NCERT Class 10 Mathematics Chapter 2 Polynomial

NCERT Class 10 Mathematics Chapter 2 Polynomial Solutions, NCERT Solutions For Class 10 Maths, CBSE Solutions For Class 10 Mathematics to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 10 Mathematics Chapter 2 Polynomial Notes and select needs one.

NCERT Class 10 Mathematics Chapter 2 Polynomial

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 10 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 10 Mathematics Chapter 2 Polynomial Solutions for All Subject, You can practice these here.

Polynomial

Chapter โ€“ 2

Exercise 2.1

The graphs of v = p(x) are given in fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Ans: (i) No zeroes.

(ii) 1

(iii) 3

(iv) 2 

(v) 4

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(vi) 3

Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients

(i) Xยฒ โ€“ 2x โ€“ 8

Ans: xยฒ-2x-8=xยฒ-4x+2x-8=x(x-4)+2(x-4) =(x+2)(x-4)

Zeroes are -2 and 4.

Sum of the zeroes = (-2)+(4) = 2 = 

-(-2)/1 = -(Coefficient of x) / (Coefficient of xยฒ

Product of the zeroes = ( โ€“ 2) (4) = (- 8)/1 = (Coefficient term)/Coefficient of xยฒ.

(ii) 4sยฒ โ€“ 4 s +1

Ans: 4sยฒ โ€“ 4 s +1 = ( 2s โ€“ 1)ยฒ

โˆด The two zeroes are ยฝ, ยฝ 

Sum of the zeroes = ยฝ + ยฝ = -(-4)/4 = (Coefficient of ร—) /(Coefficient of ร—ยฒ)

Product of two zeroes = (ยฝ) (โ…“) =ยผ  (Coefficient)/(coefficient of ร—ยฒ) .l

(iii) 6ร—ยฒ-3-7x = 6xยฒ-7x-3 = 6x-9x + 2x-3 = (2x-3) (3x + 1) Putting 2x-3=0 and 3x + 1 = 0

Ans: We get x = 3/2 and x= -1)2 Zeroes of the quadratic polynomial 2 p(x) = 6xยฒ-7x-3

Sum of the two zeroes = (3/2) + (โ…“) = 3/2 -โ…“ =7/6

= -(-7)/6 = (Coefficient of ร—)/ (coefficient of ร—ยฒ)

Product of the two zeroes = (3/2) ร— -(โ…“) = -ยฝ = (-3/6)

=(Constant term) / (coefficient of ร—ยฒ) 

(iv) 4uยฒ + 8u = 4u (u + 2)

Ans: It gives the two zeroes, 0 and 2 of the polynomial p(u) = 4uยฒ + 8u + u 

Sum of the two zeroes  0 + (-) = -2 =(8)/4 = =(Constant term) / (coefficient of ร—ยฒ) 

Product of the two zeroes = (0) (-2) = 0 = (Constant term) / (coefficient of ร—ยฒ) 

(v) tยฒ-15 = tยฒ (โˆš15) =( tยฒโˆš15) (t-โˆš15)

Ans: If gives the two zeroes of the polynomial p(t) = tยฒ +0t -15 are โˆš15 and โˆš15. 

-(Coefficient of t)/Coefficient of tยฒ)

 Product of the two zeroes

= (-โˆš15) ร— (โˆš15) = -15 = (-15) = constant Sum of the two zeroes = Product of the two zeroes (-โˆš15) + โˆš15=0= Coefficient of term)/(Coefficient of tยฒ)

(vi) 3xยฒ โ€“ ร— โ€“ 4  3xยฒ = โ€“ 4ร— + 3ร—- 4

Ans: 3xยฒ โ€“ ร— โ€“ 4  3xยฒ = โ€“ 4ร— + 3ร—- 4

= 3x(ร— + 1) โ€“ 4 (ร— + 1) = (ร— + ) ( 3x โ€“ 4)

Putting ร— + 1 = 0 and 3ร— โ€“ 4 = 0, we get = โ€“ 1 and ร— = 4/3 

i.e ., the two zeroes of the quadratic polynomial p(ร—) = 3xยฒ โ€“ ร— โ€“ 4 are โ€“ 1 and 4/3 

Sum of the two zeroes = (-1) + (4/3) = โ…“ = (-1) /3 = (Coefficient of ร—) (Coefficient of xยฒ) 

Product of the two zeroes = (-) ร— (constant term)/(coefficient of xยฒ)

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4- 1

Ans: Let the polynomial be a xยฒ + b ร— c and itโ€™s Zeroes be a  and b 

Here,a ร— b = ยผ and ab = โ€“ 1

Thus the polynomial formed = xยฒ โ€“ ( sum of zeroes ) x + product of Zeroes.

= xยฒ โ€“ ( ยผ) ร— โ€“ 1 = xยฒ โ€“ ร—/4 โ€“ 1

The other polynomial are = k ( xยฒ- ร—/4-1)

If k = 4, then polynomial is = 4xยฒ-ร—-4.

(ii) โˆš2,โ…“

Ans: Here a+b = โˆš2 ab=โ…“

Thus the polynomial formed = xยฒ( sum of Zeroes) ร— + product of zeroes.

= xยฒ โ€“ (โˆš2) ร— + โ…“ or xยฒ โ€“ (โˆš2) ร— + โ…“

Other polynomial are k (xยฒ โ€“ โˆš2ร—+โ…“)

If k = 3, then the polynomial is 3xยฒ โ€“ 3โˆš2 ร— + 1.

(iii) 0,โˆš5

Ans: Here a + b = 0 and a .b = โˆš5

Thus the polynomial formed formed = xยฒ- ( sum of zeroes) ร— + product of Zeroes = xยฒ (0) xยฒ (0)x + (โˆš5 = x + โˆš5

(vi) 1,1

Ans: Here a+b = a: b=1 this the polynomial formed 

= xยฒ ( sum of Zeroes) ร— + product of Zeroes = xยฒ โ€“ (1) ร— +1 = xยฒ โ€“ ร— + 1 

(v) ยผ,ยผ

Ans: Here a+b = -ยผ ab = ยผ

Thus the polynomial formed = xยฒ -(sum of zeroes) ร— product of Zeroes.

= xยฒ โ€“ (-ยผ) ร— + ยผ = xยฒ -1/4ร— + ยผ

The other polynomial are k (xยฒ + ยผ ร—+ยผ)

If k = 4 ,then the polynomial is 4xยฒ+ ร— + 1

(vi) 4,1

Ans: Here a+b = ,ab = 1

Thus the polynomial formed = xยฒ โ€“ (sum of zeroes) ร— + product of zeroes.

= xยฒ โ€“ (4) ร— + 1 = xยฒ โ€“ 4 ร— + 1

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