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 – 2

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

(vi) 3

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

(i) X² – 2x – 8

Ans: x²-2x-8=x²-4x+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 x²

Product of the zeroes = ( – 2) (4) = (- 8)/1 = (Coefficient term)/Coefficient of x².

(ii) 4s² – 4 s +1

Ans: 4s² – 4 s +1 = ( 2s – 1)²

∴ The two zeroes are ½, ½ 

Sum of the zeroes = ½ + ½ = -(-4)/4 = (Coefficient of ×) /(Coefficient of ײ)

Product of two zeroes = (½) (⅓) =¼  (Coefficient)/(coefficient of ײ) .l

(iii) 6ײ-3-7x = 6x²-7x-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) = 6x²-7x-3

Sum of the two zeroes = (3/2) + (⅓) = 3/2 -⅓ =7/6

= -(-7)/6 = (Coefficient of ×)/ (coefficient of ײ)

Product of the two zeroes = (3/2) × -(⅓) = -½ = (-3/6)

=(Constant term) / (coefficient of ײ) 

(iv) 4u² + 8u = 4u (u + 2)

Ans: It gives the two zeroes, 0 and 2 of the polynomial p(u) = 4u² + 8u + u 

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

Product of the two zeroes = (0) (-2) = 0 = (Constant term) / (coefficient of ײ) 

(v) t²-15 = t² (√15) =( t²√15) (t-√15)

Ans: If gives the two zeroes of the polynomial p(t) = t² +0t -15 are √15 and √15. 

-(Coefficient of t)/Coefficient of t²)

 Product of the two zeroes

= (-√15) × (√15) = -15 = (-15) = constant Sum of the two zeroes = Product of the two zeroes (-√15) + √15=0= Coefficient of term)/(Coefficient of t²)

(vi) 3x² – × – 4  3x² = – 4× + 3×- 4

Ans: 3x² – × – 4  3x² = – 4× + 3×- 4

= 3x(× + 1) – 4 (× + 1) = (× + ) ( 3x – 4)

Putting × + 1 = 0 and 3× – 4 = 0, we get = – 1 and × = 4/3 

i.e ., the two zeroes of the quadratic polynomial p(×) = 3x² – × – 4 are – 1 and 4/3 

Sum of the two zeroes = (-1) + (4/3) = ⅓ = (-1) /3 = (Coefficient of ×) (Coefficient of x²) 

Product of the two zeroes = (-) × (constant term)/(coefficient of x²)

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 x² + b × c and it’s Zeroes be a  and b 

Here,a × b = ¼ and ab = – 1

Thus the polynomial formed = x² – ( sum of zeroes ) x + product of Zeroes.

= x² – ( ¼) × – 1 = x² – ×/4 – 1

The other polynomial are = k ( x²- ×/4-1)

If k = 4, then polynomial is = 4x²-×-4.

(ii) √2,⅓

Ans: Here a+b = √2 ab=⅓

Thus the polynomial formed = x²( sum of Zeroes) × + product of zeroes.

= x² – (√2) × + ⅓ or x² – (√2) × + ⅓

Other polynomial are k (x² – √2×+⅓)

If k = 3, then the polynomial is 3x² – 3√2 × + 1.

(iii) 0,√5

Ans: Here a + b = 0 and a .b = √5

Thus the polynomial formed formed = x²- ( sum of zeroes) × + product of Zeroes = x² (0) x² (0)x + (√5 = x + √5

(vi) 1,1

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

= x² ( sum of Zeroes) × + product of Zeroes = x² – (1) × +1 = x² – × + 1 

(v) ¼,¼

Ans: Here a+b = -¼ ab = ¼

Thus the polynomial formed = x² -(sum of zeroes) × product of Zeroes.

= x² – (-¼) × + ¼ = x² -1/4× + ¼

The other polynomial are k (x² + ¼ ×+¼)

If k = 4 ,then the polynomial is 4x²+ × + 1

(vi) 4,1

Ans: Here a+b = ,ab = 1

Thus the polynomial formed = x² – (sum of zeroes) × + product of zeroes.

= x² – (4) × + 1 = x² – 4 × + 1