Hots Questions - Polynomials Class 9 Notes | EduRev

Mathematics (Maths) Class 9

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Class 9 : Hots Questions - Polynomials Class 9 Notes | EduRev

The document Hots Questions - Polynomials Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. If  p(x) = x2 – 2√2 x + 1, then find  p 2√2 .

Solution: Since, p(x) = x– 2 √2 x + 1   

Hots Questions - Polynomials Class 9 Notes | EduRev

= 4 (2) – 4 (2) + 1 = 8 – 8 + 1 = 1

Question 2. If  a + b + c = 9, and ab + bc + ca = 26, find a+ b2 + c2

Solution: (a + b + c)2 =(a+ b+ c2) + 2 (ab + bc + ca)

⇒ [9]=(a2 + b+ c2) + 2 (26)  = (a+ b2 + c2) + 52

⇒ a2 + b+ c2 =92– 52 = 81– 52 = 29

Question 3. Factorise :Hots Questions - Polynomials Class 9 Notes | EduRev

Solution:   Hots Questions - Polynomials Class 9 Notes | EduRev

Hots Questions - Polynomials Class 9 Notes | EduRevHots Questions - Polynomials Class 9 Notes | EduRev
Hots Questions - Polynomials Class 9 Notes | EduRevHots Questions - Polynomials Class 9 Notes | EduRev
Hots Questions - Polynomials Class 9 Notes | EduRev

Question 4. If a, b, c are all non-zero and a + b + c = 0, prove that Hots Questions - Polynomials Class 9 Notes | EduRev

Solution: ∵ a + b + c = 0  ⇒  a3 + b3 + c3 – 3abc = 0
or a+ b+ c3 = 3abc ⇒ Hots Questions - Polynomials Class 9 Notes | EduRev

Question 5. Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b + c) (c + a)

Solution: L.H.S. = (a + b + c)3 – a3 – b3 – c3 =  [(a + b + c)3– a3] – [b3 + c3]                 ...(1)
(a + b + c)– a= (b + c) [3a2 + b+ c2 + 3ab + 2bc + 3ca]                  ...(2)
                                                                    [using x– y3 = (x – y) (x2 + y2 + xy)]
and b3 + c3 = (b + c) [b2 + c2 – bc]            [using x3 + y= (x + y) (x2 + y2 – xy)]                        ...(3)

From (1), (2) and (3), we get L.H.S. = (b + c) (3a+ b2 + c2 + 3ab + 2bc + 3ca) – (b + c) (b2 + c2 – bc)
= (b + c) [3a2 + b2 + c2 + 3ab + 2bc + 3ca – b2 – c2 + bc]
= (b + c) [3a2 + (b2 – b2) + (c2 – c2) + 3ab + (2bc + bc) + 3ca]
= (b + c) [3a2 + 0 + 0 + 3ab + 3bc + 3ca]
= (b + c) [3 (a2 + ab + bc + ca)]
= 3 (b + c) [(a + b) (c + a)]
= 3 (a + b) (b + c) (c + a)
= RHS

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