Factorization of Cubic Polynomial 

In this section, we will discuss Factorization of Cubic Polynomial.

The degree of the cubic (highest exponent) polynomial is 3.
There are different identities in Factorization of Cubic Polynomial .
They are as follows :
1) a3 + b3 + 3a2b + 3b2a = (a + b)3

2) a3 - b3 - 3a2b + 3b2a = (a - b)3

3) a3 + b3 = (a + b)(a2 - ab + b2)

4) a3- b3 = (a - b)(a2 + ab + b2)

5) a3 + b3+ c3- 3abc
= (a + b + c)(a2 + b2+ c2 - ab - bc - ac)


6) If a + b + c = 0 then a3 + b3+ c3=3abc

Using Identities

Examples :

1)27x
3 - 8

Solution :
As first term and the 2nd term are perfect cube, we use the identity of cube

(a3 - b3) = ( a - b)(a2 +ab + b2)

a = 3x and b = 2

27x
3 - 8 = (3x -2)[(3x) 2 + 6x + 2 2 )]

= (3x -2)(9x
2 + 6x + 4)are the factors.

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2) 8x
3 +1

Solution :
As first term and the 2nd term are perfect cube, we use the identity of cube

(a3 + b3) = ( a + b)(a2 -ab + b2)

a = 2x and b = 1

8x
3 + 1 = (2x + 1)[(2x) 2 - 2x + 1 2 )]

= (2x +1 )(4x
2 - 2x + 1)are the factors.

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3) 27x
3 + y 3 + 9x 2 y + 3xy 2

Solution :
As in the given equation there are two perfect cubes, so we can use 1st identity.

27x
3 + y 3 + 27x 2 y + 9xy 2

= (3x)
3 + (y) 3 + 3(3x) 2 y + 3(3x)y 2

= (3x + y)
3 [ here a = 3x and b= y;a 3 + b 3 + 3a 2 b + 3b 2 a = (a + b) 3 ]

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4) (1.5)
3 - (0.9) 3 - (0.6) 3

Solution :
(1.5)
3 - (0.9) 3 + (-0.6) 3

Let a = 1.5, b = - 0.9 and c = -0.6

As a + b + c = 1.5 + 0.9 - 0.6 = 0

a
3 + b 3 + c 3 = 3abc

= 3(1.5)(- 0.9)(-0.6)

= 2.430




Factoring

Factorization by common factor
Factorization by Grouping
Factorization using Identities
Factorization of Cubic Polynomial
Solved Examples on Factorization

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