Skip to main content
\(\newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

Section A.2 Algebra Review

Subsection A.2.1 Lines and Linear Functions

Lines are perhaps the most important elementary geometric object. A line captures the idea of following a given direction without turning. In ordinary language, we sometimes think of a line as a smooth curve that we could draw. Mathematically, a line would then be a straight line or a straight curve that does not bend. Algebraically, we can define a line using an equation involving two variables. This section reviews the basic principles of the algebraic properties of lines.

Definition A.2.1 General Equation of a Line

Every line in the \((x,y)\) plane can be described as the set of points \((x,y)\) that satisfy an equation

\begin{equation} Ax + By = C\tag{A.2.1} \end{equation}

where \(A\text{,}\) \(B\) and \(C\) are constants.

There are some special cases that describe horizontal and vertical lines. The \((x,y)\) plane uses \(x\) as the horizontal axis (independent variable) and \(y\) as the vertical axis (dependent variable). So a horizontal line is a line where the dependent variable is constant while a vertical line is a line where the independent variable constant.

Definition A.2.2 Horizontal Line

A horizontal line in the \((x,y)\) plane is the set of points that satisfy an equation

\begin{equation} y=k\tag{A.2.2} \end{equation}

where \(k\) is a constant.

Definition A.2.3 Vertical Line

A vertical line in the \((x,y)\) plane is the set of points that satisfy an equation

\begin{equation} x=h\tag{A.2.3} \end{equation}

where \(h\) is a constant.

All other lines have an equation that involves both variables. We often wish to think of the line as describing the dependent variable as a function of the independent variable. These equations involve the calculation of the slope, which represents a rate (or ratio) of change.

Definition A.2.4 Slope as Rate of Change

Given any two points \((x_1,y_1)\) and \((x_2,y_2)\) on a non-vertical line, the change in the dependent variable \(\Delta y = y_2 - y_1\) is proportional to the change in the independent variable \(\Delta x = x_2 - x_1\text{,}\) written \(\Delta y = m \cdot \Delta x\text{.}\) The proportionality constant \(m\) is called the slope of the line, calculated as the ratio of changes (rate of change)

\begin{equation} m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}.\tag{A.2.4} \end{equation}

Knowing the slope and one point is enough to quickly find an equation of a line.

Definition A.2.5 Point–Slope Equation of Line

Given that a line has slope \(m\) and passes through a point \((x,y)=(h,k)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=m \, (x-h)+k.\tag{A.2.5} \end{equation}

We interpret \(k\) as the starting value for \(y\) and the expression \(\Delta y = m\,(x-h)\) as the change in \(y\) given the change in \(x\text{,}\) \(\Delta x = x-h\text{.}\)

A special case of the point–slope equation of the line occurs when the point is on the \(y\)-axis or, in other words, is a \(y\)-intercept.

Definition A.2.6 Slope–Intercept Equation of Line

Given that a line has slope \(m\) and passes through a \(y\)-intercept \((x,y)=(0,b)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=mx+b.\tag{A.2.6} \end{equation}
Remark A.2.7

Preparatory mathematics courses often emphasize the slope–intercept equation of a line as if it were the most important. However, the point–slope equation is the preferred equation to use in almost every circumstance.

Another special case of the point–slope equation of a line is when we know the slope and the \(x\)-intercept.

Definition A.2.8 Slope and X-Intercept Equation of Line

Given that a line has slope \(m\) and passes through an \(x\)-intercept \((x,y)=(a,0)\text{,}\) every point on the line satisfies the equation

\begin{equation} y=m(x-a).\text{.}\tag{A.2.7} \end{equation}

Subsection A.2.2 Quadratic Polynomials

Definition A.2.9

A quadratic polynomial in a variable \(x\) is an algebraic function that is equal to a formula of the form

\begin{equation} f(x) = ax^2+bx+c,\tag{A.2.8} \end{equation}

where \(a\text{,}\) \(b\) and \(c\) are constants called coefficients.

The graph of a quadratic function, \(y=ax^2+bx+c\text{,}\) is a parabola. Such a parabola has a mirror symmetry across a vertical line that passes through its vertex \(x=-\frac{b}{2a}\text{.}\) Depending on whether the vertex is above, on or below the \(x\)-axis and whether the parabola opens up or down, the graph can cross the \(x\)-axis twice, once or never. The location of these points are called \(x\)-intercepts, roots or zeros of the function. The values of the roots can always be found using the quadratic formula.

Zeros are closely related to factoring. If we know the zeros, then we can immediately rewrite the polynomial in a factored form. On the other hand, if we know the factors, then we can quickly solve for the zeros without using the quadratic formula. This is a consequence of the fundamental properties of numbers in Theorem Theorem A.1.6.

A quadratic polynomial that has complex roots is called irreducible because it can not be rewritten in a factored form involving only real roots.

There are some tricks to factoring that can be useful to know. Factoring is the reverse process of multiplying by distribution, so we start by noticing what happens when you multiply out two simple factors:

\begin{equation*} (x+a)(x+b) = x^2+(a+b)x+ab. \end{equation*}

Notice that the coefficient of \(x\) is the sum \(a+b\) and the constant term is the product \(ab\text{.}\) When trying to factor a quadratic, look for numbers that multiply to give the product term and add to give the coefficient of \(x\text{.}\) This is often a matter of trial and error.

Knowing one root \(x=r\) of a quadratic \(f(x)=ax^2+bx+c\) so that \(f(r)=0\text{,}\) we know that \(x-r\) is a factor. The other factor can be determined easily.

Synthetic division is a procedure that works for quadratics as well as higher order polynomials. This procedure uses a table that starts with the coefficients on the first row. For a more thorough discussion for higher-order polynomials, see Algorithm A.2.19.

Every quadratic can be rewritten in a form \(y=a(x-h)^2+k\) where \((h,k)\) is the vertex of the parabola and \(a\) is the leading coefficient and scaling factor. The process of rewriting a quadratic \(y=ax^2+bx+c\) in this vertex form is called completing the square. It is based on noticing what happens with expanding the square of a binomial, \((x+a)^2 = x^2+2ax+a^2\text{.}\) The strategy involves adding a term to form a perfect square and subtracting the same term to guarantee the expression does not change.

Example A.2.14

Complete the square for \(3x^2-4x+1\text{.}\)

Solution
  1. Group the non-constant terms and factor out the leading coefficient.

    \begin{align*} 3x^2-4x+1 &= (3x^2-4x) + 1\\ & = 3(x^2 - \frac{4}{3}x) + 1 \end{align*}
  2. Recognize the coefficient \(-\frac{4}{3}\) as double \(-\frac{2}{3}\) and use this to complete the square.

    \begin{align*} 3x^2-4x+1 & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x \Big) + 1 \\ & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x + (\frac{-2}{3})^2 - (\frac{-2}{3})^2 \Big) + 1 \\ & = 3 \Big(x^2 - \frac{4}{3} x + \frac{4}{9}\Big) - 3(\frac{4}{9}) + 1\\ & = 3 \Big(x - \frac{2}{3}\Big)^2 - \frac{4}{3} + 1 = 3\Big(x-\frac{2}{3}\Big)^2 - \frac{1}{3}. \end{align*}
  3. Interpret the results as saying \(h=\frac{2}{3}\) (because the completed square is always of the form \((x-h)^2\)) and \(k=-\frac{1}{3}\text{.}\) Thus the vertex of the parabola is at \((\frac{2}{3}, -\frac{1}{3})\text{.}\) The leading coefficient \(a=3\) indicates that the parabola opens up and is three times steeper than the standard parabola \(y=x^2\text{.}\)

Subsection A.2.3 Polynomials

Linear and quadratic formulas are special cases of polynomials. This section gives an overview of principles about polynomials that are likely to appear in calculus. First, we introduce some basic definitions.

Definition A.2.15 Monomials

A monomial is an expression that is a constant multiple of a variable raised to a non-negative integer power, \(a x^k\text{,}\) where \(k=0,1,2,3,\ldots\) and \(a \in \mathbb{R}\text{.}\)

Examples include \(4x^2\) (with \(a=4\) and \(k=2\)), \(\frac{1}{3}x^7\) (with \(a=\frac{1}{3}\) and \(k=7\)) and \(3\) (where \(a=3\) and \(k=0\)). The following are not monomials: \(3\sqrt{x} = 3x^{1/2}\) (since not an integer power) and \(\frac{3}{x^2} = 3x^{-2}\) since the power is a negative integer.

Definition A.2.16 Polynomials

An algebraic expression that can be rewritten as a sum of monomials is called a polynomial. The monomials are called the terms of the polynomial. The monomial with the highest power is called the leading term and its power is called the degree of the polynomial. The constant multiples in the monomials are called coefficients and the coefficient in the leading term is called the leading coefficient.

We usually write a polynomial with terms ordered by decreasing powers, called standard form. An abstract representation of a polynomial with degree \(n\) is written

\begin{equation*} p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \end{equation*}

where the symbols \(a_n, a_{n-1}, \ldots, a_0\) represent the coefficients. A missing term is represented by a coefficient zero.

Example A.2.17

\(x^4-2x^2+3x+1\) is a polynomial with degree \(n=4\text{.}\) The coefficients are \(a_4=1\text{,}\) \(a_3=0\) (since no \(x^3\) term), \(a_2=-2\text{,}\) \(a_1=3\) and \(a_0=1\text{.}\)

\(2x^2(3x+1)(x-2)\) is a polynomial, but must be expanded (multiply out) to find the coefficients.

\begin{align*} 2x^2(3x+1)(x-2) &= 2x^2(3x^2-6x+x-2) \\ &= 2x^2(3x^2-5x-2) \\ & = 6x^4-10x^3-4x^2 \end{align*}

We see that the polynomial has degree \(n=4\) and coefficients \(a_4=6\text{,}\) \(a_3=-10\text{,}\) \(a_2=-4\) and \(a_1=a_0=0\text{.}\)

Every polynomial \(p(x)\) is a function whose domain is all real numbers \((-\infty,\infty)\text{.}\) Values of \(x\) for which \(p(x)=0\) are called zeros or roots of the polynomial. These roots are related to factors.

Synthetic division is an algorithm that can both test if a value \(x=r\) is a root and determine the coefficients of the factored polynomial \(q(x)\) at the same time. Synthetic division for quadratic polynomials (degree \(n=2\)) is a special case of this process, described in Algorithm A.2.12.

Example A.2.20

Use synthetic division with the polynomial \(p(x)=x^3-6x+2\) with the test value \(x=2\) and interpret the result.

Solution

Start by identifying the coefficients. Any missed terms have a coefficient of zero,

\begin{equation*} p(x) = x^3 + 0 x^2 + -6x + 2. \end{equation*}

We start the synthetic division table using the coefficients in the first row.

\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \\ \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \end{matrix} \end{equation*}

We then finish filling the table. To find values in the second row, we use the previous result in the third row and multiply by the test value 2. To find the values in the third row, we add the values in the column. The first value in the second row is always 0. The completed table is shown below.

\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & 2 & 4 & -4 \\ 1 & 2 & -2 & -2 \end{matrix} \end{equation*}

Once the table is complete, we interpret the values in the third row as coefficients and a remainder. The last value is the remainder, \(r=-2\text{,}\) and the other values are the coefficients of a polynomial whose degree is one smaller than the original, in this case \(n-1=2\text{.}\) That is, the quotient polynomial is \(q(x) = x^2 + 2x - 2\text{.}\) The original polynomial can be written

\begin{align*} p(x) &= (x-2)q(x) + r\\ x^3-6x+2 &= (x-2)(x^2+2x-2) + -2 \end{align*}

The non-zero remainder means that \(x-2\) is not a factor and also tells us that \(p(2) = -2\text{.}\)

How do we know which numbers to try? If you have access to a graph of the polynomial, you should use the values for roots that you see. If you do not have access to a graph, then you might be able to use the results of the Integer Root Theorem or Rational Root Theorem so long as all of the coefficients of your polynomial are integers.

Example A.2.22

The polynomial \(p(x) = x^3-6x+2\) has only integer coefficients and a constant coefficient \(a_0=2\text{.}\) The only factors of \(a_0\) are \(\pm 1\) and \(\pm 2\text{.}\) So the Integer Root Theorem guarantees that these four integers are the only four numbers that we need to check if they are roots. A quick test of each of those values (below) shows that \(p(x)\) has no integer roots. (Without the theorem, we wouldn't know how many points we had to check.)

\(x\) \(p(x)\)
1 -3
-1 -5
2 -2
-2 6

The Integer Root Theorem is a special case of the Rational Root Theorem where \(s=1\) (which is always a factor of \(a_n\text{.}\)

Subsection A.2.4 Absolute Value

The absolute value operation takes a number and finds its magnitude (or distance from zero). Because magnitude is a non-negative value and positive and negative pairs are the same distance from zero, we often imagine that the role of absolute value is to remove a negative sign, \(|-3| = 3\text{.}\) However, when a variable is involved, a negative sign means finding the inverse of a value for which we may not know if it is positive or negative. So it is incorrect to say that \(|-x| = x\) (FALSE). The proper definition of absolute value uses a piecewise formula.

Definition A.2.24 Absolute Value
\begin{equation*} |x| = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases} \end{equation*}

As a function, the graph of absolute value \(y=|x|\) gives two lines: \(y=x\) when \(x \ge 0\) and \(y=-x\) when \(x \lt 0\text{.}\)

graph of the absolute value
Figure A.2.25 Graph of the absolute value, \(y=|x|\)

It is sometimes useful to take advantage of an identity between the square root of a square and the absolute value. This is the source of the plus/minus when solving an equation with a square.

Example A.2.27

Solve the equation \(x^2 = 16\text{.}\)

Solution

Applying a square root to both sides of the equation, we then get to use the absolute value identity.

\begin{equation*} \sqrt{x^2} = \sqrt{16} \quad \Leftrightarrow \quad |x| = \sqrt{16} = 4 \end{equation*}

The source of plus/minus is that there are two numbers with magnitude 4,

\begin{equation*} x = \pm 4. \end{equation*}

The absolute value splits nicely with multiplication (and division). However, addition of two values with opposite signs shows that absolute values do not add: \(|3+-4| = |-1| \ne |3| + |-4| = 7\text{.}\) Instead, we have an inequality called the triangle inequality.

The triangle inequality is used to show that the absolute value of a sum (or difference) is bounded by the sum of the magnitudes of the individual terms.

Occasionally we need to apply the triangle in reverse, showing that the absolute value of a sum (or difference) must be bigger than the difference in magnitudes of the parts.

Absolute value and subtraction is often used to describe the distance between two values. For example, the graph \(y=|x-2|\) represents a shift of the graph \(y=|x|\) two units to the right, so that instead of measuring the distance of \(x\) from 0 it measure the distance of \(x\) from 2.

graph of y=abs(x-2)

Note that \(|x+3| = |x--3|\) so that it represents the distance between \(x\) and the value \(-3\text{.}\)