Page 80Page 80Page 80
SOLVING CUBIC EQUATIONS
A cubic expression is an expression of the form ax
3
+ bx
2
+cx + d. The following
are all examples of expressions we will be working with:
2x
3
– 16, x
3
– 2x
2
– 3x, x
3
+ 4x
2
– 16, 2x
3
+ x – 3.
Remember that some quadratic expressions can be factorised into two linear
factors:
e.g. 2x
2
– 3x + 1 = (2x – 1)(x – 1)
Now, a cubic expression may be factorised into
(i) a linear factor and a quadratic factor or (ii) three linear factors.
For example, you can easily verify, by multiplying out the right hand side that:
(i) x
3
– 8 = (x – 2)(x
2
+ 2x + 4)
(ii) 4x
3
– 4x
2
– x + 1 = (x – 1)(2x – 1)(2x + 1)
There are three types of factorisation methods we will consider:
• Common factor
• Grouping terms
• Factor theorem
Type 1 - Common factor
In this type there would be no constant term.
Example 1
Solve for x: x
3
+ 5x
2
– 14x = 0
Solution
x(x
2
+ 5x – 14) = 0
\ x(x + 7)(x – 2) = 0
\ x = 0, x = 2, x = –7
Type 2 - Grouping terms
With this type, we must have all four terms of the cubic expression. We then
pair terms with a common factor and see if a common bracket emerges.
Example 2
Solve for x: x
3
+ 2x
2
– 9x – 18 = 0
Solution:
(x
3
+ 2x
2
) – (9x + 18) = 0
\ x
2
(x + 2) – 9(x + 2) = 0
5
LESSON
5
LESSON
LinearLinearLinearLinearQuadraticQuadratic
QuadraticQuadraticLinearLinear
LinearLinearLinearLinear LinearLinear
ExampleExample
SolutionSolution
ExampleExample
SolutionSolution
Lesson 1 | Algebra
Page 1
2009
Page 81
\ (x + 2)(x
2
– 9) = 0
\ (x + 2)(x – 3)(x + 3) = 0
\ x = –2, x = 3, x = –3
Type 3 - Using the factor theorem
N.B. If (x – a) is a factor of the cubic expression, then f(a) = 0.
So, we substitute in values of x = ±1, ±2 … etc until we find a value which makes
the expression equal to 0.
Example 3
Solve for x: x
3
– 5x + 2 = 0
Solution
Try x = 1: 1
3
– 5(1) + 2 = –2
Try x = –1: (–1)
3
– 5(–1) + 2 = 6
Try x = 2: 2
3
– 5(2) + 2 = 0 \ (x – 2) is a factor
\ (x – 2)(quadratic) = x
3
– 5x + 2
(x – 2)(x
2
+ kx – 1) = x
3
– 5x + 2 by inspection.
Compare x terms on LHS and RHS: –5x = –x – 2kx
\ –5 = –1 – 2k
\ k = 2
\ x
3
– 5x + 2 = (x – 2)(x
2
+ 2x – 1) = 0
x = 2 or x = –1 ±
√
_
2 (using the quadratic formula)
Example 4
Solve for x: 2x
3
– 3x
2
– 8x – 3 = 0
Solution
Try x = 1: 2(1)
3
– 3(1)
2
– 8(1) – 3 = –12
Try x = –1: 2(–1)
3
– 3(1)
2
– 8(–1) – 3 = 0 \ (x + 1) is a factor
\ (x + 1)(2x
2
+ kx – 3) = 2x
3
– 3x
2
– 8x – 3
Compare x
2
terms on both sides:
(N.B. It does not matter whether you compare x
2
or x terms)
–3x
2
= 2x
2
+ kx
2
\ –3 = 2 + k
\ k = –5
\ (x + 1)(2x
2
– 5x – 3) = 2x
3
– 3x
2
– 8x – 3 = 0
∴ (x + 1)(2x + 1)(x – 3) = 0
\ x = –1, x = –
1
_
2
, x = 3
ExampleExample
SolutionSolution
Alternatively, you can
use long division to get
the factors of x
3
– 5x + 2
Alternatively, you can
use long division to get
the factors of x
3
– 5x + 2
ExampleExample
SolutionSolution
Page 82Page 82
Activity 1
Solve for x:
1. 2x
3
– x
2
– x = 0
2. x
3
– x = 0
3.
2
_
3
x
3
– 18 = 0
4. x
3
+ 3x
2
– 4x – 12 = 0
5. x
3
– 3x – 2 = 0
6. 2x
3
+ 5x
2
– 14x – 8 = 0
7. x
3
+ 7x
2
– 36 = 0
ActivityActivity
Lesson 1 | Algebra
Page 1
2009
Page 83
8. 4x
3
+ 12x
2
+ 9x + 2 = 0
9. x
3
− 2x
2
− 4x + 3 = 0
Activity 2
1. Given that: f(x) = 6x
3
– 37x
2
+ 5x + 6 and f(6) = 0, solve for x, if f(x) = 0
2. Solve for x and y if:
y = x
3
+ 9x
2
+ 26x + 16 and y – 3x = 1
3. In the diagram: f(x) = x
3
and g(x) = –3x
2
+ x + 3
x
y
A
B
g
ƒ
ActivityActivity
Page 84Page 84
Determine the coordinates of A and B, the points of intersection of f and g.
4. In the diagram: f(x) = x
2
– 7 and g(x) =
6
_
x
y
x
f
g
Make use of the diagram, and a cubic equation, to solve the inequality:
6
_
x
³ x
2
– 7
Solutions to Activities
Activity 1
1. 2x
3
− x
2
− x = 0
\ x(2x
2
− x − 1) = 0
\ x(2x +1)(x − 1) = 0
\ x = 0 or x = –
1
_
2
or x = 1
2. x
3
− x = 0
\ x(x
2
−1) = 0
\ x(x − 1)(x +1) = 0
\ x = 0 or x = ±1
Lesson 1 | Algebra
Page 1
2009
Page 85
3.
2
_
3
x
3
− 18 = 0
\ 2x
3
− 54 = 0
\ 2x
3
= 54
\ x
3
= 27
\ x = 7
4. x = 2 is a solution since 2
3
+ 3(2)
2
− 4(2) − 12 = 0
\ (x − 2)(x
2
+ kx + 6) = x
3
+3x
2
− 4x − 12
Compare x terms on LHS and RHS:
−2kx + 6x = −4x
\ -2k + 6 = −4
\ -2k = −10
\ k = 5
\ x3 + 3x2 − 4x − 12 = (x − 2)(x
2
+ 5x + 6)
\ (x − 2)(x + 3)(x + 2) = 0
\ x = –3 or x = ±2
5. x = −1 is a solution since (−1)
3
− 3(−1) − 2 = 0
\ (x + 1)(x
2
+ kx − 2) = x
3
− 3x − 2
Compare x terms on LHS and RHS:
−2x + kx = −3x
\ −2 + k = −3
\ k = −1
\ (x + 1)(x
2
+ x − 2) = x
3
− 3x − 2
\ (x + 1)(x − 2)(x + 1) = 0
\ x = –1 or x = 2
6. x = 2 is a solution since 2(2)
3
+ 5(2)
2
− 14(2) − 8 = 0
\ (x − 2)(2x
2
+ kx + 4) = 2x
3
+ 5x
2
− 14x − 8
Compare x terms on LHS and RHS:
−2kx + 4x = −14x
\ −2k + 4 = −14
\ −2k = −18
\ k = 9
\ (x − 2)(2x
2
+ 9x + 4) = 2x
3
+ 5x
2
− 14x − 8
\ (x − 2)(2x + 1)(x + 4) = 0
\ x = 2 or x = –
1
_
2
or x = –4
Page 86Page 86
7. x = 2 is a solution since (2)
3
+ 7(2)
2
− 36 = 0
\ (x − 2)(x
2
+ kx + 18) = x
3
+ 7x
2
− 36
Compare x terms on LHS and RHS:
−2kx + 18x = 0x
\ −2k + 18 = 0
\ −2k = −18
\ k = 9
\ (x − 2)(x
2
+ 9x + 18) = x
3
+ 7x
2
− 36
\ (x − 2)(x + 3)(x + 6) = 0
\x = –2 or x = –3 or x = –6
8. x = –2 is a solution since 4(–2)
3
+ 12(–2)
2
+ 9(–2) + 2 = 0
\ (x + 2)(4x
2
+ kx + 1) = 4x
3
+ 12x
2
+ 9x + 2
Compare x terms on LHS and RHS:
x + 2kx = 9x
\ 1 + 2k = 9
\ 2k = 8
\ k = 4
\ (x + 2)(4x
2
+ 4x + 1) = 4x
3
+ 12x
2
+ 9x + 2
\ (x + 2)(2x + 1)(2x + 1) = 0
\ x = –
1
_
2
or x = –2
9. x = 3 is a solution since (3)
3
+ 2(3)
2
− 4(3) + 3 = 0
\ (x − 3)(x
2
+ kx − 1) = x
3
+ 2x
2
− 4x + 3
Compare x terms on LHS and RHS:
−3kx − x = −4x
\ −3k − 1 = −4
\ −3k = −3
\ k = 1
\ (x − 3)(x
2
+ x − 1) = x
3
+ 2x
2
− 4x + 3
\ x − 3 = 0 or x2 + x − 1 = 0
\ x = 3 x =
−1 ±
√
________
1 − 4(1)(−1)
__
2(1)
x =
−1 ±
√
_
5
_
2
Activity 2
1. f(6) = 0
\ (x − 6) is a factor of f(x)
\ f(x) = (x − 6)(6x
2
− x − 1)
\ f(x) = (x − 6)(3x + 1)(2x − 1)
\ If f(x) = 0, then
\ x = 6 or x = –
1
_
3
or x =
1
_
2
Lesson 1 | Algebra
Page 1
2009
Page 87
2. x
3
+ 9x
2
+ 26x + 16 = 3x + 1
\ x
3
+ 9x
2
+ 23x + 15 = 0
x = −1 is a solution since (−1)3 + 9(−1)2 + 23(−1) + 15 = 0
\ (x + 1)(x2 + 8x + 15) = 0
\ (x + 1)(x + 5)(x + 3) = 0
\ x = −1 or x = −5 or x = −3
3. For co-ordinates of A and B, we have
\ x
3
= -3x
2
+ x + 3
\ x
3
+ 3x
2
– x – 3 = 0
\ x
2
(x + 3) – (x + 3) = 0
\ (x + 3)(x
2
– 1) = 0
\ x = ±1 or x = –3
4. First, we must find the points of intersection. Therefore:
x
2
– 7 =
6
_
x
\ x
3
– 7x = 6
\ x
3
– 7x – 6 = 0
Now, x = –1 is a solution since (–1)
3
–7(–1) – 6 = 0
\ (x + 1)(x
2
– x – 6) = 0
\ (x + 1)(x + 2)(x – 3) = 0
\ x = –1 or x = –2 or x = 3
\ Reading solution to
6
_
x
³ x
2
– 7 from graph, we get
0 < x £ 3 or –2 £ x £ –1
y
x
f
g
y
x
f
g
