Solving Polynomial Equations

Introduction

Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations,
called polynomial equations. These have the general form:

in which x is a variable and are given constants. Also n must be a positive
integer and Examples include and
In this Section you will learn how to factorise some polynomial expressions and solve some polynomial
equations.

Prerequisites

Before starting this Section you should . . .

• be able to solve linear and quadratic
equations

Learning Outcomes

On completion you should be able to . . .

• recognise and solve some polynomial
equations

1. Multiplying polynomials together

Key Point 7

A polynomial expression is one of the form

where are known coefficients (numbers), and x is a variable.
n must be a positive integer.

For example x³ −17x² +54x−8 is a polynomial expression in x. The polynomial may be expressed
in terms of a variable other than x. So, the following are also polynomial expressions:

Note that only non-negative whole number powers of the variable (usually x) are allowed in a polynomial
expression. In this Section you will learn how to factorise simple polynomial expressions and
how to solve some polynomial equations. You will also learn the technique of equating coefficients.
This process is very important when we need to perform calculations involving partial fractions which
will be considered in Section 6.

The degree of a polynomial is the highest power to which the variable is raised. Thus x³ + 6x + 2
has degree 3, t⁶ − 6t⁴ + 2t has degree 6, and 5x + 2 has degree 1.

Let us consider what happens when two polynomials are multiplied together. For example

(x + 1)(3x − 2)

is the product of two first degree polynomials. Expanding the brackets we obtain

(x + 1)(3x − 2) = 3x² + x − 2

which is a second degree polynomial.

In general we can regard a second degree polynomial, or quadratic, as the product of two first degree
polynomials, provided that the quadratic can be factorised. Similarly

(x − 1)(x² + 3x − 7) = x³ + 2x² − 10x + 7

is a third degree, or cubic, polynomial which is thus the product of a linear polynomial and a quadratic
polynomial.

In general we can regard a cubic polynomial as the product of a linear polynomial and a quadratic
polynomial or the product of three linear polynomials. This fact will be important in the following
Section when we come to factorise cubics.

Key Point 8

A cubic expression can always be formulated as a linear expression times a quadratic expression.

If x³ − 17x² + 54x − 8 = (x − 4) × (a polynomial), state the degree of the
undefined polynomial.

Your solution

Answer
second.

(a) If 3x² + 13x + 4 = (x + 4) × (a polynomial), state the degree of the
undefined polynomial.

(b) What is the coefficient of x in this unknown polynomial ?

Your solution

(a)

(b)

Answer
(a) First.

(b) It must be 3 in order to generate the term 3x² when the brackets are removed.

If 2x² + 5x + 2 = (x + 2)× (a polynomial), what must be the coefficient of x in
this unknown polynomial ?

Your solution

Answer
It must be 2 in order to generate the term 2x² when the brackets are removed.

Two quadratic polynomials are multiplied together. What is the degree of the
resulting polynomial?

Your solution

Answer
Fourth degree.

2. Factorising polynomials and equating coefficients

We will consider how we might find the solution to some simple polynomial equations. An important
part of this process is being able to express a complicated polynomial into a product of simpler
polynomials. This involves factorisation.
Factorisation of polynomial expressions can be achieved more easily if one or more of the factors
is already known. This requires a knowledge of the technique of ‘equating coefficients’ which is
illustrated in the following example.

Example 23

Factorise the expression x³−17x²+54x−8 given that one of the factors is (x−4).

Solution

Given that x − 4 is a factor we can write
x³ − 17x² + 54x − 8 = (x − 4) × (a quadratic polynomial)

The polynomial must be quadratic because the expression on the left is cubic and x − 4 is linear.
Suppose we write this quadratic as ax² + bx + c where a, b and c are unknown numbers which we
need to find. Then

x³ − 17x² + 54x − 8 = (x − 4)(ax² + bx + c)

Removing the brackets on the right and collecting like terms together we have
x³ − 17x² + 54x − 8 = ax³ + (b − 4a)x² + (c − 4b)x − 4c

Like terms are those which involve the same power of the variable (x).

Equating coefficients means that we compare the coefficients of each term on the left with the
corresponding term on the right. Thus if we look at the x³ terms on each side we see that x³ = ax³
which implies a must equal 1. Similarly by equating coefficients of x² we find −17 = b−4a With
a = 1 we have −17 = b − 4 so b must equal −13. Finally, equating constant terms we find
−8 = −4c so that c = 2.

As a check we look at the coefficient of x to ensure it is the same on both sides. Now that we know
a = 1, b = −13, c = 2 we can write the polynomial expression as

x³ − 17x² + 54x − 8 = (x − 4)(x² − 13x + 2)

Exercises

Factorise into a quadratic and linear product the given polynomial expressions
1. x³ − 6x² + 11x − 6, given that x − 1 is a factor
2. x³ − 7x − 6, given that x + 2 is a factor
3. 2x³ + 7x² + 7x + 2, given that x + 1 is a factor
4. 3x³ + 7x² − 22x − 8, given that x + 4 is a factor

Answers

1. (x − 1)(x² − 5x + 6), 2. (x + 2)(x² − 2x − 3), 3. (x + 1)(2x² + 5x + 2),
4. (x + 4)(3x² − 5x − 2).