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Preeti Dabral 3 years, 3 months ago
All the exponents in the algebraic expression must be non-negative integers in order for the algebraic expression to be a polynomial. As a general rule of thumb if an algebraic expression has a radical in it then it isn't a polynomial. Thus it is not a polynomial.
Posted by Harshvivek Lakhera 3 years, 3 months ago
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Preeti Dabral 3 years, 3 months ago
Yes, now we have 5 rational no. betwen them. =11/90, 12/90, 13/90, 14/90, 15/90.
Posted by Anshika Tiwari 3 years, 3 months ago
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Harshvivek Lakhera 3 years, 3 months ago
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Preeti Dabral 3 years, 3 months ago
For example, √120/ 25 and √122/25 are irrational, while still being between 2 and √5. Thus, between 2 and √5 ≈2.236 we have 2.1 as one rational number and e−7/10 as one irrational number.
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Ayushman Nayak 3 years, 3 months ago
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Sia ? 3 years, 3 months ago
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Posted by Md Sameer Sameer 3 years, 3 months ago
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Posted by R V 3 years, 3 months ago
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Girija Paru 3 years, 3 months ago
Posted by Himavarshini V 3 years, 3 months ago
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Preeti Dabral 3 years, 3 months ago
Let p(x) = x3 + 13x2+ 32x + 20
p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20
= -1 + 13 - 32 + 20 = 0
∴ By Factor Theorem, x - (-1), i.e., (x + 1) is a factor of p(x).
x3 + 13x2 + 32x + 20
= x2(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x2 + 12x + 20)
= (x + 1)(x2 + 2x + 10x + 20)
= (x + 1){x(x + 2) + 10(x + 2)}
= (x + 1)(x + 2)(x + 10).
Posted by Akshita Mishra 3 years, 3 months ago
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Rohitraj Rai 3 years, 3 months ago
Posted by Samapika Chakravorty 3 years, 3 months ago
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Preeti Dabral 3 years, 3 months ago
Given that x3 − 2x2 − x+2
= x3 − x− 2x2 +2
Grouping the 1st 2 terms together and the 2nd 2 together:
= x(x2 − 1) − 2(x2 − 1)
= (x2 − 1) (x−2)
Using the identity: a2 − b2 = (a+b) (a−b)
= (x2−12) (x−2)
= (x−1) (x+1) (x−2)
Posted by Devindra Chansta 3 years, 3 months ago
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Preeti Dabral 3 years, 3 months ago
27 - 125a3 - 135a + 225a2
= (3)3 - (5a)3 - 3(3)(5a)(3 - 5a)
= (3 - 5a)3
(Using Identity (a – b)3= a3– b3– 3ab (a – b))
= (3 - 5a)(3 - 5a)(3 - 5a)
Posted by Akriti Yadav 3 years, 3 months ago
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