# Polynomial Identities

Polynomial Identities : An algebraic expression in which the variables involved have only non negative integral powers is called polynomial.

For factorization or for the expansion of polynomial we use the following identities.

 Important Polynomial Identities : 1) ( x + y )2 = x2 + 2xy +y2 2) ( x – y) 2 = x2 – 2xy + y2 3) (x + y)(x – y) = x2 – y2 4) (x + a)(x + b) = x2 +(a + b)x + ab 5) (x + y) 3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3 +3xy(x +y) 6) (x - y) 3 = x3 - 3x2y + 3xy2 - y3 = x3+ y3 -3xy(x –y) 7) (x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx 8) x3 + y3 = (x + y)(x2 – xy + y2) 9) x3 - y3 = (x - y)(x2 + xy + y2) 10) x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) If x + y + z = 0 , then x3 + y3 + z3 = 3xyz

Some solved examples :

Expand the following using Polynomial Identities.
1) (2a + 5)
2
Solution : (2x + 5) 2 = (2a) 2 + 2(2a)(5) + 5 2
[ using the identity ( x + y )
2 = x 2 + 2xy +y 2 ]
= 4a
2 + 20 a + 25
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2) ( b + 6)(b - 6)
[ using the identity (x + y)(x – y) = x
2 – y 2 ]
( b + 6)(b - 6) = b
2 - 6 2
= b
2 - 36
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3) ( 3a - 4)
3
[using the identity(x - y)
3 = x 3 - 3x 2 y + 3xy 2 - y 3 ]
( 3a - 4)
3 = (3a) 3 - (3a) 2 (4) + 3(3a)(4) 2 - 4 3
= 27a
3 -72a 2 + 36a - 64
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Factorize the following using Polynomial Identities :
1) Factorize: 64a
3 - 27b 3 - 144a 2 b + 108ab 2 .
Solution :
64a
3 - 27b 3 - 144a 2 b + 108ab 2
= (4a)
3 - (3b) 3 - 36ab(4a- 3b)
= (4a)
3 - (3b) 3 -3(4a)(3b)(4a -3b)
= (4a - 3b)
3 [ using x 3 + y 3 -3xy(x –y) ]
= (4a -3b)(4a -3b)(4a -3b)
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2)Evaluate : (104)
3
(104)
3 = ( 100 + 4) 3
= = (100)
3 + (4) 3 + 3(100)(4)(100 + 4) [Using Identity V]
= 1000000 + 64 + 124800
= 1124864
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3) Evaluate : (–12)
3 + (7) 3 + (5) 3
Solution : (–12) 3 + (7) 3 + (5) 3
From the above we can see that -12 + 7 + 5 = 0
(–12)
3 + (7) 3 + (5) 3 = 3(-12)(7)(5) [Using identity 10]
= -1260

Polynomial

Degree of the Polynomial
Zeros of Polynomial
Remainder Theorem
Find remainder by Synthetic Division
Rational root test in Polynomial
Solved Examples on Polynomial identities