Mathematics

Quadratic Expressions

First Published: | Last Modified:

Learning Objectives

After completing this chapter, you will be able to

  • factorise quadratic expressions of the form x2 + bx + c;
  • factorise quadratic expressions of the form ax2 + bx + c;;
  • distinguish between a quadratic equation and a quadratic expression;

Mini Glossary

trinomial
is an algebraic expression or equation which has three terms.
coefficient
a value or a number multiplying the variable in an algebraic expression or equation. e.g 8 in 8p; where p is the variable. 8p means 8 × p
factors
a factor of a number x are the set of whole numbers which could divide the number x without any remainder.
2 is a factor of every even number while 1 is a factor for every number
factor of 24 = 1, 2, 4, 6, 8, 12 and 24.

Introduction

Quadratic equation or expression is one which has 2 has its highest power. A quadratic equation or expression is also known as polynomial of the second degree.

A quadratic expressions has a form of ax2 + bx + c
where a, b, c represent any number
a is the coefficient of x2;
b is the coefficient of x and
c is a constant term

A quadratic equation has a form of ax2 + bx + c = 0

Note a ≠ 0
a which is the coefficient of x2 cannot be equal to zero, if the equation or expression remains quadratic.

Table 1: Contrast between a quadratic equation and a quadratic expression
ExpressionsEquations
ax2 + bx + c ax2 + bx + c = 0
ax2 + bx ax2 + bx = 0
ax2 + c ax2 + c = 0
ax2 ax2 = 0
Exercise 1

Tell if the following is an expression or equation.

  1. 5x2 + 15
  2. x3 + 3x + 4
  3. 15x2 = 5
  4. 20 + 6x + x2
Answers
  1. expression
  2. expression
  3. equation
  4. expression

Factorising a trinomial of form x2 + bx + c

A trinomial is an algebraic expression which has three terms. ax2 + bx + c is an example of trinomial because it has three terms ax2, bx and c.
If a = 1, the expression becomes x2 + bx + c, this is a simple trinomial.

Example 1

Factorise x2 + 7x + 6

Solution

a = 1; b = 7; c = 6
Using trial and improvement method, two numbers (factors) which can be multiplied to give 6 that add up to 7
Factors of 6 are 1, 2, 3 and 6

6 = 2 × 3 and 2 + 3 = 5 Wrong ✗
6 = 1 × 6 and 1 + 6 = 7 Correct ✔
The required factors are 6 and 1

The next step is to split the middle term (7x) into 6x and x
Note: 6x + x = 7x
∴ x2 + 7x + 6 = x2 + 6x + x + 6

Factorise by grouping
Find the HCF of the first two term x2 and 6x

x x2 6x
x 6

The HCF of x2 and 6x is x

The HCF of the next two terms x and 6

1 x 6
x 6

The HCF of x and 6 is 1

∴ x2 + 6x + x + 6
= (x2 + 6x) + (x + 6)
= x(x + 6) + 1(x + 6)
= (x + 6)(x + 1)

∴ therefore, x2 + 7x + 6 = (x + 6)(x + 1)

Example 2

Factorise x2 + 3x - 4

Solution

b = 3; c = -4
factors of -4 are -4, -2, -1, 1, 2, 4

Zero is not the factor of any number except zero itself because if we divide a number by zero, the value is not defined.

-4 = -2 × 2 and -2 + 2 = 0 Wrong ✗
-4 = -4 × 1 and -4 + 1 = -3 Wrong ✗
-4 = -1× 4 and -1 + 4 = 3 Correct ✔
The required factors are -1 + 4

The next step is to split the middle term (3x)
∴ x2 + 3x - 4 = x2 - x + 4x -4

Factorise by grouping
Find the HCF of the first two term x2 and -x

x x2 -x
x -1

The HCF of x2 and -x is x

The HCF of the next two terms 4x and -4

4 4x -4
x -1

The HCF of 4x and -4 is 4

∴ x2 - x + 4x - 4
= (x2 - x) + (4x - 4)
= x(x - 1) + 4(x - 1)
= (x - 1)(x + 4)

∴ therefore, x2 + 3x - 4 = (x - 1)(x + 4)

The HCF of the pair of the terms grouped must be equal
x(x - 1) + 4(x - 1)
if not, recheck your workings, there should be a mistake hiding.

Exercise 2

Find the HCF of the following pairs of term:

  1. 12x2 and 28x2
  2. 8x2y2 and 24xy2
Answers
  1. 4x2
  2. 8xy2
Exercise 3

Factorise the following quadratic expression

  1. 6 - 5x + x2
  2. y2 + 9y + 8
Answers
  1. (x - 1) (x - 6)
  2. (y + 1) (y + 8)

Factorisation of a trinomial of form ax2 + bx + c

Steps to take when factorising a trinomials of form ax2 + bx + c

  1. Multiply the first and last terms;
  2. Look for two factors of the product;
  3. Factorise as usual.
Example 3

Factorise 4d2 - 13d - 12

Solution

first-term = 4d2; second-term = -13d; third-term = -12.

Multiply the first and the last terms
4d2 × (-12) = -48d2

Finding the required pair of factor

factors of -48d2 sum
-24d and 2d -22d Wrong ✗
-2d and 24d +22d Wrong ✗
-12d and 4d -8d Wrong ✗
-4d and 12d +8d Wrong ✗
16d and -3d +13d Wrong ✗
-16d and 3d -13d Correct ✔

∴ the required factors are -16d and 3d

Split the middle term
4d2 - 13d - 12
= 4d2 - 16d + 3d -12

Factorise by grouping by using their HCF
4d(d - 4) + 3(d - 4)
= (4d + 3)(d - 4)

Example 4

Factorise -15 + 7a + 2a2

Solution

Re-arrange the expression
2a2 + 7a - 15

first-term = 2a2;
second-term = 7a;
third-term = -15.

Multiply the first and the last terms
2a2 × (-15) = -30a2

Finding the required pair of factor

factors of -30a2 sum
-30a and a -29a Wrong ✗
30a and -a +29a Wrong ✗
-15a and 2a -13a Wrong ✗
15a and -2a +13a Wrong ✗
-10a and 3a -7a Wrong ✗
10a and -3a +7a Correct ✔

∴ the required factors are 10a and -3a

Replace the middle term (7a) with +10a - 3a
2a2 + 7a - 15
= 2a2 + 10a + -3a -15

Factorise by grouping by using their HCF
2a(a + 5) - 3(a + 5)
= (a + 5)(2a - 3)

In the above examples, all possibilities were written until we find the suitable pair of factor.
When making your workings, you don't need to write them all, you can just write down only the suitable pair of factors.

Exercise 4

Factorise the following:

  1. -9 - 14m + 8m2
  2. 3x2 + x - 2

    3. Challenge!
  3. c2 + 9cd - 39d2
    Answer
  1. (4m - 9) (2m - 1)
  2. (3x - 2) (x + 1)
  3. (c - 3d) (c + 12d)

Summary

  • A quadratic expression is an equation which 2 is the highest power of the unknown;
  • To factorise a quadratic expression:
    1. Re-arrange the expression
      ax2 + bx + c
    2. Identify the first, second and third term of the trinomial
    3. Multiply the first and last term, that is, a × c
    4. Find the pair of factors which when multiplied equal to the product of the first and last term and when added equal to the middle term, b
      If the suitable factors are x and y;
      x × y = a × c;
      x + y = b
    5. Replace the middle term with the suitable factors
    6. factorise the resulting expression by grouping and finding the HCF of each grouped terms
    7. See Example 3 and Example 4 for more info.
Next
Quadratic Equation