Try the Free Math Solver or Scroll down to Tutorials!

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# 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.

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 ?

(a)

(b)

(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 ?

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?

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