The document Hots Questions - Polynomials Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Question 1. If p(x) = x ^{2} – 2**√2

**Solution:** Since, p(x) = x^{2 }– 2 √2 x + 1

= 4 (2) – 4 (2) + 1 = 8 – 8 + 1 = 1

**Question 2. If a + b + c = 9, and ab + bc + ca = 26, find a ^{2 }+ b^{2} + c^{2}**

**Solution:** (a + b + c)^{2} =(a^{2 }+ b^{2 }+ c^{2}) + 2 (ab + bc + ca)

⇒ [9]^{2 }=(a^{2} + b^{2 }+ c^{2}) + 2 (26) = (a^{2 }+ b^{2} + c^{2}) + 52

⇒ a^{2} + b^{2 }+ c^{2} =92– 52 = 81– 52 = 29

**Question 3. Factorise :**

**Solution: **

**Question 4. If a, b, c are all non-zero and a + b + c = 0, prove that **

**Solution: **∵ a + b + c = 0 ⇒ a^{3} + b^{3} + c^{3} – 3abc = 0

or a^{3 }+ b^{3 }+ c^{3} = 3abc ⇒

**Question 5. Prove that (a + b + c) ^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b + c) (c + a)**

**Solution: **L.H.S. = (a + b + c)3 – a^{3} – b^{3} – c^{3} = [(a + b + c)^{3}– a^{3}] – [b^{3} + c^{3}] ...(1)

(a + b + c)^{3 }– a^{3 }= (b + c) [3a^{2} + b^{2 }+ c^{2} + 3ab + 2bc + 3ca] ...(2)

[using x^{3 }– y^{3} = (x – y) (x^{2} + y^{2} + xy)]

and b^{3} + c^{3} = (b + c) [b^{2} + c^{2} – bc] [using x^{3} + y^{3 }= (x + y) (x^{2} + y^{2} – xy)] ...(3)

From (1), (2) and (3), we get L.H.S. = (b + c) (3a^{2 }+ b^{2} + c^{2} + 3ab + 2bc + 3ca) – (b + c) (b^{2} + c^{2} – bc)

= (b + c) [3a^{2} + b^{2} + c^{2} + 3ab + 2bc + 3ca – b^{2} – c^{2} + bc]

= (b + c) [3a^{2} + (b^{2} – b^{2}) + (c^{2} – c^{2}) + 3ab + (2bc + bc) + 3ca]

= (b + c) [3a^{2} + 0 + 0 + 3ab + 3bc + 3ca]

= (b + c) [3 (a^{2} + ab + bc + ca)]

= 3 (b + c) [(a + b) (c + a)]

= 3 (a + b) (b + c) (c + a)

= RHS

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