5.4: Algebraic Simplifications Necessary for Calculus

Math 370 Learning Objectives

• Focus on simplifying expressions (similar in rigor to those encountered in calculus after the quotient or product rule has been applied).

Before we move on to transcendental functions, which dominate the rest of this course, we need to make sure that our algebraic simplification skills are "up to snuff." In this section, we briefly restate some common factorization formulas from your Intermediate Algebra courses. We also review simplifying compound rational expressions and rationalizing (both numerators and denominators). These skills combine to be foundational for all of Calculus and it will be assumed that you, the reader, have mastered such activities before entering into Calculus.

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.

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Nontrivial Factoring Techniques

Recalling the Old Factoring Formulas

Each of the following factoring techniques will be required at many different points throughout Calculus. They will often be needed as required peripheral skills while working on your homework, quizzes, and exams. As a brief summary, the common factoring formulas are listed below with the exception of factoring trinomials (there are too many "specialty methods" that the reader might already know and adding to that list will only cause confusion). Again, it is assumed the reader has mastered these factoring skills prior to entering this course.

Theorem \(\PageIndex{1}\): Difference of Squares

\[ F^2 - L^2 = \left( F - L \right)\left( F + L \right) \label{diffofsquares} \]

Theorem \(\PageIndex{2}\): Sums and Differences of Cubes

\[ F^3 \pm L^3 = (F \pm L)(F^2 \mp FL + L^2) \label{sumdiffcubes} \]

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

\[ \begin{array}{rcl}
\dfrac{8x^3 - 27}{3 - 2x} & = & \dfrac{\left( 2x \right)^2 - (3)^3}{3 - 2x} \\
& = & \dfrac{\left( 2x - 3 \right)\left( 4x^2 + 6x + 9 \right)}{3 - 2x} \\
& = & \dfrac{-\left( 3 - 2x \right)\left( 4x^2 + 6x + 9 \right)}{(3 - 2x)} \\
& = & \dfrac{-\cancel{\left( 3 - 2x \right)}\left( 4x^2 + 6x + 9 \right)}{\cancelto{1}{(3 - 2x)}} \\
& = & -\left( 4x^2 + 6x + 9 \right) \\
\end{array} \nonumber \]

As an aside on non-trivial factoring techniques, we have already seen (several times) the need to identify and factor trinomials that are "quadratic in form." These do creep up in Calculus, so keep an eye out for them. 

Factoring the GCF Out of Expressions Involving Negative Rational Exponents

A common factoring technique required in Calculus comes up when the result of some Calculus manipulations 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 our Elementary Algebra course - factoring out the GCF.

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 \( 3 \) off of 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 \( a^m b^n \).

\[ a^M b^n + a^m b^N = a^m b^n \left( \dfrac{a^M b^n}{a^m b^n} + \dfrac{a^m b^N}{a^m b^n} \right) = a^m b^n \left( a^{M-m} + b^{N - n} \right). \nonumber \]

Of course, to memorize this formula is perposterous, so we will focus on the concept rather than the rote memorization.

Example \(\PageIndex{2}\): Factor Out a Common Expression Having a Negative Rational Exponent

It's the third exam in your Differential Calculus course (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 \( \left(1 + x\right)^{-2/3} \).

\[ \begin{array}{rclcr}
3(1 + x)^{1/3} - x(1 + x)^{-2/3} & = & (1 + x)^{-2/3} \left(\dfrac{3(1 + x)^{1/3}}{(1 + x)^{-2/3}} - \dfrac{x (1 + x)^{-2/3}}{(1 + x)^{-2/3}} \right) & \quad & \left( \text{Algebra: Factoring out the GCF is division} \right) \\
  & = & (1 + x)^{-2/3} \left(3(1 + x)^{1/3 + 2/3} - x (1 + x)^{-2/3 + 2/3} \right) & \quad & \left( \text{Algebra: Laws of Exponents} \right) \\
  & = & (1 + x)^{-2/3} \left(3(1 + x) - x (1 + x)^{0} \right) & \quad & \left( \text{Arithmetic} \right) \\
  & = & (1 + x)^{-2/3} \left(3(1 + x) - x \right) & \quad & \left( \text{Algebra: Laws of Exponents} \right) \\
  & = & (1 + x)^{-2/3} \left(3 + 2x \right) & \quad & \left( \text{Algebra: Distribution and combining like terms} \right) \\
\end{array} \nonumber \]

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Knowing When to Distribute

Traditional Elementary Algebra and Intermediate Algebra courses spend months covering factoring techniques. Why? It turns out that many of our functions we work with in Mathematics can be written as polynomials (yes, including exponential and some trigonometric functions) and factoring a polynomial can lead to faster computations in the long run. Are there times when you don't want or need to factor? Sure! However, if you have an expression that is already factored, then distributing things out is almost always a bad idea.

In Calculus, there are going to be many times when you are going to multiply the numerator and denominator of a given expression by a fraction equivalent to \( 1 \). You might be multiplying both the numerator and denominator by the LCD of all fractions within a compound fraction,¹ or you might be multiplying the numerator and denominator by the conjugate of the numerator, for example.

In these cases, you are 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

When you have no choice but to multiply the numerator and denominator of a rational expression by either

1. a conjugate (to clear radicals), or
2. 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 \( \PageIndex{3} \).

Subsection Footnotes

¹ Also known as a complex fraction.

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Simplifying Compound Fractions

The first half of Calculus will be littered with "fractions containing fractions." These are commonly known as compound fractions. For example, it is inevitable that you will 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 need to find the LCD of all fractions in the numerator and denominator. If you multiply both numerator and denominator of the compound fraction by this LCD, the entire compound fraction will unravel.

Example \(\PageIndex{3}\): Simplify a Compound Rational Expression

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

\[ \begin{array}{rclcr}
\dfrac{(1+x+h)^{-1} - (1 + x)^{-1}}{h} & = & \dfrac{\dfrac{1}{1+x+h} - \dfrac{1}{1 + x}}{h} & \quad & \left( \text{Algebra: Laws of Exponents} \right) \\
  & = & \dfrac{\left( \dfrac{1}{1+x+h} - \dfrac{1}{1 + x} \right)}{h} \cdot \dfrac{(1+x+h)(1+x)}{(1+x+h)(1+x)} & \quad & \left( \text{Algebra: Multiply numerator and denominator by the LCD} \right) \\
  & = & \dfrac{\dfrac{(1+x+h)(1+x)}{1+x+h} - \dfrac{(1+x+h)(1+x)}{1 + x}}{h (1+x+h)(1+x)} & \quad & \left( \text{Algebra: Distribute}^2 \right) \\
  & = & \dfrac{\dfrac{\cancel{(1+x+h)}(1+x)}{\cancel{(1+x+h)}} - \dfrac{(1+x+h)\cancel{(1+x)}}{\cancel{(1 + x)}}}{h (1+x+h)(1+x)} & \quad & \left( \text{Algebra: Cancel like factors} \right) \\
  & = & \dfrac{(1+x) - (1+x+h)}{h (1+x+h)(1+x)} & \quad &  \\
  & = & \dfrac{1+x - 1 - x - h}{h (1+x+h)(1+x)} & \quad & \left( \text{Algebra: Distribution} \right) \\
  & = & \dfrac{- h}{h (1+x+h)(1+x)} & \quad & \left( \text{Algebra: Combine like terms} \right) \\
  & = & \dfrac{- \cancel{h}}{\cancel{h} (1+x+h)(1+x)} & \quad & \left( \text{Algebra: Cancel like factors} \right) \\
  & = & \dfrac{- 1}{(1+x+h)(1+x)} & \quad & \left( \text{Algebra: Cancel like factors} \right) \\
\end{array} \nonumber \]

Subsection Footnotes

² In the distribution step (third line), notice that we did not distribute the denominator. This was because the entire reason we were forced to multiply by the LCD was the fractions in the numerator. Remember the Rule of Thumb: do not use distribution on the non-offending piece.

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Rationalizing Numerators and Denominators

As with simplifying compound fractions, you will need to instinctively know when to rationalize a numerator or denominator at many points throughout Calculus (without explicit instruction to do so from your instructor). My Rule of Thumb on knowing when to distribute and when not to distribute will be helpful here.

Example \(\PageIndex{4}\): Rationalize to Avoid Division By Zero

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 and try to remove \( h \) from the denominator.

Solution

The first step to simplifying this expression is to simplify the compound fraction.

\[ \begin{array}{rclcr}
-\dfrac{ \dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x + h}} }{h} & = & -\dfrac{ \left(\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x + h}} \right) }{h} \cdot \dfrac{\sqrt{x} \sqrt{x + h}}{\sqrt{x} \sqrt{x + h}} & \quad & \left( \text{Algebra: Multiply numerator and denominator by the LCD} \right) \\
  & = & -\dfrac{ \sqrt{x + h} - \sqrt{x} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{Algebra: Distribute}^3 \right) \\
  & = & \dfrac{ -\sqrt{x + h} + \sqrt{x} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{Algebra: Distribute} \right) \\
  & = & \dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & \quad & \left( \text{Arithmetic: Commutative property of addition} \right) \\
\end{array} \nonumber \]

Now that the compound fraction has been simplified, we are faced with an expression that still has the \( h \) in the denominator. Remember, there is a hidden, underlying reason that will compel us to somehow remove that through creative uses of Algebra. One such use is conjugate multiplication!

\[ \begin{array}{rclcr}
\dfrac{ \sqrt{x} - \sqrt{x + h} }{h \sqrt{x} \sqrt{x + h}} & = & \dfrac{ \left( \sqrt{x} - \sqrt{x + h} \right)}{h \sqrt{x} \sqrt{x + h}} \cdot \dfrac{ \left( \sqrt{x} + \sqrt{x + h} \right) }{ \left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{Algebra: Multiply numerator and denominator by the conjugate of the numerator} \right) \\
 & = & \dfrac{ x - (x + h)}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{Algebra: Distribute}^4 \right) \\
 & = & \dfrac{-h}{h \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{Algebra: Distribute} \right) \\
 & = & \dfrac{-\cancel{h}}{\cancel{h} \sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } & \quad & \left( \text{Algebra: Cancel like factors} \right) \\
 & = & \dfrac{-1}{\sqrt{x} \sqrt{x + h}\left( \sqrt{x} + \sqrt{x + h} \right) } &  &  \\
\end{array} \nonumber \]

Our final expression is no longer a compound fraction nor does it have that pesky \( h \) in the denominator!

Subsection Footnotes

³ In this distribution step, notice that we did not (and, technically, could not) distribute the denominator. The entire reason we were forced to multiply by the LCD was the fractions in the numerator. Remember the Rule of Thumb: do not use distribution on the non-offending piece.

⁴ In this distribution step, notice that we did not distribute the denominator. The entire reason we were forced to multiply by the conjugate was the difference of radicals in the numerator. Remember the Rule of Thumb: do not use distribution on the non-offending piece.