \dfrac{\left( 2x \right)^2 - (3)^3}{3 - 2x} \\
\\
& = & \dfrac{\left( 2x - 3 \right)\left( 4x^2 + 6x + 9 \right)}{3 - 2x} \\
\\
& = & \dfrac{-\left( 3 - 2x \right)\left( 4x^2 + 6x + 9 \right)}{(3 - 2x)} \\
\\
& = & \dfrac{-\cancelto{1}{\left( 3 - 2x \right)}\left( 4x^2 + 6x + 9 \right)}{\cancelto{1}{(3 - 2x)}} \\
\\
& = & -\left( 4x^2 + 6x + 9 \right) \\
\end{array} \nonumber \]
Factoring the GCF Out of Expressions Involving Negative Rational Exponents
A common factoring technique required in Calculus comes up when the result of some manipulation leads to an expression of the form\[ 3(1 + x)^{1/3} - x(1 + x)^{-2/3}. \nonumber \]It will be necessary to factor this expression. But how? The answer comes from Algebra - factor out the GCF. While we already know this concept from our previous math classes, our approach here will seem slightly higher in level.
The critical concept to remember when factoring out the GCF is that factoring is a division process. That is, when we factor \( 3x - 6 \), we divide out the \( 3 \) from both \( 3x \) and \( -6 \). Hence, \( 3x - 6 = 3\left( \frac{3x}{3} - \frac{6}{3} \right) = 3 (x - 2) \). Of course, most people skip that middle step. With this being said, we can focus on factoring more complex algebraic expressions.
When factoring the GCF out of an algebraic expression, we were taught to factor out the smallest-powered versions of the shared factors. For example, given\[ a^M b^n + a^m b^N, \nonumber \]where \( m \lt M \) and \( n \lt N \), we identify the common factors as \( a \) (to a power) and \( b \) (to a power). Since the smallest power on \( a \) is \( m \) and the smallest power on \( b \) is \( n \), we factor out \( a^m b^n \).\[ a^M b^n + a^m b^N = a^m b^n \left( \dfrac{a^M b^n}{a^m b^n} + \dfrac{a^m b^N}{a^m b^n} \right) = a^m b^n \left( a^{M-m} + b^{N - n} \right). \nonumber \]To memorize this formula is preposterous, so we will focus on the concept rather than the rote memorization.
Example \(\PageIndex{2}\)
It's the third exam in your Differential Calculus class (also known as Calculus I) and you are trying to use Calculus to graph the function\[ f(x) = \dfrac{3}{2} \left( 1 + x \right)^{4/3} + 3 \left( 1 + x \right)^{1/3} + 2. \nonumber \]You have done some great work so far, but at some point you arrive at the expression\[ 3(1 + x)^{1/3} - x(1 + x)^{-2/3}, \nonumber \]and you need to factor this to make the rest of your mathematics easy.
- Solution
-
Both terms in the expression share the factor \( (1 + x) \); however, the second term has the smallest-powered version of this factor. Therefore, we factor out \( \left(1 + x\right)^{-2/3} \).\[ \begin{array}{rclcr}
3(1 + x)^{1/3} - x(1 + x)^{-2/3} & = & (1 + x)^{-2/3} \left(\dfrac{3(1 + x)^{1/3}}{(1 + x)^{-2/3}} - \dfrac{x (1 + x)^{-2/3}}{(1 + x)^{-2/3}} \right) & \quad & \left( \text{factoring out the GCF (which is just division)} \right) \\
\\
& = & (1 + x)^{-2/3} \left(3(1 + x)^{1/3 + 2/3} - x (1 + x)^{-2/3 + 2/3} \right) & \quad & \left( \text{Laws of Exponents} \right) \\
\\
& = & (1 + x)^{-2/3} \left(3(1 + x) - x (1 + x)^{0} \right) & \quad & \left( \text{Arithmetic} \right) \\
\\
& = & (1 + x)^{-2/3} \left(3(1 + x) - x \right) & \quad & \left( \text{Laws of Exponents} \right) \\
\\
& = & (1 + x)^{-2/3} \left(3 + 2x \right) & \quad & \left( \text{distributing and combining like terms} \right) \\
\end{array} \nonumber \]
Knowing When to Distribute
You might have wondered why your Algebra instructors spent months covering factoring techniques. It is a fair question and one that is often answered with a terse, "because you will need it." The truth, however, is much more complex.
It turns out that many functions in Mathematics can be written as polynomials (albeit, very long polynomials), and factoring a polynomial can lead to faster computations in the long run.1 While incredibly simplified, this is, in a nutshell, why factoring is important. So, if the ultimate goal is to get expressions into a completely factored form,2 should we give up on the "undoing" of factoring (called distribution)?
Just as subtraction (the "undoing" of addition) is as important as addition, and division (the "undoing" of multiplication) is as important as multiplication, distribution is still an essential action despite our ultimate desire to factor. If you have an expression that is already factored (for example, an expression like \( \left( 3x - 1 \right)\left( 2x + 5 \right) \)), distributing things out is almost always a bad idea; however, that word "almost" is what we are concerned with. How do we know when distribution is a bad idea? In Calculus, the answer to this question boils down to two scenarios, each involving rational expressions.
There are many times when you are going to multiply the numerator and denominator of a rational expression by a fraction equivalent to \( 1 \). For example, you might be multiplying both the numerator and denominator by the LCD of all fractions within a compound fraction, as seen below.3\[ \dfrac{1/2}{1/3} = \dfrac{1/2}{1/3} \cdot \dfrac{6}{6} = \dfrac{1/\cancelto{1}{2}}{1/\cancelto{1}{3}} \cdot \dfrac{\cancelto{3}{6}}{\cancelto{2}{6}} = \dfrac{3}{2} \nonumber \]In this case, we multiplied the given expression, which was a bit complex-looking, by \( 1 \) in the form of \( \frac{6}{6} \). The result was a much cleaner-looking expression.
Another common need in Calculus is to multiply the numerator and denominator of an expression by a conjugate. For example, given \( \frac{2}{\sqrt{3} + 1} \), we could rewrite the expression by multiplying it by something equivalent to \( 1 \). The "something" we would use is the conjugate of the denominator, which is \( \sqrt{3} - 1 \).\[ \dfrac{2}{\sqrt{3} + 1} = \dfrac{2}{\sqrt{3} + 1} \cdot \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} = \dfrac{2\left( \sqrt{3} - 1 \right)}{\left( \sqrt{3} + 1 \right)\left( \sqrt{3} - 1 \right)} = \dfrac{2 \sqrt{3} - 2}{3 - \sqrt{3} + \sqrt{3} - 1} = \dfrac{2 \sqrt{3} - 2}{2} = \dfrac{\cancelto{1}{2}\left( \sqrt{3} - 1 \right)}{\cancelto{1}{2}} = \sqrt{3} - 1 \nonumber \]Notice, however, that we distributed the \( 2 \) through the numerator after the third equals sign, but then factored it back out later. That was a wasted step! In fact, in these cases, you are often faced with a simple decision - do you distribute everything out? Do you only distribute out the numerator? Do you only distribute out the denominator? Do you leave things factored?
The following is a good rule of thumb to follow:
Rule of Thumb: Knowing When to Distribute
When you have no choice but to multiply the numerator and denominator of a rational expression by either
- a conjugate (to clear radicals), or
- an LCD (to clear a compound fraction),
only perform distribution on the part (numerator or denominator) that caused the need for such an operation. That is, do not use distribution on the non-offending piece.
That Rule of Thumb probably doesn't make sense, so we will clarify it in Example \( \PageIndex{3} \).
Simplifying Compound Fractions
The first half of Calculus will be littered with "fractions containing fractions." These are commonly known as compound fractions. For example, you will inevitably have to deal with a compound fraction of the form\[ \dfrac{(1+x+h)^{-1} - (1 + x)^{-1}}{h}. \nonumber \]When faced with such an expression, you must find the LCD of all fractions contained in the numerator and the denominator. If you multiply both the numerator and denominator of the compound fraction by this LCD, the entire compound fraction will unravel.
Example \(\PageIndex{3}\)
On your first Calculus exam, you end up having to work with the following rational expression.\[ \dfrac{(1+x+h)^{-1} - (1 + x)^{-1}}{h} \nonumber \]However, you cannot move past the first step of the given exam problem without having to simplify the compound rational expression (a fact that will not be explicitly stated).
- Solution
- \[ \begin{array}{rclcr}
\dfrac{(1+x+h)^{-1} - (1 + x)^{-1}}{h} & = & \dfrac{\dfrac{1}{1+x+h} - \dfrac{1}{1 + x}}{h} & \quad & \left( \text{Laws of Exponents} \right) \\
\\
& = & \dfrac{\left( \dfrac{1}{1+x+h} - \dfrac{1}{1 + x} \right)}{h} \cdot \dfrac{(1+x+h)(1+x)}{(1+x+h)(1+x)} & \quad & \left( \text{multiplying numerator and denominator by the LCD} \right) \\
\\
& = & \dfrac{\dfrac{(1+x+h)(1+x)}{1+x+h} - \dfrac{(1+x+h)(1+x)}{1 + x}}{h (1+x+h)(1+x)} & \quad & \left( \text{distributing} \right) \\
\\
& = & \dfrac{\dfrac{\cancel{(1+x+h)}(1+x)}{\cancel{(1+x+h)}} - \dfrac{(1+x+h)\cancel{(1+x)}}{\cancel{(1 + x)}}}{h (1+x+h)(1+x)} & \quad & \left( \text{canceling like factors} \right) \\
\\
& = & \dfrac{(1+x) - (1+x+h)}{h (1+x+h)(1+x)} & \quad & \\
\\
& = & \dfrac{1+x - 1 - x - h}{h (1+x+h)(1+x)} & \quad & \left( \text{distributing} \right) \\
\\
& = & \dfrac{- h}{h (1+x+h)(1+x)} & \quad & \left( \text{combining like terms} \right) \\
\\
& = & \dfrac{- \cancel{h}}{\cancel{h} (1+x+h)(1+x)} & \quad & \left( \text{canceling like factors} \right) \\
\\
& = & \dfrac{- 1}{(1+x+h)(1+x)} & \quad & \left( \text{canceling like factors} \right) \\
\end{array} \nonumber \]
Notice that we did not distribute the denominator in the distribution step on the third line. This was because we were forced to multiply by the LCD because of the fractions in the numerator. Remember the Rule of Thumb: only use distribution on the offending piece (which is the numerator, in the case of Example \( \PageIndex{ 3 } \)).
Rationalizing Numerators and Denominators
As with simplifying compound fractions, you must instinctively know when to rationalize a numerator or denominator throughout Calculus (without explicit instruction from your instructor). The Rule of Thumb on knowing when to distribute (and, more importantly, when not to distribute) will be helpful here.
Example \(\PageIndex{4}\)
Early on in Calculus, you will be working with expressions like\[ -\dfrac{ \dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x + h}} }{h}. \nonumber \]You must immediately identify the need to simplify the compound fraction. Moreover, the \( h \) in the denominator is going to be a problem (the reason for which will be discussed in Calculus). Simplify this expression as much as possible, and make sure your final simplification does not have a factor of \( h \) in the denominator.
- Solution
-
The first step to simplifying this expression is to simplify the compound fraction.\[ \begin{array}{rclcr}
-\dfrac{ \dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x + h}} }{h} & = & -\dfrac{ \left(\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x + h}} \right) }{h} \cdot \dfrac{\sqrt{x} \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}} & \quad & \left( \text{multiplying numerator and denominator by the LCD} \right) \\
\\
& = & -\dfrac{ \sqrt{x + h} - \sqrt{x} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{distributing}^* \right) \\
\\
& = & \dfrac{ -\sqrt{x + h} + \sqrt{x} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{distributing} \right) \\
\\
& = & \dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{Commutative Property of Addition} \right) \\
\end{array} \nonumber \]Now that the compound fraction has been simplified, we are faced with an expression that still has the \( h \) in the denominator. Remember, a hidden, underlying reason will compel us to somehow remove that through creative uses of Algebra. One such use is conjugate multiplication!\[ \begin{array}{rclcr}
\dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & = & \dfrac{ \left( \sqrt{x} - \sqrt{x + h} \right)}{h \sqrt{x} \sqrt{x + h}} \cdot \dfrac{ \left( \sqrt{x} + \sqrt{x + h} \right) }{ \left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{multiplying numerator and denominator by the conjugate of the numerator} \right) \\
\\
& = & \dfrac{ x - (x + h)}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{distributing}^* \right) \\
\\
& = & \dfrac{-h}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{distributing} \right) \\
\\
& = & \dfrac{-\cancelto{1}{h}}{\cancelto{1}{h} \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{canceling like factors} \right) \\
\\
& = & \dfrac{-1}{\sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & & \\
\end{array} \nonumber \]Our final expression is no longer a compound fraction, nor does it have that pesky \( h \) in the denominator!
In the solution to Example \( \PageIndex{ 4 } \), there were a couple of times when we did not (and, technically, could not) distribute the denominator (I marked these with an asterisk). We were forced to multiply by the LCD because of the fractions in the numerator. Remember the Rule of Thumb: only use distribution on the offending piece (the numerator in both cases).
1 Exponential functions, logarithmic functions, and even some trigonometric functions can be represented as infinitely long polynomials!
2 This statement is an overgeneralization. While the factored form of an expression is nice to have, there are times (especially in applications) when working with the unfactored expression is preferable.
3 This type of expression (a fraction involving more fractions) is known as a compound fraction (also known as a complex fraction).
