NCERT Solutions Class 10 Maths Chapter 2 Polynomials

NCERT Solutions Class 10 Maths Chapter 2 Polynomials

NCERT Solutions Class 10 Maths Chapter 2 Polynomials:- National Council of Educational Research and Training Class 10 Maths Chapter 2 Solutions – Polynomials. NCERT Solutions Class 10 Maths Chapter 2 PDF Download.

 

NCERT Solutions Class 10 Maths Chapter 2: Overview

Board NCERT
Class 10
Subject Maths
Chapter 2
Chapter Name Polynomials
Topic Exercise Solutions

 

NCERT Solutions

Class 10 Maths

Chapter 2Polynomials

 

Exercise – 2.1

 

(1) The graphs of y = 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.

Solution:

i)

The number of zeroes is O. As the graph does not intersects at any point of x axis.

(ii)

The number of zeroes is 1. As the graph intersects the x-axis at one point only.

 

(iii)

The number of zeroes is 3 as the graph interests the x-axis at 3 points.

 

(iv)

The number of zeroes is 2 as the graph intersects the x-axis at two points.

 

(v)

The number of zeroes is (four) as the graph intersects the x-axis at four points.

 

(vi)

The number of zeroes is 3 as the graph intersects the x-axis at the three points.

 

Exercise 2.2

 

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

(i) x2 – 2x – 8

Solution:

x2 – 2x – 8 = 0

Or, x2 – (4 – 2) x – 8 = 0

Or, x2 – 4x + 2x – 8 = 0

Or, x (x – 4) + 2 (x – 4) = 0

Or, (x – 4 (x + 2) = 0

x – 4 = 0

Or, x = 4

 

x + 2 = 0

Or, x = -2

 

The zeroes of x2 – 2x – 8 are 4 and -2.

Now, Sum of zeroes = 4 + (-2) = 2/1 = – (Coefficient of)/(coefficient of)

= – (-2)/1 = 2

Product of zeroes = (4) × (-2) = -8

= constant term/coefficient of x2

= -8/1 =-8

 

(ii) 4s2 – 4s + 1

Solution:

4s2 – 4S + 1 = 0

Or, 4s2 – (2 + 2) S + 1 = 0

Or, 4s2 – 2s – 2s + 1 = 0

Or, 2s (2s – 1) -1 (2s – 1) = 0

Or, (2s – 1) (2s – 1) = 0

2s – 1 = 0

S = 1/2

2s – 1 = 0

s = 1/2

So, the zeroes of 4s2 – 4s + 1 are 1/2 and 1/2.

Now, sum of zeroes = 1/2 + 1/2 = 1 = ½

= – (coefficient of s)/Coefficient of s2

= – (-4)/4 = 1

Product of zeroes = ½ × 1/2 = 1/4 = constant terms/coefficient of s2

= 1/4

 

(iii) 6x2 – 3 – 7x

Solution:

6x2 – 3 – 7x = 0

Or, 6x2 – 7x – 3 = 0

Or, 6x2 – (9 – 2) x – 3 = 0

Or, 6x2 – 9x + 2x – 3 = 0

Or, 3x (2x – 3) + 1 (2x – 3) = 0

Or, (3x + 1) (2x – 3) = 0

2x – 3 = 0

2x = 3

x = 3/2

3x + 1 = 0

3x = -1

x = – 1/3

The zeroes of 6x2 – 3 – 7x are (3/2) and (-1/3)

Sum of zeroes = 3/2 + (-1/3) = 9-2/6 = 7/6

= (coefficient of x)/coefficient of x2

= – (-7)/6 = 7/6

Product of zeroes = (3/2) × (-1/3)

= – 1/2 constant term/coefficient of x2 = -3/6 = – 1/2

 

(iv) 4u2 + 8u

Solution:

4u2 + 8u = 0

Or, u2 + 2u = 0

Or, u (u + 2) = 0

u = 0

u + 2 = 0

u = -2

 

The zeroes of 4u2 + 8u are 0 and –2

Now, Sum of zeroes = 0 + (-2) = -2/1 = – (Coefficient of u)/(Coefficient of u2)

= -8/4 = -2

Product of zeroes = 0×(-2 = 0

= constant term/coefficient of u2

= 0/8 = 0

 

(v) t2 – 15

Solution:

t2 – 15 = 0

Or, t2 = 15

Or, t = ± √15

Or, t = + √15

t = – √15

The zeroes of t2 – 15 are √15 and (-√15)

Now, Sum of zeroes = + √15 – √15 = 0 – (coefficient t)/(Coefficient of t2)

= 0/1 = 0

Product of zeroes = (√15) × (-√15)

= -15 = constant term/coefficient of t2 = -15/1 = -15

 

(vi) 3x2 – x – 4

Solution:

3x2 – x – 4 = 0

Or, 3x2 – (4 – 3) x – 4 = 0

Or, 3x2 – 4x + 3x – 4 = 0

Or, x (3x – 4) + 1 (3x – 4) = 0

Or, (x + 1) (3x – 4) = 0

x + 1 = 0

x = -1

3x – 4 = 0

3x = 4

x = 4/3

The zeroes of (3x2 – x – 4) are (-1) and 4/3

Now, Sum of zeroes = -1 + 4/3 = -3+4/3

= 1/3 = – (coefficient of x)/coefficient of x2

= – (-1)/3 = 1/3

Product of zeroes = (-1) × 4/3

= – 4/3 constant term/coefficient of x2

= -4/3

 

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

(i) 1/4, -1

Solution:

Let, the quadratic polynomial be x2 – (α + β) x + αβ = 0 and its zeroes be α and β

We have, α + β = 1/4 and α.β = -1

Now, x2 – (α + β) x + α.β = 0

Or, x2 – 1/4 x + (-1) = 0

Or, 4x2 – x – 4 = 0

∴ The quadratic polynomial be 4x2 – x – 4

 

(ii) √2, 1/3

Solution:

Let, the quadratic polynomial be x2 – (α+β) x + αβ = 0 and it’s zeroes be α and β.

We have α + β = √2 and αβ = 1/3

Now, x2 – (α + β) x + α.β = 0

Or, x2 – √2 x + 1/3 = 0

Or, 3x2 – 3√2 x + 1 = 0

∴ The quadratic polynomial be 3x2 – 3√2 x + 1

 

(iii) 0, √5

Solution:

Let, the quadratic polynomial be x2 – (α + β) x + α.β = 0 and it’s zeroes be α and β.

We have α + β = 0 and α.β = √5

Now, x2 – (α + β) x + α.β = 0

Or, x2 – 0.x + √5 = 0

Or, x2 + √5 = 0

The quadratic polynomial be x2 + √5 = 0

 

(iv) 1, 1

Solution:

Let, the quadratic polynomial be

x2 – (α + β) x + α.β = 0, and it’s zeroes be α and β.

We have (α + β) = 1 and α.β = 1

Now, x2 – (α + β) x + α.β = 0

x2 – 1.x – 1 = 0

x2 – x + 1 = 0

∴ The quadratic polynomial be x2 – x + 1

 

(v) – ¼, 1/4 

Solution:

Let, the quadratic polynomial be x2 – (α + β) x + α.β = 0. and it zeroes be α and β

We have α + β = – 1/4 and α.β = 1/4

Now, x2 – (α + β) x + α.β = 0

x2 + x/4 + ¼ = 0

4x2 + x + 1 = 0

∴ The quadratic polynomial is 4x2 + x + 1

 

(vi) 4, 1

Solution:

Let, the quadratic polynomial be x2 – (α + β) x + α.β = 0 and it zeroes be α and β.

We have α + β = 4 and α.β = 1

Now, x2 – (α + β) x + α.β = 0

Or, x2 – 4x + 1 = 0

The quadratic polynomial be x2 – 4x + 1.

 

EXERCISE 2.3

 

(1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

 

(i) p(x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2

Solution:

Given, p(x) = x3 – 3x2 + 5x – 3, g (x) = x2 – 2

Hence, Quotient = x – 3, Remainder = 7x – 9

 

(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x

Solution:

Given, p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x

Hence, we get Quotient = x2 + x – 3

Remainder = 8.

 

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Solution:

p(x) = x4 – 5x + 6, g(x) = 2-x2

Hence, we get Quotient = -x2 – 2

Remainder = -5x + 10

 

(2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12

Solution:

Given, first polynomial = t2 – 3

Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12

Hence, Remainder is zero. So the first polynomial is a factor of the second polynomial.

 

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

Solution:

Given, first polynomial = x2 + 3x + 1

Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2

Hence, remainder is zero. So the first polynomial is a factor of the second polynomial.

 

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Solution:

Given, First polynomial = x3 – 3x + 1

2nd polynomial = x5 – 4x3 + x2 + 3x + 1

Hence, remainder is 2. So the first polynomial is not a factor of the second polynomial.

 

(3) Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, If two of its zeroes are √5/3 and -√5/3.

Solution:

3x4 + 6x3 – 2x2 – 10x – 5, If two of its zeroes are √5/3 and -√5/3

So, x = √5/3 and x = – √5/3

Or, (x – √5/3) = 0 or, (x + v5/3) = 0 –

 

Now, (x – √5/3) (x + √5/3) = 0

(x)2 – (√5/3)2 = 0  [(a – b) (a + b) = a2 – b2]

Or, x2 – 5/3 = 0

Or, 3x2 – 5 = 0

(3x2 – 5) = 0

(3x2 – 5) a factor of given polynomial

Hence, x2 + 2x + 1 is the other factor.

x2 + 2x + 1 = 0

x2 + x + x + 1 = 0

x (x + 1) +1 (x + 1) = 0

(x + 1) (x + 1)

x + 1 = 0

x = -1

x + 1 = 0

x = -1

Other zeroes are -1 and -1

 

(4) On dividing x3 – 3x + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x +4, respectively. Find g(x).

Solution:

Dividend = Divisor × quotient + remainder

or divisor × quotient = dividend – remainder

or divisor = dividend – Remainder/quotient

∴ Hence g(x) = x2 – x + 1

 

(5) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

 

(i) deg p(x) = deg q(x)

Solution:

Example:-

Let, p(x) = 2x2 – 2x + 2 is divided by 2

∴ p(x) = 2x2 – 2x + 2

q (x) = x2 – x + 2

deg p(x) = deg q(x) = 2

 

(ii) deg q(x) = deg r(x)

Solution:

q (x) = x

r(x) = x

degq (x) = degr (x) = 1

 

(iii) deg r(x) = 0

Solution:

So, degr9x) = 0

 

EXERCISE 2.4 (Optional)

 

(1) Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

 

(i) 2x3 + x2 – 5x + 2; 1/2, 1, -2

Solution:

p(x) = 2x3 + x2 – 5x + 2

p (1/2) = 2 × (1/2)3 + (1/2)2 – 5× ½ + 2

= 2 × 1/8 + 1/4 – 5/2 + 2

= 1/4 + 1/4 – 5/2 + 2

= 1+1-10+8/4 = 0/4 = 0

p(1) = 2× (1)3 + (1)2 -5×1 + 2

= 2 × 1 + 1 – 5 + 2

= 2 + 1 – 5 + 2 = 5 – 5 = 0

p (-2) = 2 (-2)3 + (-2)2 – 5 (-2) + 2

= 2 × (-8) + 4 + 10 + 2

= – 16 + 16 = 0

Here, α = – ½, β = 1, ð = -2

p(x) = 2x3 + x2 – 5x + 2,

∴ a = 2, b = 1, c = -5, d = 2

Now, α + β + ð = – b/a

1/2 + 1 – 2 = -1/2

– 1/2 = -1/2 (Verify)

 

Now, αβ + βð + ðα = c/a

½ × 1 + 1 × (-2) + (-2) × ½ = -5/2

Or, 1/2 – 2 – 1 = -5/2

Or, 1-4-2/2 = -5/2

Or, – 5/2 = – 5/2 (Verify)

α.βð = -d/a

½. 1 × (-2) = -2/2

-1 = -1 (Verify)

 

(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

Solution:

p(x) = x3 – 4x2 + 5x – 2

p(2) = 23 – 4 × (2)2 + 5 × 2 – 2

= 8 – 16 + 10 – 2 = 18 – 18 = 0

p(1) = 13 – 4 (1)2 + 5×1 – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

p(1) = 13 – 4(1)2 + 5×1 – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

Here, α = 2, β = 1, ð = 1

p(x) = x3 – 4x2 + 5x – 2

∴ a = 1, b = -4, c = 5, d = -2

 

Now, α+β+ð = – b/a

Or, 2 + 1 + 1 = – (-4)/1

Or, 4 = 4 (verify)

 

αβ + βð + ðα = c/a

2×1 + 1×1 + 1×2 = 5/1

Or, 2 + 1 + 2 = 5

Or, 5 = 5 (Verify)

 

α.β.ð = – d/a

α.β.ð = – (-c)/d

Or, 2×1×1 = 2

Or, 2 = 2 Verify.

 

(2) Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution:

Let the cubic polynomial be

ax3 + bx2 + cx + d = 0

And the zeroes are α, β, ð.

∴ α + β + ð = – b/a = 2

∴ – a/b = 2

Or, b/a = -2/1

a = 1, b = -2

 

αβ + βð + ðα – c/a = -7

c/a = -7

Or, c/a = -7/1

C = -7 a = 1

 

α.β.ð = – d/a

∴ – d/a = -14

Or, d/a = 14/1

d = 14, a = 1

 

The cubic polynomial is

1.x3 + (-2) x2 + (-7) x + 14

= x3 – 2x2 – 7x + 14

 

(3) If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.

Solution:

p(x) = x3 – 3x2 + x + 1 and the zeroes are (a-b), a, (a+b)

p(x) = x3 – 3x2 + x + 1

a = 1, b = -3, c = 1, d = 1

α+β+ð = – b/a

a+b+a+a+b = – (-3)/1

3a = 3

a = 1

 

α.β.ð = – d/a

(a-b) × a × (a+b) = -1/1

a (a2 – b2) = 1

1 (12 – b2) = 1

b2 = 1 + 1

b2 = 2

b = ± √2

Hence, (1 – √2), 1, (1 + √2) are the zeroes of x3 – 3x2 + x + 1.

 

(4) If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3, find other zeroes.

Solution:

p(x) = x4 – 6x3 – 26x2 + 138x – 35 and two zeroes are 2 ± √3.

So, x = 2 + √3 and x = 2 – √3

Or, x – (2 + √3) = 0 or, x – (2 – √3) = 0

Now, {x – (2 – √3)} (x – (2 – √3)}

= {(x – 2) – √3} {(x – 2) + √3)}

= {(x – 2)2 – (√3)2}

= x2 – 4x + 4 = 3 = x2 – 4x + 1

So, x2 – 2x – 35 is the other factor.

x2 – 2x – 35 = 0

x2 – (7-5) x – 35 = 0

x2 – 7x – 5x – 35 = 0

x (x – 7) -5 (x – 7) = 0

(x – 5) (x – 7) = 0

x – 5 = 0

Or, x = 5

x – 7 = 0

x = 7

Hence, the other zeroes are 5 & 7.

 

(5) If the polynomial x4 – 6x3 + 16x2 – 25x + 10x is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:

Remainder = (2K – 9) x + (10 – 8k + k2)

(2k – 9) x + (10 – 8k + k2) = x + a

∴ Comparing this….

2K – 9 = 1

Or, 2k = 1+9

2K = 10

k = 5

 

10 – 8K + k2 = a

Or, a = 10 – 8.5 + 52

= 10 – 40 + 25

= 5

Hence, k = 5 and a = 5

Updated: November 24, 2021 — 10:52 am

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