# Calculus/Algebra

 ← Precalculus Calculus Functions → Algebra

This section is intended to review algebraic manipulation. It is important to understand algebra in order to do calculus. If you have a good knowledge of algebra, you should probably just skim this section to be sure you are familiar with the ideas.

## Rules of arithmetic and algebra

The following laws are true for all a, b, and c, whether a, b, and c are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions.

• Commutative Law: $a+b=b+a \,$.
• Associative Law: $(a+b)+c=a+(b+c)\,$.
• Additive Identity: $a+0=a\,$.
• Additive Inverse: $a+(-a)=0\,$.

### Subtraction

• Definition: $a-b = a+(-b)\,$.

### Multiplication

• Commutative law: $a\times b=b\times a$.
• Associative law: $(a\times b)\times c=a\times (b\times c)\,$.
• Multiplicative identity: $a\times 1=a\,$.
• Multiplicative inverse: $a\times \frac{1}{a}=1$, whenever $a \neq 0\,$
• Distributive law: $a\times (b+c)=(a\times b)+(a\times c)\,$.

### Division

• Definition: $\frac{a}{b}=a\times \frac{1}{b}$, whenever $b \neq 0\,$.

Let's look at an example to see how these rules are used in practice.

 $\frac{(x+2)(x+3)}{x+3}$ = $\left[(x+2)\times (x+3)\right]\times \left( \frac{1}{x+3}\right)$ (from the definition of division) = $(x+2)\times \left[(x+3)\times \left(\frac{1}{x+3} \right) \right]$ (from the associative law of multiplication) = $((x+2)\times (1)),\qquad x \neq -3 \,$ (from multiplicative inverse) = $x+2, \qquad x \neq -3.$ (from multiplicative identity)

Of course, the above is much longer than simply cancelling $x+3$ out in both the numerator and denominator. But, when you are cancelling, you are really just doing the above steps, so it is important to know what the rules are so as to know when you are allowed to cancel. Occasionally people do the following, for instance, which is incorrect:

$\frac{2\times (x + 2)}{2}= \frac{2}{2} \times \frac{x+2}{2}=1 \times \frac{x+2}{2}= \frac{x+2}{2}$.

The correct simplification is

$\frac{2\times (x + 2)}{2}= \left( 2 \times \frac{1}{2} \right) \times (x+2)=1 \times (x+2)=x+2$,

where the number $2$ cancels out in both the numerator and the denominator.

## Interval notation

There are a few different ways that one can express with symbols a specific interval (all the numbers between two numbers). One way is with inequalities. If we wanted to denote the set of all numbers between, say, 2 and 4, we could write "all x satisfying 2<x<4." This excludes the endpoints 2 and 4 because we use < instead of $\leq$. If we wanted to include the endpoints, we would write "all x satisfying $2 \leq x \leq 4$." This includes the endpoints.

Another way to write these intervals would be with interval notation. If we wished to convey "all x satisfying 2<x<4" we would write (2,4). This does not include the endpoints 2 and 4. If we wanted to include the endpoints we would write [2,4]. If we wanted to include 2 and not 4 we would write [2,4); if we wanted to exclude 2 and include 4, we would write (2,4].

Thus, we have the following table:

Endpoint conditions Inequality notation Interval notation
Including both 2 and 4 all x satisfying $2 \leq x \leq 4$
$[2,4] \,\!$
Not including 2 nor 4 all x satisfying $2
$(2,4) \,\!$
Including 2 not 4 all x satisfying $2 \leq x < 4$
$[2,4) \,\!$
Including 4 not 2 all x satisfying $2 < x \leq 4$
$(2,4] \,\!$

In general, we have the following table:

Meaning Interval Notation Set Notation
All values greater than or equal to $a$ and less than or equal to $b$ $\left[a,b\right]$ $\left\{x:a\le x\le b\right\}$
All values greater than $a$ and less than $b$ $\left(a,b\right)$ $\left\{x:a < x < b\right\}$
All values greater than or equal to $a$ and less than $b$ $\left[a,b\right)$ $\left\{x:a\le x < b\right\}$
All values greater than $a$ and less than or equal to $b$ $\left(a,b\right]$ $\left\{x:a < x\le b\right\}$
All values greater than or equal to $a$. $\left[a,\infty\right)$ $\left\{x:x\ge a\right\}$
All values greater than $a$. $\left(a,\infty\right)$ $\left\{x:x > a\right\}$
All values less than or equal to $a$. $\left(-\infty,a\right]$ $\left\{x:x\le a\right\}$
All values less than $a$. $\left(-\infty,a\right)$ $\left\{x:x < a\right\}$
All values. $\left(-\infty,\infty\right)$ $\left\{x: x\in\mathbb{R}\right\}$

Note that $\infty$ and $-\infty$ must always have an exclusive parenthesis rather than an inclusive bracket. This is because $\infty$ is not a number, and therefore cannot be in our set. $\infty$ is really just a symbol that makes things easier to write, like the intervals above.

The interval (a,b) is called an open interval, and the interval [a,b] is called a closed interval.

Intervals are sets and we can use set notation to show relations between values and intervals. If we want to say that a certain value is contained in an interval, we can use the symbol $\in$ to denote this. For example, $2\in[1,3]$. Likewise, the symbol $\notin$ denotes that a certain element is not in an interval. For example $0\notin(0,1)$.

There are a few rules and properties involving exponents and radicals that you'd do well to remember. As a definition we have that if n is a positive integer then $a^n$ denotes n factors of a. That is,

$a^n = a\cdot a \cdot a \cdots a \qquad (n~ \mbox{times}).$

If $a \not= 0$ then we say that $a^0 =1 \,$.

If n is a negative integer then we say that $a^{-n} = \frac{1}{a^n} .$

If we have an exponent that is a fraction then we say that $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m .$

In addition to the previous definitions, the following rules apply:

Rule Example
$a^n \cdot a^m = a^{n+m}$ $3^6 \cdot 3^9 = 3^{15}$
$\frac{a^n}{a^m} = a^{n-m}$ $\frac{x^3}{x^2} = x^{1} = x$
$(a^n)^m = a^{n\cdot m}$ $(x^4)^5 = x^{20} \,\!$
$(ab)^n = a^n b^n \,\!$ $(3x)^5 = 3^5 x^5 \,\!$
$\bigg(\frac{a}{b}\bigg)^n = \frac{a^n}{b^n}$ $\bigg(\frac{7}{3}\bigg)^3 = \frac{7^3}{3^3}.$

## Factoring and roots

Given the expression $x^2 + 3x + 2$, one may ask "what are the values of x that make this expression 0?" If we factor we obtain

$x^2 + 3x + 2 = (x + 2)(x + 1). \,\!$

If x=-1 or -2, then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of x that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial $px^2 + qx + r$ that factors as

$px^2 + qx + r = (ax + c)(bx + d) \,\!$

then we have that x = -c/a and x = -d/b are roots of the original polynomial.

A special case to be on the look out for is the difference of two squares, $a^2 - b^2$. In this case, we are always able to factor as

$a^2 - b^2 = (a+b)(a-b). \,\!$

For example, consider $4x^2 - 9$. On initial inspection we would see that both $4x^2$ and $9$ are squares ($(2x)^2 = 4x^2$ and $3^2 = 9$). Applying the previous rule we have

$4x^2 - 9 = (2x+3)(2x-3). \,\!$

The following is a general result of great utility.

Given any quadratic equation $ax^2+bx+c=0, a\neq0$, all solutions of the equation are given by the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
 Example: Find all the roots of $4x^2+7x-2$ Finding the roots is equivalent to solving the equation $4x^2+7x-2=0$. Applying the quadratic formula with $a=4, b=7, c=-2$, we have: $x=\frac{-7\pm\sqrt{7^2-4(4)(-2)}}{2(4)}$ $x=\frac{-7\pm\sqrt{49+32}}{8}$ $x=\frac{-7\pm\sqrt{81}}{8}$ $x=\frac{-7\pm9}{8}$ $x=\frac{2}{8}, x=\frac{-16}{8}$ $x=\frac{1}{4}, x=-2$

The quadratic formula can also help with factoring, as the next example demonstrates.

 Example: Factor the polynomial $4x^2+7x-2$ We already know from the previous example that the polynomial has roots $x=\frac{1}{4}$ and $x=-2$. Our factorization will take the form $C(x+2)(x-\frac{1}{4})$ All we have to do is set this expression equal to our polynomial and solve for the unknown constant C: $C(x+2)(x-\frac{1}{4})=4x^2+7x-2$ $C(x^2+(-\frac{1}{4}+2)x-\frac{2}{4})=4x^2+7x-2$ $C(x^2+\frac{7}{4}x-\frac{1}{2})=4x^2+7x-2$ You can see that $C=4$ solves the equation. So the factorization is $4x^2+7x-2=4(x+2)(x-\frac{1}{4})=(x+2)(4x-1)$

Note that if $4ac>b^2$ then the roots will not be real numbers.

## Simplifying rational expressions

Consider the two polynomials

$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

and

$q(x) = b_m x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0.$

When we take the quotient of the two we obtain

$\frac{p(x)}{q(x)} = \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0}{b_m x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0}.$

The ratio of two polynomials is called a rational expression. Many times we would like to simplify such a beast. For example, say we are given $\frac{x^2-1}{x+1}.$ We may simplify this in the following way:

$\frac{x^2-1}{x+1} = \frac{(x+1)(x-1)}{x+1} = x-1, \qquad x \neq -1 \,\!$

This is nice because we have obtained something we understand quite well, $x-1$, from something we didn't.

## Formulas of multiplication of polynomials

Here are some formulas that can be quite useful for solving polynomial problems:

$(a+b)^2=a^2+2ab+b^2$
$(a-b)^2=a^2-2ab+b^2$
$(a-b)(a+b)=a^2-b^2$
$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$
$a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$

## Polynomial Long Division

Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:

### Example

 Divide $x^2-2x-15$ (the dividend or numerator) by $x+3$ (the divisor or denominator) Similar to long division of numbers, we set up our problem as follows: $\begin{array}{rl}\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\end{array}$ First we have to answer the question, how many times does $x+3$ go into $x^2$? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in $x$ times. We record this above the leading term of the dividend: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ \end{array}$ , and we multiply $x+3$ by $x$ and write this below the dividend as follows: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ \end{array}$ Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ &\!\!\!\!~~~~~~-5x-15~~~\\ \end{array}$ Now we repeat, treating the bottom line as our new dividend: $\begin{array}{rl}&~~\,x-5\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ &\!\!\!\!~~~~~~-5x-15~~~\\ &\!\!\!\!~~~-\underline{(-5x-15)~~~}\\ &\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\ \end{array}$ In this case we have no remainder.

### Application: Factoring Polynomials

We can use polynomial long division to factor a polynomial if we know one of the factors in advance. For example, suppose we have a polynomial $P(x)$ and we know that $r$ is a root of $P$. If we perform polynomial long division using P(x) as the dividend and $(x-r)$ as the divisor, we will obtain a polynomial $Q(x)$ such that $P(x)=(x-r)Q(x)$, where the degree of $Q$ is one less than the degree of $P$.

### Exercise

1. Factor $x-1$ out of $6x^3-4x^2+3x-5$.

$(x-1)(6x^2+2x+5)$

Solution

### Application: Breaking up a rational function

Similar to the way one can convert an improper fraction into an integer plus a proper fraction, one can convert a rational function $P(x)$ whose numerator $N(x)$ has degree $n$ and whose denominator $D(x)$ has degree $d$ with $n\geq d$ into a polynomial plus a rational function whose numerator has degree $\nu$ and denominator has degree $\delta$ with $\nu<\delta$.

Suppose that $N(x)$ divided by $D(x)$ has quotient $Q(x)$ and remainder $R(x)$. That is

$N(x)=D(x)Q(x)+R(x)$

Dividing both sides by $D(x)$ gives

$\frac{N(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$

$R(x)$ will have degree less than $D(x)$.

#### Example

 Write $\frac{x-1}{x-3}$ as a polynomial plus a rational function with numerator having degree less than the denominator. $\begin{array}{rl}&~~\,1\\ x-3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x-1 \end{array}\\ &\!\!\!\!-\underline{(x-3)~~~}\\ &\!\!\!\!~~~~~~~~~2~~~\\ \end{array}$ so $\frac{x-1}{x-3}=1+\frac{2}{x-3}$
 ← Precalculus Calculus Functions → Algebra