Factorization of Cubic Polynomial 

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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)27x3- 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

27x3- 8 = (3x -2)[(3x)2 + 6x + 22)]

= (3x -2)(9x2 + 6x + 4)are the factors.

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2) 8x3 +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

8x3+ 1 = (2x + 1)[(2x)2 - 2x + 12)]

= (2x +1 )(4x2 - 2x + 1)are the factors.

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3) 27x3 + y3 + 9x2y + 3xy2

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

27x3 + y3 + 27x2y + 9xy2

= (3x)3 + (y)3 + 3(3x)2y + 3(3x)y2

= (3x + y)3 [ here a = 3x and b= y;a3 + b3 + 3a2b + 3b2a = (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

a3 + b3+ c3 = 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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