Polynomial Identities
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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} = x^{2} + 2xy +y^{2} 2) ( x – y) ^{2} = x^{2} – 2xy + y^{2} 3) (x + y)(x – y) = x^{2 }– y^{2} 4) (x + a)(x + b) = x^{2} +(a + b)x + ab 5) (x + y) ^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3} = x^{3} + y^{3} +3xy(x +y) 6) (x - y) ^{3 }= x^{3} - 3x^{2}y + 3xy^{2} - y^{3} = x^{3}+ y^{3 }-3xy(x –y) 7) (x + y + z) ^{2} = x^{2 }+ y^{2 }+ z^{2} + 2xy + 2yz + 2zx 8) x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2)} 9) x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2)} 10) x^{3} + y^{3} + z^{3 }– 3xyz = (x + y + z)(x^{2} + y^{2 }+ z^{2} – xy – yz – zx) If x + y + z = 0 , then x^{3} + y^{3 }+ z^{3} = 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
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