Lesson 2

Functions, and Equations

Functions and their algebra, partial fractions, quadratic equations, logarithms, dividing polynomials, and the binomial expansion.

Taught 04 Sept 2026

Everything here rests on the arithmetic and algebra of the first lesson: the rules of arithmetic, powers and roots, surds, and the difference between an equation and an identity. What is new is the idea of a function, and the algebra that idea makes possible.

Functions

Definition 2.1 (Function).

A function, defined for all numbers, is an association which to each number associates another number. If we denote the function by $f$ , the association is written

f : x \mapsto f(x),

and $f(x)$ is called the value of the function at $x$ , or the image of $x$ under $f$ .

Example 2.2.

The association $f : x \mapsto x^2$ is a function, called the square. The association $g : x \mapsto x + 1$ is another. Their values are read off by substituting:

f(2) = 4, \quad f(3) = 9, \quad f(-1) = 1, \quad f(-2) = 4, \qquad g(1) = 2, \quad g(2) = 3, \quad g(50) = 51 .

Thinking of the value given to $x$ as an input and the corresponding value of the function as an output, we may read $f : r \mapsto 2r + 1$ as “the function which, when we input a value for $r$ , gives as output the value $2r + 1$ “.

Figure 2.1. A function as a rule turning an input into an output. What makes it a function is that each input has exactly one output.

Using $f$ as the symbol for a function, we write $f(x)$ for a function of $x$ and $f(r)$ for a function of $r$ , so the two associations above may equally be written $f(x) = (x+2)^2$ and $f(r) = 2r + 1$ . To represent the value when $x = 1$ we write $f(1)$ . So for $f : x \mapsto (x+2)^2$ ,

f(1) = (1+2)^2 = 9, \qquad f(3) = (3+2)^2 = 25, \qquad f(-2) = (-2+2)^2 = 0 .

Both $x$ and $f(x)$ vary, but $x$ may be given any value while the value of $f(x)$ depends on it. So $x$ is called the independent variable and $f(x)$ the dependent variable.

Example 2.3.

The association which to each number $x$ associates the number $4$ is the constant function with value $4$ . More generally, for a given number $c$ , the association

x \mapsto c \quad \text{for all numbers } x

is the constant function with value $c$ .

Remark (A warning about language).

It is convenient, and slightly incorrect, to write a sentence like “let $f(x)$ be such and such a function”. The trouble is that $x$ is not quantified: what is meant is the function whose value at a number $x$ is such and such. We shall try to avoid the loose phrasing at least while the idea is new.

The examples so far have been given by formulas. Functions may be defined quite arbitrarily, and no formula is required: to describe a function amounts to giving its values at all numbers for which it is defined.

Remark (What the values are).

We have adopted the convention that the values of a function are numbers. It is not a universal convention, and it is a useful one here.

The Algebra of Functions

Functions defined on the same set can be added and multiplied, and when they are they obey the rules of arithmetic from the first lesson.

Let $f$ and $g$ be functions defined on the same set $S$ . Their sum $f + g$ is the function whose value at an element $x$ of $S$ is

(f+g)(x) = f(x) + g(x).

Associativity of addition for numbers gives associativity for functions at once: for any $f$ , $g$ , $h$ defined on $S$ ,

(f + g) + h = f + (g + h),

and commutativity for numbers gives $f + g = g + f$ in the same way.

There is a zero function, whose value at every $x$ in $S$ is $0$ , and which we also denote by $0$ . For every $f$ defined on $S$ ,

f + 0 = 0 + f = f .

If $f$ is defined on $S$ then minus $f$ , written $-f$ , is the function whose value at $x$ is $-f(x)$ .

Example 2.4.

If $f(x) = x^2$ then $(-f)(x) = -x^2$ , so $(-f)(5) = -25$ , and

f + (-f) = 0,

the zero function.

So functions satisfy the same basic rules for addition as numbers do, and the same holds for multiplication. The product $fg$ of two functions defined on $S$ is the function whose value at $x$ is

(fg)(x) = f(x)\,g(x),

and this product is commutative and associative for the same reason as before. If $1$ denotes the constant function with value $1$ for all $x$ in $S$ , then

1f = f \qquad\text{and}\qquad 0f = 0 .

Multiplication distributes over addition. Reading the definitions off one after another,

\begin{aligned} \bigl((f+g)h\bigr)(x) &= (f+g)(x) \cdot h(x) \\ &= \bigl(f(x) + g(x)\bigr)h(x) \\ &= f(x)h(x) + g(x)h(x) \\ &= (fh)(x) + (gh)(x) \\ &= (fh + gh)(x), \end{aligned}

and since the two sides agree at every $x$ in $S$ ,

(f+g)h = fh + gh .

Functions of numbers occur in physical life whenever one quantity is described in terms of another.

Problem 2.1.

Let $f(x) = x^2$ and $g(x) = x + 1$ , both defined for all numbers.

Give the values of $(f+g)(3)$ , $(fg)(3)$ and $(-f)(3)$ .
Write down formulas for $(f+g)(x)$ and $(fg)(x)$ .
Show that $fg$ and $gf$ have the same value at every $x$ .

Problem 2.2.

Find $f(0)$ , $f(1)$ , $f(-2)$ and $f(5)$ for each of the following.

$f : x \mapsto x^2 - 3x$ ;
$f : x \mapsto (x-7)(x+2)$ ;
$f : x \mapsto \dfrac{1}{x^2 + 1}$ .

Polynomials

Definition 2.5 (Polynomial).

If every term of a function has the form $ax^n$ , where $a$ is a constant and $n$ is a non-negative integer, the function is a polynomial. The highest power of $x$ occurring in it is its degree.

Example 2.6.

The functions $x^2$ , $2x$ , $3x^5 - 7x + 6$ and $(x-4)^2$ are polynomials; $\sqrt{x}$ , $\sqrt{x-2}$ and $1/x$ are not, since their exponents are not non-negative integers (Definition 1.29 gives the meaning of $x^{1/2}$ , and it is not a whole power).

The polynomial $5x^6 - 7x^3 + 6x$ has degree $6$ .

Fractional Functions

A fractional function is one of the form

\frac{p(x)}{q(x)}

with $p$ and $q$ polynomials. It is called proper if the degree of the numerator is less than the degree of the denominator, and improper if the degree of the numerator is greater than or equal to that of the denominator.

An improper numerical fraction such as $\tfrac{9}{7}$ may be written as a whole number plus a proper fraction,

\frac{9}{7} = \frac{7+2}{7} = 1 + \frac{2}{7},

and an improper algebraic fraction splits the same way. The fraction $\dfrac{x^2-1}{x^2+1}$ is improper, its numerator and denominator having equal degree, and

\frac{x^2 - 1}{x^2 + 1} = \frac{(x^2 + 1) - 2}{x^2 + 1} = 1 - \frac{2}{x^2+1}.

Problem 2.3.

State the degree of each polynomial, and say which of the following fractional functions are proper and which improper.

$\dfrac{x+3}{(x-2)(x+4)}$ ;
$\dfrac{x^3}{x^2 - 1}$ ;
$\dfrac{2x^2 - 1}{x^2 + x + 1}$ .

Partial Fractions

Two fractions with different denominators may be combined into one. Starting from

f(x) = \frac{2}{x+1} + \frac{x}{x^2+1},

putting both over a common denominator gives

f(x) = \frac{2(x^2+1) + x(x+1)}{(x+1)(x^2+1)} = \frac{3x^2 + x + 2}{(x+1)(x^2+1)} .

It is often useful to run this backwards: to take a fractional function and write it as a sum of separate fractions with simpler denominators. This is called decomposing the function into partial fractions. If the original fraction is proper then the partial fractions are proper too.

The shape of the decomposition is fixed by the factors of the denominator, and the constants in it are found by substitution or by comparing coefficients.

Linear Factors

A linear denominator carries a single constant on top.

Example 2.7 (A proper fraction with linear factors).

Decompose $\dfrac{x+3}{(x-2)(x+4)}$ into partial fractions.

The factors are linear, so we look for constants $A$ and $B$ with

\frac{x+3}{(x-2)(x+4)} = \frac{A}{x-2} + \frac{B}{x+4} = \frac{A(x+4) + B(x-2)}{(x-2)(x+4)} .

The denominators are identical, so the numerators must be as well:

x + 3 = A(x+4) + B(x-2),

and this holds for every value of $x$ . Substituting $x = 2$ kills the $B$ term and gives $5 = 6A$ , so $A = \tfrac{5}{6}$ . Substituting $x = -4$ kills the $A$ term and gives $-1 = -6B$ , so $B = \tfrac{1}{6}$ . Therefore

\frac{x+3}{(x-2)(x+4)} = \frac{5}{6(x-2)} + \frac{1}{6(x+4)} .

A Quadratic Factor

A quadratic denominator that does not factorise carries a linear numerator, so that the partial fraction is still proper.

Example 2.8 (A quadratic factor in the denominator).

Express $\dfrac{x-3}{(x-1)(x^2+1)}$ in partial fractions.

The second factor is quadratic, so we look for

\frac{x-3}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1},

each numerator chosen so that its fraction is proper. Clearing denominators,

x - 3 = A(x^2+1) + (Bx+C)(x-1). \tag{$*$}

Substituting $x = 1$ gives $-2 = 2A$ , so $A = -1$ . There is no value of $x$ making $x^2 + 1$ zero, so $A$ cannot be eliminated the same way, and the remaining constants come from comparing coefficients. On the left the coefficient of $x^2$ is $0$ ; on the right it is $A + B$ . Hence $B = -A = 1$ . Substituting $x = 0$ in $(*)$ gives $-3 = A - C = -1 - C$ , so $C = 2$ . Therefore

\frac{x-3}{(x-1)(x^2+1)} = \frac{-1}{x-1} + \frac{x+2}{x^2+1} .

Either route reaches the constants: substituting values chosen to eliminate terms, or comparing the coefficients of each power. In practice a mixture of the two is quickest.

Repeated Factors

Example 2.9 (A repeated factor).

Express $\dfrac{x-1}{(x+1)(x-2)^2}$ in partial fractions.

Since $(x-2)^2$ is a repeated factor we might think of writing $\dfrac{Bx+C}{(x-2)^2}$ over it. That is not the simplest form. Putting $C = -2B + D$ ,

\frac{Bx + C}{(x-2)^2} = \frac{Bx - 2B + D}{(x-2)^2} = \frac{B(x-2) + D}{(x-2)^2} = \frac{B}{x-2} + \frac{D}{(x-2)^2},

so a repeated linear factor is better written as two fractions. We therefore look for

\frac{x-1}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{D}{(x-2)^2},

that is, $x - 1 = A(x-2)^2 + B(x+1)(x-2) + D(x+1)$ . Substituting $x = 2$ gives $1 = 3D$ , so $D = \tfrac13$ . Substituting $x = -1$ gives $-2 = 9A$ , so $A = -\tfrac29$ . Comparing the coefficients of $x^2$ gives $0 = A + B$ , so $B = \tfrac29$ . Therefore

\frac{x-1}{(x+1)(x-2)^2} = -\frac{2}{9(x+1)} + \frac{2}{9(x-2)} + \frac{1}{3(x-2)^2} .

In general a repeated factor $(ax+b)^2$ in a denominator gives rise to two partial fractions,

\frac{A}{ax+b} \qquad\text{and}\qquad \frac{B}{(ax+b)^2},

and a factor $(ax+b)^3$ gives rise to three, with denominators $ax+b$ , $(ax+b)^2$ and $(ax+b)^3$ .

Improper Fractions

An improper fraction has to be divided out first, leaving a whole part and a proper remainder.

Example 2.10 (An improper fraction).

Express $\dfrac{x^2+1}{(x+1)(x-3)}$ in partial fractions.

The numerator and the denominator both have degree $2$ , so the fraction is improper. Since $(x+1)(x-3) = x^2 - 2x - 3$ ,

\frac{x^2+1}{x^2-2x-3} = \frac{(x^2 - 2x - 3) + (2x + 4)}{x^2 - 2x - 3} = 1 + \frac{2x+4}{(x+1)(x-3)},

and the remaining fraction is proper. Writing

\frac{2x+4}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}, \qquad\text{that is}\qquad 2x + 4 = A(x-3) + B(x+1),

and substituting $x = -1$ gives $2 = -4A$ , so $A = -\tfrac12$ ; substituting $x = 3$ gives $10 = 4B$ , so $B = \tfrac52$ . Therefore

\frac{x^2+1}{(x+1)(x-3)} = 1 - \frac{1}{2(x+1)} + \frac{5}{2(x-3)} .

Problem 2.4.

Express each of the following in partial fractions.

$\dfrac{5}{(x-1)(x+4)}$ ;
$\dfrac{x}{(x-1)(x+1)^2}$ ;
$\dfrac{x^2}{(x-1)(x-2)}$ .

Quadratic Equations

Definition 2.11 (Quadratic Equation).

An equation of the form

ax^2 + bx + c = 0, \qquad a \neq 0,

with $a, b, c$ real numbers, is a quadratic equation. Its solutions are called its roots.

Solving by Factorising

If the left-hand side factorises, the equation is solved by setting each factor to zero, since a product of two numbers is zero only when one of them is.

Example 2.12.

Solve $2x^2 - 7x + 3 = 0$ .

The left-hand side factorises, so the equation becomes

(2x-1)(x-3) = 0,

from which either $2x - 1 = 0$ , giving $x = \tfrac12$ , or $x - 3 = 0$ , giving $x = 3$ .

Losing a Solution

Solutions can be lost if a step is taken carelessly. Consider $2x^2 - 14x = 0$ and two ways of handling it.

Dividing through by $2x$ gives $x - 7 = 0$ , so $x = 7$ .

Factorising instead gives $2x(x-7) = 0$ , so either $2x = 0$ or $x - 7 = 0$ , giving $x = 0$ or $x = 7$ .

The first route lost the solution $x = 0$ , and it lost it because the equation was divided by the common factor $x$ . Dividing by a constant factor is correct and desirable; dividing by a factor containing the unknown throws away every solution that makes the factor zero. This is worth remembering when solving any equation, quadratic or otherwise.

Problem 2.5.

Solve each equation, taking care not to lose a solution.

$x^2 + 5x - 6 = 0$ ;
$x(x-3) = 6$ ;
$x(1-x) = x(2x-1)$ .

Completing the Square

Not every quadratic factorises over the rationals, and one that does not can still be rearranged into a square.

Example 2.13.

Solve $2x^2 - 5x + 1 = 0$ .

Dividing by $2$ gives

x^2 - \tfrac52 x + \tfrac12 = 0, \qquad\text{that is}\qquad x^2 - \tfrac52 x = -\tfrac12 .

Half the coefficient of $x$ is $-\tfrac54$ , so adding $\left(\tfrac54\right)^2 = \tfrac{25}{16}$ to both sides makes the left-hand side a perfect square:

\left(x - \tfrac54\right)^2 = \tfrac{25}{16} - \tfrac12 = \tfrac{17}{16} .

Taking square roots,

x - \tfrac54 = \pm\tfrac{\sqrt{17}}{4}, \qquad\text{so}\qquad x = \frac{5 \pm \sqrt{17}}{4} .

Since $\sqrt{17} = 4.123$ to four significant figures, $x = 2.28$ or $x = 0.22$ .

The Quadratic Formula

Completing the square on the general equation gives a formula for the roots once and for all.

Let $a, b, c$ be real with $a \neq 0$ . The solutions of $ax^2 + bx + c = 0$ are

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

provided $b^2 - 4ac$ is positive or zero. If $b^2 - 4ac$ is negative the equation has no solution in the real numbers.

We obtain this by completing the square as in the example. Solving the equation amounts to solving $ax^2 + bx = -c$ , and dividing by $a$ makes this

x^2 + \frac{b}{a}x = -\frac{c}{a} .

To complete the square on the left we want $x^2 + \frac{b}{a}x = x^2 + 2sx$ , so $s = \dfrac{b}{2a}$ . Adding $s^2 = \dfrac{b^2}{4a^2}$ to both sides gives

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2} .

If $b^2 - 4ac$ is negative the right-hand side is negative and cannot be the square of a real number, so the equation has no real solution. If $b^2 - 4ac$ is positive or zero we may take the square root, and

x + \frac{b}{2a} = \pm\frac{\sqrt{b^2-4ac}}{2a},

which rearranges into the formula.

Remark.

The formula is worth committing to memory. Read it aloud like a line of verse: ” $x$ equals minus $b$ , plus or minus the square root of $b$ squared minus four $ac$ , all over two $a$ .”

The Discriminant

The quantity under the root decides how many roots there are.

Definition 2.14 (Discriminant).

For the equation $ax^2 + bx + c = 0$ the number $b^2 - 4ac$ is called the discriminant.

If the discriminant is positive, $\sqrt{b^2-4ac}$ can be evaluated and the two signs give two different values: the equation has two real distinct roots. If it is zero, both signs give the same value $x = -\tfrac{b}{2a}$ : the equation has one repeated root, also called equal roots. If it is negative, $\sqrt{b^2-4ac}$ has no real value and the equation has no real roots. To summarise, $ax^2 + bx + c = 0$

has two real distinct roots if $b^2 - 4ac > 0$ ;
has equal roots if $b^2 - 4ac = 0$ ;
has no real roots if $b^2 - 4ac < 0$ .

Figure 2.2. The three cases. The roots are the points where the curve meets the horizontal axis, and the discriminant records how many such points there are.

Example 2.15 (Determining the nature of the roots).

For $4x^2 - 7x + 3 = 0$ the discriminant is $(-7)^2 - 4(4)(3) = 49 - 48 = 1 > 0$ , so there are two distinct real roots.
For $x^2 + ax + a^2 = 0$ the discriminant is $a^2 - 4a^2 = -3a^2$ , which is negative for every $a \neq 0$ , so there are no real roots; when $a = 0$ the discriminant is $0$ and the equation $x^2 = 0$ has the repeated root $x = 0$ .
For $x^2 - px - q^2 = 0$ the discriminant is $(-p)^2 - 4(1)(-q^2) = p^2 + 4q^2$ , which is positive unless $p$ and $q$ are both zero, so there are two distinct real roots.

Example 2.16.

Find the value of $k$ for which $2x^2 - kx + 8 = 0$ has equal roots.

Equal roots means the discriminant vanishes, so $(-k)^2 - 4(2)(8) = 0$ , that is $k^2 = 64$ and $k = \pm 8$ .

Problem 2.6.

Determine the nature of the roots of the following, without solving them.

$x^2 - 6x + 9 = 0$ ;
$3x^2 + 4x + 2 = 0$ ;
$x^2 + kx - 1 = 0$ , for a real constant $k$ .

Problem 2.7.

Show that the roots of $ax^2 + (a+b)x + b = 0$ are real for all values of $a$ and $b$ , and find a relationship between $p$ and $q$ for which the roots of $px^2 + qx + 1 = 0$ are equal.

The Sum and the Product of the Roots

The roots of a quadratic can be described without being found.

Let $\alpha$ and $\beta$ be the roots of $ax^2 + bx + c = 0$ . Then $(x-\alpha)(x-\beta) = 0$ has the same solutions, and expanding it gives

x^2 - (\alpha + \beta)x + \alpha\beta = 0 .

Dividing the original equation by $a$ gives

x^2 + \frac{b}{a}x + \frac{c}{a} = 0 .

Both are quadratics in which the coefficient of $x^2$ is $1$ and which are satisfied by exactly $\alpha$ and $\beta$ , so the remaining coefficients agree:

\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} .

An equation may therefore be written as

x^2 - (\text{sum of roots})\,x + (\text{product of roots}) = 0 .

Example 2.17.

The roots of $2x^2 - 3x + 6 = 0$ have sum $-\left(-\tfrac32\right) = \tfrac32$ and product $\tfrac62 = 3$ . Conversely, a quadratic whose roots have sum $7$ and product $10$ may be written $x^2 - 7x + 10 = 0$ .

Example 2.18 (Building a new equation from an old one).

The roots of $2x^2 - 7x + 4 = 0$ are $\alpha$ and $\beta$ . Find $\dfrac1\alpha + \dfrac1\beta$ and $\dfrac{1}{\alpha\beta}$ , and write down the equation whose roots are $\dfrac1\alpha$ and $\dfrac1\beta$ .

From the equation, $\alpha + \beta = \tfrac72$ and $\alpha\beta = 2$ . Expressing the first quantity in terms of these,

\frac1\alpha + \frac1\beta = \frac{\alpha+\beta}{\alpha\beta} = \frac{7/2}{2} = \frac74, \qquad \frac{1}{\alpha\beta} = \frac12 .

The required equation has roots summing to $\tfrac74$ and multiplying to $\tfrac12$ , so it is

x^2 - \tfrac74 x + \tfrac12 = 0, \qquad\text{that is}\qquad 4x^2 - 7x + 2 = 0 .

Remark.

This method works only when each new root depends in the same way on each old one. It applies to roots $\alpha^2$ and $\beta^2$ , or $\tfrac1\alpha$ and $\tfrac1\beta$ ; it does not apply to roots $\alpha + \beta$ and $\alpha - \beta$ .

Example 2.19.

If $\alpha$ and $\beta$ are the roots of $x^2 + 3x - 2 = 0$ , find $\alpha^3 + \beta^3$ and $\alpha^3\beta^3$ , and write down the equation whose roots are $\alpha^3$ and $\beta^3$ .

Here $\alpha + \beta = -3$ and $\alpha\beta = -2$ . Expanding $(\alpha+\beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha+\beta)$ and rearranging,

\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = (-3)^3 - 3(-2)(-3) = -27 - 18 = -45,

and $\alpha^3\beta^3 = (\alpha\beta)^3 = (-2)^3 = -8$ . So the required equation is $x^2 - (-45)x + (-8) = 0$ , that is

x^2 + 45x - 8 = 0 .

Example 2.20.

Find the range of values of $k$ for which $x^2 - 2x - k = 0$ has real roots. If the roots differ by one, find $k$ .

Real roots require $b^2 - 4ac \geqslant 0$ , and $(-2)^2 - 4(1)(-k) = 4 + 4k \geqslant 0$ gives $k \geqslant -1$ .

If the roots differ by one, write them as $\alpha$ and $\alpha + 1$ . Their sum is $2\alpha + 1 = -(-2) = 2$ , so $\alpha = \tfrac12$ . Their product is $\alpha(\alpha+1) = -k$ , so $-k = \tfrac12 \cdot \tfrac32 = \tfrac34$ and $k = -\tfrac34$ .

Problem 2.8.

The roots of $3x^2 + 2x - 1 = 0$ are $\alpha$ and $\beta$ . Without solving the equation, write down and simplify the equation whose roots are

$\dfrac1\alpha$ and $\dfrac1\beta$ ;
$2\alpha$ and $2\beta$ ;
$\alpha^2$ and $\beta^2$ .

Logarithms

A logarithm is another word for an index, read backwards. Since $2^3 = 8$ , we may say that $3$ is the power to which the base $2$ must be raised to obtain $8$ , or that $3$ is the logarithm which, with base $2$ , gives $8$ . This is written

3 = \log_2 8 .

Definition 2.21 (Logarithm).

Let $a > 0$ with $a \neq 1$ , and let $b > 0$ . The logarithm of $b$ to base $a$ , written $\log_a b$ , is the number $x$ for which

a^x = b .

The powers here are those of Definition 1.29.

Example 2.22.

Since $3^2 = 9$ , the number $2$ is the logarithm which with base $3$ gives $9$ , so $\log_3 9 = 2$ . Since $\left(\tfrac15\right)^{-2} = 25$ , we have $\log_{1/5} 25 = -2$ .

The base may be any admissible number. Common logarithms have base $10$ , and it is usual to omit the base and write $\lg$ for them, so that

\lg 5 = 0.6990, \qquad \lg 100 = 2, \qquad \lg 0.01 = -2,

these saying respectively that $10^{0.6990} = 5$ , that $10^2 = 100$ and that $10^{-2} = 0.01$ . Any base other than $10$ must be stated.

The Laws of Logarithms

Three rules carry all the manipulation, and each is a rule for indices in disguise.

Let $\log_a b = x$ and $\log_a c = y$ , so that $a^x = b$ and $a^y = c$ .

For the first rule, $bc = a^x a^y = a^{x+y}$ , so $x + y = \log_a bc$ , that is

\log_a b + \log_a c = \log_a bc . \tag{1}

For the second, $\dfrac{b}{c} = \dfrac{a^x}{a^y} = a^{x-y}$ , so $x - y = \log_a \dfrac{b}{c}$ , that is

\log_a b - \log_a c = \log_a \frac{b}{c} . \tag{2}

For the third, put $z = \log_a b^{\,n}$ , so that $a^z = b^{\,n} = (a^x)^n = a^{nx}$ , whence $z = nx$ , that is

n \log_a b = \log_a b^{\,n} . \tag{3}

The identities $(1)$ , $(2)$ and $(3)$ are the three laws of logarithms. Written with a single symbol for the base they read

\log xy = \log x + \log y, \qquad \log \frac{x}{y} = \log x - \log y, \qquad \log x^{\,y} = y \log x .

Example 2.23.

Using the laws, an expression may be broken into pieces:

\log \frac{a^2}{b^3} = \log a^2 - \log b^3 = 2\log a - 3\log b .

The base is not specified here; it may be anything, provided it is the same in every term.

Conversely, several logarithms may be gathered into one:

\log 100 - 2\log 50 = \log 100 - \log 50^2 = \log \frac{100}{2500} = \log \frac{1}{25} .

Problem 2.9.

Express each of the following in terms of $\log a$ , $\log b$ and $\log c$ .

$\log abc$ ;
$\log \dfrac{ab}{c}$ ;
$\log \dfrac{a^3}{\sqrt{bc}}$ .

Changing the Base

Tables and calculators supply logarithms to base $10$ , so a logarithm to another base has to be converted before it can be evaluated. Take $\log_7 2$ and call it $x$ . Then $7^x = 2$ , and taking common logarithms of both sides,

x \lg 7 = \lg 2, \qquad\text{so}\qquad x = \frac{\lg 2}{\lg 7} = \frac{0.3010}{0.8451} = 0.3562 .

The same argument in general changes base $a$ to base $b$ . If $\log_a c = x$ then $a^x = c$ , and taking logarithms to base $b$ gives $x \log_b a = \log_b c$ , that is

\log_a c = \frac{\log_b c}{\log_b a} .

Taking $c = b$ in this identity, and using $\log_b b = 1$ , gives the special case

\log_a b = \frac{1}{\log_b a} .

Exponential Equations

An exponential equation is one in which the unknown appears as an index. The third law brings the index down where it can be reached.

Example 2.24.

Solve $5^x = 10$ .

Taking common logarithms of both sides and using the third law,

x \lg 5 = \lg 10, \qquad\text{so}\qquad x = \frac{\lg 10}{\lg 5} = \frac{1}{0.6990} = 1.43

to three significant figures.

Taking logarithms is no help when the unknown appears in a sum of powers, since $\log\bigl(2^{2x} + 3 \cdot 2^x\bigr)$ cannot be simplified. What works instead is a substitution which turns the equation into a quadratic.

Example 2.25.

Solve $2^{2x} + 3\left(2^x\right) - 4 = 0$ .

Since $2^{2x} = \left(2^x\right)^2$ , putting $y = 2^x$ makes the equation

y^2 + 3y - 4 = 0, \qquad\text{that is}\qquad (y+4)(y-1) = 0,

so $y = -4$ or $y = 1$ . There is no real $x$ with $2^x = -4$ , since a power of $2$ is positive. From $2^x = 1$ we get $x = 0$ , which is the only solution.

Example 2.26.

Solve the simultaneous equations $xy = 80$ and $\lg x - 2\lg y = 1$ .

By the second and third laws the second equation is $\lg \dfrac{x}{y^2} = 1$ , so $\dfrac{x}{y^2} = 10$ and $x = 10y^2$ . Substituting into the first,

10y^3 = 80, \qquad\text{so}\qquad y^3 = 8, \quad y = 2, \quad x = 40 .

Example 2.27.

Solve $\log_3 x - 4\log_x 3 + 3 = 0$ .

By the special case of the change-of-base identity, $\log_x 3 = \dfrac{1}{\log_3 x}$ , so the equation becomes

\log_3 x - \frac{4}{\log_3 x} + 3 = 0 .

Putting $y = \log_3 x$ and multiplying through by $y$ gives $y^2 + 3y - 4 = 0$ , that is $(y+4)(y-1) = 0$ , so $y = -4$ or $y = 1$ . Hence

x = 3^{-4} = \tfrac{1}{81} \qquad\text{or}\qquad x = 3 .

Problem 2.10.

Solve the following.

$2^x = 5$ ;
$3^{2x} - 4\left(3^x\right) + 3 = 0$ ;
the simultaneous equations $\log_2 y = 2$ and $xy = 8$ .

Polynomials and Division

The polynomials of the first chapter deserve a closer look. A function $f$ defined for all numbers is a polynomial if there are numbers $a_0, a_1, \ldots, a_n$ such that for all $x$

f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 . \tag{1}

Example 2.28.

The function $f$ with $f(x) = 3x^5 - 2x + 1$ is a polynomial, and $f(1) = 3 - 2 + 1 = 2$ . The function $g$ with $g(x) = \tfrac12 x^4 + 3x^2 - x + 5$ is a polynomial, and

g(2) = \tfrac12 \cdot 2^4 + 3 \cdot 2^2 - 2 + 5 = 8 + 12 - 2 + 5 = 23 .

Coefficients and Degree

When a polynomial can be written as in $(1)$ we say it is of degree at most $n$ , and if $a_n \neq 0$ we should like to say it has degree $n$ . Some care is needed before we may. Could the same function also be written

f(x) = b_m x^m + \cdots + b_0

with a different top power? Could it happen, say, that

7x^5 - 5x^4 + 2x + 1 = x^6 - 17x^3 + x + 1

for every number $x$ ? Looking at the two sides settles nothing, and with large coefficients there is no easy test. If the answer were yes then the degree could not be defined at all, since in the example we would not know whether to call it $5$ or $6$ . The answer is no, and we come to it below by way of the roots.

Definition 2.29 (Root).

Let $f$ be a polynomial. A number $c$ with $f(c) = 0$ is called a root of $f$ .

Example 2.30.

Let $f(x) = x^2 - 3x + 2$ . Then $f(1) = 0$ , so $1$ is a root, and $f(2) = 0$ , so $2$ is a root as well.

Let $f(x) = ax^2 + bx + c$ with $a \neq 0$ . If $b^2 - 4ac = 0$ the polynomial has the single root $-\dfrac{b}{2a}$ , and if $b^2 - 4ac > 0$ it has the two distinct roots

\frac{-b + \sqrt{b^2-4ac}}{2a} \qquad\text{and}\qquad \frac{-b - \sqrt{b^2-4ac}}{2a} .

These are the quadratic equations of the previous chapter, said in the language of roots.

Taking Out a Root

Let $f$ be a polynomial of degree at most $n$ and let $c$ be a root of it. Then there is a polynomial $g$ of degree at most $n-1$ with

f(x) = (x-c)\,g(x) \qquad \text{for all } x .

We show this by rewriting $f$ in powers of $x - c$ rather than powers of $x$ . Write $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ and substitute

x = (x-c) + c

for $x$ throughout. Each $k$ -th power $\bigl((x-c)+c\bigr)^k$ expands into a sum of powers of $(x-c)$ multiplied by numbers, so there are numbers $b_0, b_1, \ldots, b_n$ with

f(x) = b_0 + b_1(x-c) + b_2(x-c)^2 + \cdots + b_n(x-c)^n

for all $x$ . Setting $x = c$ makes every term after the first vanish, and $f(c) = 0$ , so $b_0 = 0$ . We may therefore take the factor $(x-c)$ out of what is left:

f(x) = (x-c)\Bigl(b_1 + b_2(x-c) + \cdots + b_n(x-c)^{n-1}\Bigr),

and $g(x) = b_1 + b_2(x-c) + \cdots + b_n(x-c)^{n-1}$ is the polynomial wanted.

Remark (The leading coefficient survives).

Look more closely at that $g$ . Expanding $\bigl((x-c)+c\bigr)^k$ produces one term $(x-c)^k$ and others involving lower powers of $x-c$ , so the highest power of $x-c$ to appear anywhere is the $n$ -th, and it comes only from $\bigl((x-c)+c\bigr)^n$ , carrying the coefficient $a_n$ . Hence $b_n = a_n$ , and $g$ has the form

g(x) = a_n x^{n-1} + \text{lower terms} .

How Many Roots There Can Be

Let $f$ be a polynomial and let $a_0, \ldots, a_n$ be numbers with $a_n \neq 0$ such that $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0$ for all $x$ . Then $f$ has at most $n$ roots.

We show this by taking the roots out one at a time. Let $c_1, c_2, \ldots, c_r$ be distinct roots of $f$ and suppose $r \geqslant n$ . Write $f(x) = (x-c_1)g_1(x)$ with $g_1$ of degree at most $n-1$ . Then

0 = f(c_2) = (c_2 - c_1)\,g_1(c_2),

and $c_2 \neq c_1$ , so $g_1(c_2) = 0$ : that is, $c_2$ is a root of $g_1$ . The same argument applies to $g_1$ , giving $g_1(x) = (x-c_2)g_2(x)$ with $g_2(c_3) = 0$ and $g_2$ of degree at most $n-2$ . Continuing in this way until what is left is constant,

f(x) = (x-c_1)(x-c_2)\cdots(x-c_n)\,c

for all $x$ , and by the remark above $c = a_n \neq 0$ . So if $x$ is not one of $c_1, \ldots, c_n$ then no factor vanishes and $f(x) \neq 0$ . There is therefore no room for a further root.

The Coefficients Are Determined

Suppose a polynomial $f$ can be written both as

f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 \qquad\text{and}\qquad f(x) = b_nx^n + b_{n-1}x^{n-1} + \cdots + b_0 .

Then $a_i = b_i$ for every $i$ .

We show this by subtracting. The polynomial

0 = f(x) - f(x) = (a_n - b_n)x^n + (a_{n-1}-b_{n-1})x^{n-1} + \cdots + (a_0 - b_0)

takes the value $0$ at every number. Put $d_i = a_i - b_i$ and suppose some $d_i$ is non-zero; let $m$ be the largest index for which this happens, so that

0 = d_mx^m + \cdots + d_0

for all $x$ , with $d_m \neq 0$ . A polynomial written this way has at most $m$ roots, and this one has every number for a root. That is impossible, so every $d_i$ is zero.

So there is only one way of writing a polynomial in the form $(1)$ : the numbers $a_n, \ldots, a_0$ are determined by the function. They are called the coefficients of $f$ , the number $a_n$ is the leading coefficient when $a_n \neq 0$ , and $a_0$ is the constant term. The worry raised above is settled, and a polynomial with $a_n \neq 0$ may safely be said to have degree $n$ .

Example 2.31.

Let $f(x) = 4x^5 - 7x^3 + x - 20$ . Its coefficients are

4, \quad 0, \quad -7, \quad 0, \quad 1, \quad -20,

where we have written down the coefficient of every power up to the fifth, those of $x^4$ and $x^2$ being zero. The leading coefficient is $4$ and the constant term is $-20$ , so $f$ has degree $5$ .

Remark (Vanishing at a point, and vanishing everywhere).

It often happens that $f(x) = 0$ for some $x$ , that is, that $f$ has a root. This does not make $f$ the zero polynomial. We call $f$ the zero polynomial, or say it is identically zero, only when $f(x) = 0$ for all numbers $x$ , which happens exactly when all of its coefficients are $0$ . A polynomial with even one non-zero coefficient is not the zero polynomial, however many roots it may have.

Factorising

The previous chapter determines all the roots of a polynomial of degree $2$ . For higher degrees it is much harder. Formulas using radicals exist for degrees $3$ and $4$ , and it is a classical result that no such formula can exist in general from degree $5$ upwards.

For a polynomial with integer coefficients one may look for the rational roots, and a great deal of time is often spent factoring to find them. It is unusual for a polynomial to factorise so obligingly, and in the quadratic case the systematic answer is the formula rather than a search for factors. Two cases of factorising are worth having.

Example 2.32.

Let $f(x) = 3x^2 - 5x + 1$ . Its discriminant is $25 - 12 = 13$ , so by the formula its roots are

c_1 = \frac{5 + \sqrt{13}}{6}, \qquad c_2 = \frac{5 - \sqrt{13}}{6},

and taking each root out in turn factors $f$ into two factors of degree $1$ :

f(x) = 3(x - c_1)(x - c_2) .

The general case runs the same way. If $f(x) = ax^2 + bx + c$ with $a \neq 0$ and $b^2 - 4ac > 0$ , then $f$ has two distinct roots $c_1$ and $c_2$ and

f(x) = a(x-c_1)(x-c_2) .

Example 2.33.

Let $f(x) = x^n - 1$ . Then $f(1) = 1^n - 1 = 0$ , so $1$ is a root and $x - 1$ must be a factor:

x^n - 1 = (x-1)\,g(x)

for some polynomial $g$ . Multiplying out $(x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)$ and watching the terms cancel in pairs identifies $g$ .

Long Division

Dividing one polynomial by another is the exact analogue of dividing one positive integer by another with a remainder, so we recall that first.

Example 2.34 (Dividing integers).

Divide $327$ by $17$ . Set out as at school,

which tells us that $327 = 19 \cdot 17 + 4$ , and $4$ is the remainder. The first digit $1$ was chosen as the largest integer whose product with $17$ is at most $32$ ; multiplying, subtracting and bringing down the $7$ left $157$ ; then $9$ was chosen as the largest integer whose product with $17$ is at most $157$ , and subtracting $153$ left $4$ . Since $4 < 17$ we stop.

In general, for positive integers $n$ and $d$ there are an integer $q \geqslant 0$ and an integer $r$ with $0 \leqslant r < d$ such that $n = qd + r$ . Note that although the procedure is called long division, it uses only multiplication and subtraction.

The same procedure works for polynomials, and is called the Euclidean algorithm: for non-zero polynomials $f$ and $g$ there are polynomials $q$ and $r$ , the degree of $r$ being smaller than the degree of $g$ , such that

f(x) = q(x)\,g(x) + r(x) .

The polynomial $r$ is the remainder.

Example 2.35 (Dividing polynomials).

Let $f(x) = 4x^3 - 3x^2 + x + 2$ and $g(x) = x^2 + 1$ . Laid out as for integers,

              4x -  3
          ------------------------
 x² + 1 ) 4x³ - 3x² +  x + 2
          4x³       + 4x
          ------------------------
              - 3x² - 3x + 2
              - 3x²      - 3
          ------------------------
                    - 3x + 5

so that $q(x) = 4x - 3$ and $r(x) = -3x + 5$ , and

4x^3 - 3x^2 + x + 2 = (4x-3)(x^2+1) + (-3x+5) .

Each step goes as follows. We take $4x$ first because $4x \cdot x^2$ is the term of highest degree in $f$ ; multiplying $4x$ by $x^2 + 1$ gives $4x^3 + 4x$ , which we write underneath with matching powers aligned, and subtracting leaves $-3x^2 - 3x + 2$ . We then take $-3$ , because $(-3) \cdot x^2$ is the term of highest degree in what is left; multiplying gives $-3x^2 - 3$ , and subtracting leaves $-3x + 5$ . This has degree $1$ , smaller than the degree of $g$ , so the computation is finished.

The reason the procedure works is visible in the steps. Writing $f_3 = f$ , the multiplier $4x$ was chosen so that $f_3(x) - 4x(x^2+1)$ has degree $2$ , the term $4x^3$ cancelling; call the result $f_2$ . The multiplier $-3$ was then chosen so that $f_2(x) - (-3)(x^2+1)$ has degree $1$ , the term $-3x^2$ cancelling. Putting the two together,

f_3(x) - (4x-3)\,g(x) = -3x + 5,

which is exactly $f = qg + r$ .

Example 2.36.

Let $f(x) = 2x^4 - 3x^2 + 1$ and $g(x) = x^2 - x + 3$ . The same pattern gives

                 2x² + 2x -  7
          ---------------------------------
 x² - x + 3 ) 2x⁴       - 3x²      + 1
              2x⁴ - 2x³ + 6x²
          ---------------------------------
                  + 2x³ - 9x²      + 1
                    2x³ - 2x² + 6x
          ---------------------------------
                        - 7x² - 6x + 1
                        - 7x² + 7x - 21
          ---------------------------------
                              - 13x + 22

so $q(x) = 2x^2 + 2x - 7$ and $r(x) = -13x + 22$ .

Remark.

Long division gives a second route to the factor taken out earlier. Dividing $f$ by $x - c$ ,

f(x) = q(x)(x-c) + r(x)

where $r$ has degree smaller than $1$ and so is a constant, say $a$ . Evaluating both sides at $x = c$ gives $0 = 0 + a$ , so $a = 0$ : the remainder vanishes and $x - c$ is a factor.

The Remainder on Division

That last observation is worth stating on its own, since it evaluates a remainder without any division at all.

If $f$ is a polynomial which, on division by $x - a$ , gives quotient $Q(x)$ and remainder $R$ , then

f(x) = (x-a)\,Q(x) + R,

and substituting $a$ for $x$ makes the first term vanish and gives

R = f(a) .

So when $f(x)$ is divided by $x - a$ the remainder is $f(a)$ .

Example 2.37.

Divide $f(x) = x^3 - 7x^2 + 6x - 2$ by $x - 2$ . Long division gives quotient $x^2 - 5x - 4$ and remainder $-10$ , that is

x^3 - 7x^2 + 6x - 2 = (x-2)(x^2 - 5x - 4) - 10 .

Substituting $x = 2$ in this identity kills the quotient and leaves $f(2) = -10$ , as the rule promises.

Example 2.38.

The remainder when $x^3 - 2x^2 + 6$ is divided by $x + 3$ is

f(-3) = (-3)^3 - 2(-3)^2 + 6 = -27 - 18 + 6 = -39 .

The remainder when $6x^2 - 7x + 2$ is divided by $2x - 1$ is $f\left(\tfrac12\right) = \tfrac64 - \tfrac72 + 2 = 0$ .

Remark.

This gives the remainder only. If the quotient is wanted, the long division must still be carried out.

Recognising a Factor

If $x - a$ is a factor of $f(x)$ there is no remainder, so $R = 0$ and $f(a) = 0$ ; and conversely, as we saw, a root produces a factor. So

x - a \text{ is a factor of } f(x) \qquad\text{exactly when}\qquad f(a) = 0 .

This is what makes polynomials of degree greater than $2$ tractable by hand.

Example 2.39 (Factorising a quartic).

Factorise $x^4 - 3x^3 + 4x^2 - 8$ completely.

Put $f(x) = x^4 - 3x^3 + 4x^2 - 8$ . The constant term is $-8$ , whose divisors are $\pm1, \pm2, \pm4, \pm8$ , so those are the only whole numbers worth trying. Now

f(1) = 1 - 3 + 4 - 8 = -6 \neq 0,

so $x - 1$ is not a factor, while

f(-1) = 1 + 3 + 4 - 8 = 0,

so $x + 1$ is a factor. Taking it out, by inspection or by long division,

x^4 - 3x^3 + 4x^2 - 8 = (x+1)\bigl(x^3 - 4x^2 + 8x - 8\bigr) .

Now put $g(x) = x^3 - 4x^2 + 8x - 8$ . Since $g(-1) = -1 - 4 - 8 - 8 \neq 0$ , the factor $x+1$ does not repeat; since $g(2) = 8 - 16 + 16 - 8 = 0$ , the factor $x - 2$ appears, and

x^3 - 4x^2 + 8x - 8 = (x-2)\bigl(x^2 - 2x + 4\bigr) .

The discriminant of $x^2 - 2x + 4$ is $4 - 16 = -12 < 0$ , so it has no real roots and no linear factors. Therefore

x^4 - 3x^3 + 4x^2 - 8 = (x+1)(x-2)\bigl(x^2 - 2x + 4\bigr) .

Note that after the first factor has been taken out it must be tried again on what remains, since repeated factors are common.

The Factors of $a^3 \pm b^3$

Read as a polynomial in $a$ , the expression $a^3 - b^3$ vanishes when $a = b$ , so $a - b$ is a factor of it; and $a^3 + b^3$ vanishes when $a = -b$ , so $a + b$ is a factor of that. Carrying out the division gives

a^3 - b^3 = (a-b)\bigl(a^2 + ab + b^2\bigr), \qquad a^3 + b^3 = (a+b)\bigl(a^2 - ab + b^2\bigr),

as multiplying out confirms.

Problem 2.11.

Determine whether each linear function is a factor of the polynomial beside it.

$x - 1$ and $x^3 - 7x + 6$ ;
$x - 2$ and $x^3 - 6x^2 + 6x - 2$ ;
$x + a$ and $x^3 + ax^2 - ax - a^2$ .

Problem 2.12.

Factorise as far as possible.

$x^3 + 2x^2 - x - 2$ ;
$27x^3 - 1$ ;
$x^3 + a^3$ .

Problem 2.13.

If $x^2 - 7x + a$ leaves remainder $1$ on division by $x + 1$ , find $a$ . If $x - 2$ is a factor of $ax^3 - 12x + 4$ , find $a$ .

Binomial Expansions

Definition 2.40 (Binomial).

A binomial is the sum, or difference, of two terms. So $a+b$ , $2x+3y$ and $p - r$ are binomials.

Expanding a power of a binomial by repeated multiplication is quick enough for a square,

(x+y)^3 = (x+y)(x+y)^2 = (x^2 + 2xy + y^2)(x+y) = x^3 + 3x^2y + 3xy^2 + y^3,

and tedious from the cube upwards. There is a much faster route.

Pascal’s Triangle

Set out the first few powers of $1+x$ :

\begin{aligned} (1+x)^0 &= 1, \\ (1+x)^1 &= 1 + x, \\ (1+x)^2 &= 1 + 2x + x^2, \\ (1+x)^3 &= 1 + 3x + 3x^2 + x^3, \\ (1+x)^4 &= 1 + 4x + 6x^2 + 4x^3 + x^4 . \end{aligned}

Each expansion is the one above it multiplied by $(1+x)$ , which is to say the one above it added to the one above it shifted up a power. So the coefficient of any power of $x$ is the sum of the coefficients of the same power and the preceding power in the line before. Writing the coefficients alone as a triangular array, and continuing the rule as far as we please, gives Pascal’s triangle.

Figure 2.3. Pascal’s triangle. Each entry is the sum of the two above it, as the arrows show for

10 + 10 = 20

Row five of the triangle therefore gives

(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 .

Expanding a Power

Once the coefficients are known, any binomial power follows by substitution.

Example 2.41.

Expand $(1+3y)^3$ .

From the triangle, $(1+x)^3 = 1 + 3x + 3x^2 + x^3$ . Replacing $x$ by $3y$ ,

(1+3y)^3 = 1 + 3(3y) + 3(3y)^2 + (3y)^3 = 1 + 9y + 27y^2 + 27y^3 .

Example 2.42.

Expand $(a+b)^4$ .

From the triangle, $(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$ . Writing

(a+b)^4 = a^4\left(1 + \frac{b}{a}\right)^4

and replacing $x$ by $\dfrac{b}{a}$ gives

(a+b)^4 = a^4\left(1 + 4\frac{b}{a} + 6\frac{b^2}{a^2} + 4\frac{b^3}{a^3} + \frac{b^4}{a^4}\right) = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 .

Three things are worth noticing in that expansion.

The powers of $a$ and $b$ in each term add up to $4$ .
As the powers of $a$ decrease, the powers of $b$ increase.
The numerical coefficients are those of $(1+x)^4$ .

With these, an expansion may be written straight down from the triangle. Row six gives

(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6 .

Example 2.43.

Expand $(2x-3y)^3$ .

Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = 2x$ and $b = -3y$ ,

(2x-3y)^3 = (2x)^3 + 3(2x)^2(-3y) + 3(2x)(-3y)^2 + (-3y)^3 = 8x^3 - 36x^2y + 54xy^2 - 27y^3 .

Problem 2.14.

Expand each of the following.

$(1+2x)^4$ ;
$(1-y)^5$ ;
$(2x - 1)^3$ .

Problem 2.15.

By expanding $(1+0.01)^3$ , evaluate $(1.01)^3$ exactly without a calculator. Then evaluate $(2.1)^3$ the same way.

Exercises

Exercise 2.1.

Find the stated values.

$f(x) = 2x + x^2 - 5$ : find $f(1)$ and $f(-1)$ .
$f(x) = \sqrt[4]{x}$ : for which numbers does this formula define a function, and what is $f(16)$ ?
$f(x) = \dfrac{1}{x-2}$ : for which numbers does this formula define a function, and what is $f(0)$ ?

Exercise 2.2.

Write $|x| = \sqrt{x^2}$ , the positive square root, so that $|x|$ is $x$ when $x$ is positive or zero and $-x$ when $x$ is negative.

Find $|1|$ , $|-3|$ and $\left|-\tfrac23\right|$ .
Let $g(x) = x + |x|$ . Find $g(2)$ , $g(-4)$ and $g(-5)$ , and describe in a sentence what $g$ does to a number.

Exercise 2.3.

A function defined for all numbers is even if $f(x) = f(-x)$ for every $x$ , and odd if $f(x) = -f(-x)$ for every $x$ . Determine which of the following are even, which are odd, and which are neither.

$f(x) = x$ ;
$f(x) = x^2$ ;
$f(x) = x^3$ ;
$f(x) = 1/x$ for $x \neq 0$ , with $f(0) = 0$ .

Exercise 2.4.

Show that any function $f$ defined for all numbers can be written as the sum of an even function and an odd function, by considering

\frac{f(x) + f(-x)}{2} \qquad\text{and}\qquad \frac{f(x) - f(-x)}{2} .

Exercise 2.5.

Let $f(x) = x^2 - 1$ and $g(x) = x + 1$ , both defined for all numbers.

Write down formulas for $(f+g)(x)$ and $(fg)(x)$ .
Find $(f + g)(2)$ and $(fg)(-1)$ .
Show that $f$ has a root which $g$ also has, and factorise $f$ to display it.

Exercise 2.6.

Express each of the following in partial fractions.

$\dfrac{7}{(x-3)(x+4)}$ ;
$\dfrac{2x+1}{(x-1)^2}$ ;
$\dfrac{x^2 + x}{(x+2)(x-1)}$ .

Exercise 2.7.

Solve each equation, giving surds in exact form.

$3x^2 - 7x + 2 = 0$ ;
$x^2 - 4x + 1 = 0$ ;
$x(x+2) = 3x + 6$ .

Exercise 2.8.

Determine the nature of the roots of each equation without solving it, and in the last case find the values of $k$ .

$5x^2 - 2x + 1 = 0$ ;
$x^2 + 2\sqrt{3}\,x + 3 = 0$ ;
$kx^2 + 4x + k = 0$ has equal roots.

Exercise 2.9.

The roots of $x^2 - 5x + 2 = 0$ are $\alpha$ and $\beta$ . Without solving the equation, find

$\alpha + \beta$ and $\alpha\beta$ ;
$\alpha^2 + \beta^2$ ;
the equation whose roots are $\alpha + 1$ and $\beta + 1$ .

Exercise 2.10.

Express in terms of $\log a$ , $\log b$ and $\log c$ .

$\log \dfrac{a^2b}{c}$ ;
$\log \sqrt{ab}$ ;
$\log \dfrac{1}{a^3}$ .

Exercise 2.11.

Solve the following.

$3^x = 20$ , to three significant figures;
$4^x - 5\left(2^x\right) + 4 = 0$ ;
$\log_2 x + \log_2 (x-2) = 3$ .

Exercise 2.12.

Show that $\log_a b \times \log_b a = 1$ , and use a change of base to evaluate $\log_5 12$ to three significant figures.

Exercise 2.13.

Let $f(x) = 2x^3 - x^2 - 5x - 2$ .

Find the remainder when $f(x)$ is divided by $x - 1$ .
Show that $x + 1$ is a factor of $f(x)$ , and factorise $f(x)$ completely.
Write down all the roots of $f$ .

Exercise 2.14.

Carry out the long division of $f$ by $g$ , giving the quotient and the remainder.

$f(x) = x^3 + 2x^2 - x + 5$ and $g(x) = x + 3$ ;
$f(x) = 3x^4 - x^2 + 2$ and $g(x) = x^2 + x - 1$ .

Exercise 2.15.

Factorise as far as possible, using the factors of $a^3 \pm b^3$ where they help.

$8x^3 - 27$ ;
$x^3 + 64$ ;
$x^4 - 16$ .

Exercise 2.16.

Expand the following.

$(1 + 2x)^4$ ;
$(x - 2y)^3$ ;
$\left(a + \tfrac1a\right)^4$ , simplifying each term.

Exercise 2.17.

Use Pascal’s triangle to write down the expansion of $(1+\sqrt{2})^4$ , and simplify it to the form $p + q\sqrt{2}$ with $p$ and $q$ whole numbers. Do the same for $(1-\sqrt{2})^4$ , and add your two answers.

Check Yourself

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Functions

The Algebra of Functions

Polynomials

Fractional Functions

Partial Fractions

Linear Factors

A Quadratic Factor

Repeated Factors

Improper Fractions

Quadratic Equations

Solving by Factorising

Losing a Solution

Completing the Square

The Quadratic Formula

The Discriminant

The Sum and the Product of the Roots

Logarithms

The Laws of Logarithms

Changing the Base

Exponential Equations

Polynomials and Division

Coefficients and Degree

Taking Out a Root

How Many Roots There Can Be

The Coefficients Are Determined

Factorising

Long Division

The Remainder on Division

Recognising a Factor

The Factors of a3±b3a^3 \pm b^3a3±b3

Binomial Expansions

Pascal’s Triangle

Expanding a Power

Exercises

Check Yourself

The Factors of $a^3 \pm b^3$