Lab 1: Basic Algebraic Operations

The following material should be read prior to attending lab. You are responsible for preparing for lab so that you don't slow down your group.

In this lab, we are going to review several basic algebra concepts that all of you have seen, but you will likely need to do some reviewing. Since this lab is to help you with calculus, we are going to look at some more difficult problems using the basic concepts.

Exponent Rules

Given real numbers x, y, m, and n, then the following statements are true as long as both sides are defined.

  1. $x^{m} \cdot x^{n} = x^{m \; + \; n} $
  2. $ \frac{x^{m} }{x^{n} } = x^{m \; - \; n} = \frac{1}{x^{n \; - \; m} }$
  3. $(x^{m})^{n} = x^{m \; \cdot \; n} $
  4. $(xy)^m = x^m \cdot y^m$
  5. $ \left(\frac{x}{y}\right)^{\!m} = \frac{x^{m} }{y^{m} } $
  6. $x^{0} = 1,\; x \ne 0$
  7. $ x^{ - m} = \frac{1}{x^{m} }$

You will notice that we have allowed for both m and n to be any real numbers, which includes rational numbers. However, we need to be careful in this situation because this could place restrictions on what values x and y can be. In particular, we are not allowed to take even roots of negative numbers. For example, the function $f(x) = x^{1/3} = \sqrt[3]{x}$ has a domain of all real numbers, but $g(x) = x^{1/4} = \sqrt[4]{x}$ has a domain of all real numbers greater than or equal to zero. This leads us to two more exponent rules, involving radicals.

  1. $\sqrt[n]{x^n} = x $ if n is an odd integer
  2. $\sqrt[n]{x^n} = |x|$ if n is an even integer

Example 1.1

  1. $\sqrt[3]{\left( -3 \right)^{3} } = \sqrt[3]{ -27} = -3 $
  2. $ \sqrt[4]{(-2)^{4}} = \sqrt[4]{16} = 2 = |-2| $

Example 1.2

  1. $ \begin{array}{rcl} \left(\frac{3}{2} \right)^{4} \cdot \left( \frac{1}{3} \right)^{3} & = & \frac{3^{4} }{2^{4} } \cdot \frac{1}{3^{3} } \\ & = & \frac{1}{2^{4} } \cdot \frac{3^{4} }{3^{3} } \\ & = & \frac{1}{2^{4} } \cdot 3^{4 \; - \; 3} \\ & = & \frac{1}{2^{4} } \cdot 3^{1} = \frac{3}{2^{4} } \\ \end{array} $
  2. $ \begin{array}{rcl} \frac{\sqrt[4]{81x^{8} } }{\sqrt[3]{125y^{9} } } & = & \frac{( 3^{4} x^{8})^{1/4} }{(5^{3} y^{9})^{1/3} } \\ & = & \frac{3^{4 \, \cdot \, 1/4} \; \cdot \; x^{8 \, \, \cdot \, 1/4} }{5^{3 \, \cdot \, 1/3} \; \cdot \; y^{9 \, \cdot \, 1/3} } \\ & = & \frac{3^{1} x^{2} }{5^{1} y^{3} } = \frac{3x^{2} }{5y^{3} } \\ \end{array} $

Factoring

First, you need to recall some special factoring formulas. Some of these are as follows.

  1. Difference of squares:
    $x^{2} - y^{2} = (x + y)(x - y)$
  2. Perfect square trinomial:
    $ \Bigg\{ \begin{array}{rcl} x^{2} + 2xy + y^{2} & = & (x + y)^{2} \\ x^{2} - 2xy + y^{2} & = & (x - y)^{2} \end{array}$
  3. Difference of cubes:
    $ x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2}) $
  4. Sum of cubes:
    $x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2}) $

Another factoring scheme that is helpful for deriving the power rule for the derivative is the following.

$ \begin{array}{rcl} x^{4} - y^{4} & = & (x^{2} - y^{2})(x^{2} + y^{2}) \\ & = & (x - y)(x + y)(x^{2} + y^{2}) \\ & = & (x - y)(x^{3} + x^{2}y + xy^{2} + y^{3}) \\ \end{array} $

The first thing to do when factoring is to check to see if there is a common factor among all of the terms. Thus, you should search for the Greatest Common Factor. If a polynomial does not fit one of the special factoring formulas, a technique that can be useful is factoring by grouping. This technique is illustrated using the following video and in the example below.

Insert video here

Example 1.3:

Factoring by grouping
  1. $\begin{array}{rcl} 4a^{3}b - 16ab^{3} & = & 4ab(a^{2} - 4b^{2}) \\ & = & 4ab(a + 2b)(a - 2b) \end{array}$
  2. $ \begin{array}{rcl} x^{3} - 2x^{2} + 3x - 6 & = & x^{2}(x - 2) + 3(x - 2) \\ & = & (x - 2)(x^{2} + 3) \end{array} $

Two more concepts that are important in simplifying algebraic expressions are simplifying complex fractions and multiplying an expression by its conjugate. A complex fraction is a fraction that might look something like this, $\frac{\frac{1}{2} - \frac{1}{x}}{x \, - \, 2}$. We will illustrate how to simplify a complex fraction in the next example. Most of you are familiar with the term conjugate in reference to complex numbers. For example, the complex conjugate of $2 + i$ is $2 - i$. However, we can talk about conjugates in terms of radicals as well. Thus, we can say that the conjugate of $3 - \sqrt{x}$ is $3 + \sqrt{x}$. We often multiply an expression by its conjugate when we are rationalizing the denominator of the expression. However, in calculus sometimes we may need to rationalize a numerator instead of a denominator.

Example 1.4:

Simplify the following expressions.
  1. $ \frac{\left(\frac{1}{2} \, - \, \frac{1}{x} \right)}{x \, - \, 2} $$ = $$ \frac{\left( \frac{x}{2x} \, - \, \frac{2}{2x} \right)}{x \, - \, 2} $Get a common denominator in the numerator.
    $ = $$ \frac{\left( \frac{x \, - \, 2}{2x} \right)}{x \, - \, 2} $
    $ = $$ \frac{x \, - \, 2}{2x} \cdot \frac{1}{x \, - \, 2} $
    $ = $$ \frac{1}{2x} $
  2. $ \frac{4 \, - \, x}{2 \, - \, \sqrt{x} }$$=$$ \frac{(4 \, - \, x)}{( 2 \, - \, \sqrt{x})} \cdot \frac{(2 \, + \, \sqrt{x})}{(2 \, + \, \sqrt{x})}$ Rationalize the denominator by multiplying both numerator and denominator by the conjugate of the denominator.
    $ = $$ \frac{\left( 4 \, - \, x \right)\left( 2 \, + \sqrt{x} \, \right)}{4 \, - \, (\sqrt{x})^{2} } $We multiply by the conjugate so we get a difference of two squares and the middle term of the product disappears.
    $=$$ \frac{(4 \, - \, x)(2 \, + \, \sqrt{x})}{4 \, - \, x} $
    $ = $ $2 + \sqrt{x} $