RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

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rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
rd sharma solutions for class 10 chapter 2 ex 2.1
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rd sharma solutions for class 10 chapter 2 ex 2.3
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rd sharma solutions for class 10 chapter 2 ex 2.3

 

Chapter 2- Polynomials has three exercises and RD Sharma Solutions for Class 10 here contains the answers to the problems done in a very intelligible and detailed manner. Let’s get an insight to this chapter to get a better idea of what it’s about.

  • Polynomial and its types
  • Geometrical representation of linear and quadratic polynomials
  • The geometric meaning of the zeros of a polynomial
  • Relationship between the zeros and coefficients of a polynomial

Also, get RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials – exercise wise

Exercise 2.1 Solutions

Exercise 2.2 Solutions

Exercise 2.3 Solutions

Exercise 2.1 Page No: 2.33

1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

(i) f(x) = x– 2x – 8

Solution:

Given,

f(x) = x– 2x – 8

To find the zeros, we put f(x) = 0

⇒ x– 2x – 8 = 0

⇒  x2 – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x – 4)(x + 2) = 0

This gives us 2 zeros, for

x = 4 and x = -2

Hence, the zeros of the quadratic equation are 4 and -2.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x2

4 x (-2) = (-8) / 1

-8 = -8

Therefore, the relationship between zeros and their coefficients is verified.

(ii) g(s) = 4s– 4s + 1

Solution:

Given,

g(s) = 4s– 4s + 1

To find the zeros, we put g(s) = 0

⇒ 4s– 4s + 1 = 0

⇒  4s2 – 2s – 2s + 1= 0

⇒  2s(2s – 1) – (2s – 1) = 0

⇒ (2s – 1)(2s – 1) = 0

This gives us 2 zeros, for

s = 1/2 and s = 1/2

Hence, the zeros of the quadratic equation are 1/2 and 1/2.

Now, for verification

Sum of zeros = – coefficient of s / coefficient of s2

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s2

1/2 x 1/2 = 1/4

1/4 = 1/4

Therefore, the relationship between zeros and their coefficients is verified.

(iii) h(t)=t– 15

Solution:

Given,

h(t) = t– 15 = t+(0)t – 15

To find the zeros, we put h(t) = 0

⇒ t– 15 = 0

⇒ (t + √15)(t – √15)= 0

This gives us 2 zeros, for

t = √15 and t = -√15

Hence, the zeros of the quadratic equation are √15 and -√15.

Now, for verification

Sum of zeros = – coefficient of t / coefficient of t2

√15 + (-√15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t2

√15 x (-√15) = -15/1

-15 = -15

Therefore, the relationship between zeros and their coefficients is verified.

(iv) f(x) = 6x2 – 3 – 7x

Solution:

Given,

f(x) = 6x– 3 – 7x

To find the zeros, we put f(x) = 0

⇒ 6x– 3 – 7x = 0

⇒  6x2 – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

This gives us 2 zeros, for

x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

3/2 + (-1/3) = – (-7) / 6

7/6 = 7/6

Product of roots = constant / coefficient of x2

3/2 x (-1/3) = (-3) / 6

-1/2 = -1/2

Therefore, the relationship between zeros and their coefficients is verified.

(v) p(x) = x+ 2√2x – 6

Solution:

Given,

p(x) = x+ 2√2x – 6

To find the zeros, we put p(x) = 0

⇒ x+ 2√2x – 6 = 0

⇒  x2 + 3√2x – √2x – 6 = 0

⇒ x(x + 3√2) – √2 (x + 3√2) = 0

⇒ (x – √2)(x + 3√2) = 0

This gives us 2 zeros, for

x = √2 and x = -3√2

Hence, the zeros of the quadratic equation are √2 and -3√2.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

√2 + (-3√2) = – (2√2) / 1

-2√2 = -2√2

Product of roots = constant / coefficient of x2

√2 x (-3√2) = (-6) / 2√2

-3 x 2 = -6/1

-6 = -6

Therefore, the relationship between zeros and their coefficients is verified.

(vi) q(x)=√3x+ 10x + 7√3

Solution:

Given,

q(x) = √3x+ 10x + 7√3

To find the zeros, we put q(x) = 0

⇒ √3x+ 10x + 7√3 = 0

⇒  √3x2 + 3x +7x + 7√3x = 0

⇒ √3x(x + √3) + 7 (x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

This gives us 2 zeros, for

x = -√3 and x = -7/√3

Hence, the zeros of the quadratic equation are -√3 and -7/√3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

-√3 + (-7/√3) = – (10) /√3

(-3-7)/ √3 = -10/√3

-10/ √3 = -10/√3

Product of roots = constant / coefficient of x2

(-√3) x (-7/√3) = (7√3) / √3

7 = 7

Therefore, the relationship between zeros and their coefficients is verified.

(vii) f(x) = x– (√3 + 1)x + √3

Solution:

Given,

f(x) = x– (√3 + 1)x + √3

To find the zeros, we put f(x) = 0

⇒ x– (√3 + 1)x + √3 = 0

⇒  x2 – √3x – x + √3 = 0

⇒ x(x – √3) – 1 (x – √3) = 0

⇒ (x – √3)(x – 1) = 0

This gives us 2 zeros, for

x = √3 and x = 1

Hence, the zeros of the quadratic equation are √3 and 1.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

√3 + 1 = – (-(√3 +1)) / 1

√3 + 1 = √3 +1

Product of roots = constant / coefficient of x2

1 x √3 = √3 / 1

√3 = √3

Therefore, the relationship between zeros and their coefficients is verified.

(viii) g(x)=a(x2+1)–x(a2+1)

Solution:

Given,

g(x) = a(x2+1)–x(a2+1)

To find the zeros, we put g(x) = 0

⇒ a(x2+1)–x(a2+1) = 0

⇒ ax2 + a − a2x – x = 0

⇒ ax− a2x – x + a = 0

⇒ ax(x − a) − 1(x – a) = 0

⇒ (x – a)(ax – 1) = 0

This gives us 2 zeros, for

x = a and x = 1/a

Hence, the zeros of the quadratic equation are a and 1/a.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

a + 1/a = – (-(a2 + 1)) / a

(a2 + 1)/a = (a2 + 1)/a

Product of roots = constant / coefficient of x2

a x 1/a = a / a

1 = 1

Therefore, the relationship between zeros and their coefficients is verified.

(ix) h(s) = 2s– (1 + 2√2)s + √2

Solution:

Given,

h(s) = 2s– (1 + 2√2)s + √2

To find the zeros, we put h(s) = 0

⇒ 2s– (1 + 2√2)s + √2 = 0

⇒  2s– 2√2s – s + √2 = 0

⇒ 2s(s – √2) -1(s – √2) = 0

⇒ (2s – 1)(s – √2) = 0

This gives us 2 zeros, for

x = √2 and x = 1/2

Hence, the zeros of the quadratic equation are √3 and 1.

Now, for verification

Sum of zeros = – coefficient of s / coefficient of s2

√2 + 1/2 = – (-(1 + 2√2)) / 2

(2√2 + 1)/2 = (2√2 +1)/2

Product of roots = constant / coefficient of s2

1/2 x √2 = √2 / 2

√2 / 2 = √2 / 2

Therefore, the relationship between zeros and their coefficients is verified.

(x) f(v) = v+ 4√3v – 15

Solution:

Given,

f(v) = v+ 4√3v – 15

To find the zeros, we put f(v) = 0

⇒ v+ 4√3v – 15 = 0

⇒  v2 + 5√3v – √3v – 15 = 0

⇒ v(v + 5√3) – √3 (v + 5√3) = 0

⇒ (v – √3)(v + 5√3) = 0

This gives us 2 zeros, for

v = √3 and v = -5√3

Hence, the zeros of the quadratic equation are √3 and -5√3.

Now, for verification

Sum of zeros = – coefficient of v / coefficient of v2

√3 + (-5√3) = – (4√3) / 1

-4√3 = -4√3

Product of roots = constant / coefficient of v2

√3 x (-5√3) = (-15) / 1

-5 x 3 = -15

-15 = -15

Therefore, the relationship between zeros and their coefficients is verified.

(xi) p(y) = y+ (3√5/2)y – 5

Solution:

Given,

p(y) = y+ (3√5/2)y – 5

To find the zeros, we put f(v) = 0

⇒ y+ (3√5/2)y – 5 = 0

⇒  y2 – √5/2 y + 2√5y – 5 = 0

⇒ y(y – √5/2) + 2√5 (y – √5/2) = 0

⇒ (y + 2√5)(y – √5/2) = 0

This gives us 2 zeros, for

y = √5/2 and y = -2√5

Hence, the zeros of the quadratic equation are √5/2 and -2√5.

Now, for verification

Sum of zeros = – coefficient of y / coefficient of y2

√5/2 + (-2√5) = – (3√5/2) / 1

-3√5/2 = -3√5/2

Product of roots = constant / coefficient of y2

√5/2 x (-2√5) = (-5) / 1

– (√5)2 = -5

-5 = -5

Therefore, the relationship between zeros and their coefficients is verified.

(xii) q(y) = 7y– (11/3)y – 2/3

Solution:

Given,

q(y) = 7y– (11/3)y – 2/3

To find the zeros, we put q(y) = 0

⇒ 7y– (11/3)y – 2/3 = 0

⇒  (21y2 – 11y -2)/3 = 0

⇒ 21y2 – 11y – 2 = 0

⇒ 21y2 – 14y + 3y – 2 = 0

⇒ 7y(3y – 2) – 1(3y + 2) = 0

⇒ (3y – 2)(7y + 1) = 0

This gives us 2 zeros, for

y = 2/3 and y = -1/7

Hence, the zeros of the quadratic equation are 2/3 and -1/7.

Now, for verification

Sum of zeros = – coefficient of y / coefficient of y2

2/3 + (-1/7) = – (-11/3) / 7

-11/21 = -11/21

Product of roots = constant / coefficient of y2

2/3 x (-1/7) = (-2/3) / 7

– 2/21 = -2/21

Therefore, the relationship between zeros and their coefficients is verified.

2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.

(i) -8/3 , 4/3

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -8/3 and product of zero= 4/3

Thus,

The required polynomial f(x) is,

⇒ x2 – (-8/3)x + (4/3)

⇒ x2 + 8/3x + (4/3)

So, to find the zeros we put f(x) = 0

⇒ x2 + 8/3x + (4/3) = 0

⇒ 3x2 + 8x + 4 = 0

⇒ 3x2 + 6x + 2x + 4 = 0

⇒ 3x(x + 2) + 2(x + 2) = 0

⇒ (x + 2) (3x + 2) = 0

⇒ (x + 2) = 0 and, or (3x + 2) = 0

Therefore, the two zeros are -2 and -2/3.

(ii) 21/8 , 5/16

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = 21/8 and product of zero = 5/16

Thus,

The required polynomial f(x) is,

⇒ x2 – (21/8)x + (5/16)

⇒ x2 – 21/8x + 5/16

So, to find the zeros we put f(x) = 0

⇒ x2 – 21/8x + 5/16 = 0

⇒ 16x2 – 42x + 5 = 0

⇒ 16x2 – 40x – 2x + 5 = 0

⇒ 8x(2x – 5) – 1(2x – 5) = 0

⇒ (2x – 5) (8x – 1) = 0

⇒ (2x – 5) = 0 and, or (8x – 1) = 0

Therefore, the two zeros are 5/2 and 1/8.

(iii) -2√3, -9

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -2√3 and product of zero = -9

Thus,

The required polynomial f(x) is,

⇒ x2 – (-2√3)x + (-9)

⇒ x2 + 2√3x – 9

So, to find the zeros we put f(x) = 0

⇒ x2 + 2√3x – 9 = 0

⇒ x2 + 3√3x – √3x – 9 = 0

⇒ x(x + 3√3) – √3(x + 3√3) = 0

⇒ (x + 3√3) (x – √3) = 0

⇒ (x + 3√3) = 0 and, or (x – √3) = 0

Therefore, the two zeros are -3√3and √3.

(iv) -3/2√5, -1/2

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -3/2√5 and product of zero = -1/2

Thus,

The required polynomial f(x) is,

⇒ x2 – (-3/2√5)x + (-1/2)

⇒ x2 + 3/2√5x – 1/2

So, to find the zeros we put f(x) = 0

⇒ x2 + 3/2√5x – 1/2 = 0

⇒ 2√5x2 + 3x – √5 = 0

⇒ 2√5x2 + 5x – 2x – √5 = 0

⇒ √5x(2x + √5) – 1(2x + √5) = 0

⇒ (2x + √5) (√5x – 1) = 0

⇒ (2x + √5) = 0 and, or (√5x – 1) = 0

Therefore, the two zeros are -√5/2 and 1/√5.

3. If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1/α + 1/β – 2αβ.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4

So, we can find

Sum of the roots = α+β = -b/a = – (-5)/1 = -5

Product of the roots = αβ = c/a = 4/1 = 4

To find, 1/α +1/β – 2αβ

⇒ [(α +β)/ αβ] – 2αβ

⇒ (-5)/ 4 – 2(4) = -5/4 – 8 = -27/ 4

4. If α and β are the zeros of the quadratic polynomial p(y) = 5y2 – 7y + 1, find the value of 1/α+1/β.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1

So, we can find

Sum of the roots = α+β = -b/a = – (-7)/5 = 7/5

Product of the roots = αβ = c/a = 1/5

To find, 1/α +1/β

⇒ (α +β)/ αβ

⇒ (7/5)/ (1/5) = 7

5. If α and β are the zeros of the quadratic polynomial f(x)=x2 – x – 4, find the value of 1/α+1/β–αβ.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = – 4

So, we can find

Sum of the roots = α+β = -b/a = – (-1)/1 = 1

Product of the roots = αβ = c/a = -4 /1 = – 4

To find, 1/α +1/β – αβ

⇒ [(α +β)/ αβ] – αβ

⇒ [(1)/ (-4)] – (-4) = -1/4 + 4 = 15/ 4

6. If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of 1/α – 1/β.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = – 2

So, we can find

Sum of the roots = α+β = -b/a = – (1)/1 = -1

Product of the roots = αβ = c/a = -2 /1 = – 2

To find, 1/α – 1/β

⇒ [(β – α)/ αβ]

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.1 - 1

7. If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k.

Solution:

From the question, it’s given that:

The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9

And, for roots to be negative of each other, let the roots be α and – α.

So, we can find

Sum of the roots = α – α = -b/a = – (-8k)/1 = 8k = 0 [∵ α – α = 0]

⇒ k = 0

8.  If the sum of the zeroes of the quadratic polynomial f(t)=kt2 + 2t + 3k is equal to their product, then find the value of k.

Solution:

Given,

The quadratic polynomial f(t)=kt2 + 2t + 3k, where a = k, b = 2 and c = 3k.

And,

Sum of the roots = Product of the roots

⇒ (-b/a) = (c/a)

⇒ (-2/k) = (3k/k)

⇒ (-2/k) = 3

∴ k = -2/3

9. If α and β are the zeros of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2β+αβ2.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1

So, we can find

Sum of the roots = α+β = -b/a = – (-5)/4 = 5/4

Product of the roots = αβ = c/a = -1/4

To find, α2β+αβ2

⇒ αβ(α +β)

⇒ (-1/4)(5/4) = -5/16

10. If α and β are the zeros of the quadratic polynomial f(t)=t2– 4t + 3, find the value of α4β33β4.

Solution:

From the question, it’s given that:

α and β are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3

So, we can find

Sum of the roots = α+β = -b/a = – (-4)/1 = 4

Product of the roots = αβ = c/a = 3/1 = 3

To find, α4β33β4

⇒ α3β3 (α +β)

⇒ (αβ)3 (α +β)

⇒ (3)3 (4) = 27 x 4 = 108


Exercise 2.2 Page No: 2.43

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeros and coefficients in each of the following cases:

(i) f(x) = 2x3 + x2 – 5x + 2; 1/2, 1, -2

Solution:

Given, f(x) = 2x3 + x2 – 5x + 2, where a= 2, b= 1, c= -5 and d= 2

For x = 1/2

f(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2

= 1/4 + 1/4 – 5/2 + 2 = 0

⇒ f(1/2) = 0, hence x = 1/2 is a root of the given polynomial.

For x = 1

f(1) = 2(1)3 + (1)2 – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

⇒ f(1) = 0, hence x = 1 is also a root of the given polynomial.

For x = -2

f(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2

= -16 + 4 + 10 + 2 = 0

⇒ f(-2) = 0, hence x = -2 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1/2 + 1 – 2 = – (1)/2

-1/2 = -1/2

Sum of the products of the zeros taken two at a time = c/a

(1/2 x 1) + (1 x -2) + (1/2 x -2) = -5/ 2

1/2 – 2 + (-1) = -5/2

-5/2 = -5/2

Product of zeros = – d/a

1/2 x 1 x (– 2) = -(2)/2

-1 = -1

Hence, the relationship between the zeros and coefficients is verified.

(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1

Solution:

Given, g(x) = x3 – 4x2 + 5x – 2, where a= 1, b= -4, c= 5 and d= -2

For x = 2

g(2) = (2)3 – 4(2)2 + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

⇒ f(2) = 0, hence x = 2 is a root of the given polynomial.

For x = 1

g(1) = (1)3 – 4(1)2 + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

⇒ g(1) = 0, hence x = 1 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1 + 1 + 2 = – (-4)/1

4 = 4

Sum of the products of the zeros taken two at a time = c/a

(1 x 1) + (1 x 2) + (2 x 1) = 5/ 1

1 + 2 + 2 = 5

5 = 5

Product of zeros = – d/a

1 x 1 x 2 = -(-2)/1

2 = 2

Hence, the relationship between the zeros and coefficients is verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, -1 and -3 respectively.

Solution:

Generally,

A cubic polynomial say, f(x) is of the form ax3 + bx2 + cx + d.

And, can be shown w.r.t its relationship between roots as.

⇒ f(x) = k [x3 – (sum of roots)x2 + (sum of products of roots taken two at a time)x – (product of roots)]

Where, k is any non-zero real number.

Here,

f(x) = k [x3 – (3)x2 + (-1)x – (-3)]

∴ f(x) = k [x3 – 3x2 – x + 3)]

where, k is any non-zero real number.

3. If the zeros of the polynomial f(x) = 2x3 – 15x2 + 37x – 30 are in A.P., find them.

Solution:

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)

And given, the zeros are in A.P.

So, let’s consider the roots as

α = a – d, β = a and γ = a +d

Where, a is the first term and d is the common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

⇒ Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 ⇒ a = 5/2

⇒ Product of roots = (a – d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15

⇒ a(a2 –d2) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)2 –d2] = 15

⇒ 5[(25/4) –d2] = 30

⇒ (25/4) – d2 = 6

⇒ 25 – 4d2 = 24

⇒ 1 = 4d2

∴ d = 1/2 or -1/2

Taking d = 1/2 and a = 5/2

We get,

the zeros as 2, 5/2 and 3

Taking d = -1/2 and a = 5/2

We get,

the zeros as 3, 5/2 and 2


Exercise 2.3 Page No: 2.57

1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:

(i) f(x) = x3– 6x2 + 11x – 6, g(x) = x2 + x +1

Solution:

Given,

f(x) = x3– 6x2 +11x – 6, g(x) = x2 +x + 1

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 1

Thus,

q(x) = x – 7 and r(x) = 17x -1

(ii) f(x) = 10x4 + 17x3 – 62x2 + 30x – 3, g(x) = 2x2 + 7x + 1

Solution:

Given,

f(x) = 10x4 + 17x3 – 62x2 + 30x – 3 and g(x) = 2x2 + 7x + 1

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 2

Thus,

q(x) = 5x2 – 9x – 2 and r(x) = 53x – 1

(iii) f(x) = 4x3 + 8x2 + 8x + 7, g(x)= 2x2 – x + 1

Solution:

Given,

f(x) = 4x3 + 8x2 + 8x + 7 and g(x)= 2x2 – x + 1

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 3

Thus,

q(x) = 15x + 10 and r(x) = 3x – 32

(iv) f(x) = 15x3 – 20x2 + 13x – 12, g(x) = x2 – 2x + 2

Solution:

Given,

f(x) = 15x3 – 20x2 + 13x – 12 and g(x) = x2 – 2x + 2

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 4

Thus,

q(x) = 15x + 10 and r(x) = 3x – 32

2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

(i) g(t) = t2–3; f(t)=2t4 + 3t3 – 2t2 – 9t – 12

Solution:

Given,

g(t) = t2 – 3; f(t) =2t4 + 3t3 – 2t2 – 9t – 12

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 5

Since, the remainder r(t) = 0 we can say that the first polynomial is a factor of the second polynomial.

(ii) g(x) = x3 – 3x + 1; f(x) = x5 – 4x3 + x2 + 3x + 1

Solution:

Given,

g(x) = x3 – 3x + 1; f(x) = x5 – 4x3 + x2 + 3x + 1

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 6

Since, the remainder r(x) = 2 and not equal to zero we can say that the first polynomial is not a factor of the second polynomial.

(iii) g(x) = 2x2– x + 3; f(x) = 6x5 − x4 + 4x3 – 5x2 – x –15

Solution:

Given,

g(x) = 2x2– x + 3; f(x)=6x5 − x4 + 4x3 – 5x2 – x –15

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 7

Since, the remainder r(x) = 0 we can say that the first polynomial is not a factor of the second polynomial.

3. Obtain all zeroes of the polynomial f(x)= 2x4 + x3 – 14x2 – 19x–6, if two of its zeroes are -2 and -1.

Solution:

Given,

f(x)= 2x4 + x3 – 14x2 – 19x – 6

If the two zeros of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)

⇒ (x+2)(x+1) = x2 + x + 2x + 2 = x2 + 3x +2 …… (i)

This means that (i) is a factor of f(x). So, performing division algorithm we get,

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 8

The quotient is 2x2 – 5x – 3.

⇒ f(x)= (2x2 – 5x – 3)( x2 + 3x +2)

For obtaining the other 2 zeros of the polynomial

We put,

2x2 – 5x – 3 = 0

⇒ (2x + 1)(x – 3) = 0

∴ x = -1/2 or 3

Hence, all the zeros of the polynomial are -2, -1, -1/2 and 3.

4.  Obtain all zeroes of f(x) = x3 + 13x2 + 32x + 20, if one of its zeros is -2.

Solution:

Given,

f(x)= x3 + 13x2 + 32x + 20

And, -2 is one of the zeros. So, (x + 2) is a factor of f(x),

Performing division algorithm, we get

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 9

⇒ f(x)= (x2 + 11x + 10)( x + 2)

So, putting x2 + 11x + 10 = 0 we can get the other 2 zeros.

⇒ (x + 10)(x + 1) = 0

∴ x = -10 or -1

Hence, all the zeros of the polynomial are -10, -2 and -1.

5. Obtain all zeroes of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6, if the two of its zeroes are −√3 and √3.

Solution:

Given,

f(x) = x4 – 3x3 – x2 + 9x – 6

Since, two of the zeroes of polynomial are −√3 and √3 so, (x + √3) and (x–√3) are factors of f(x).

⇒ x2 – 3 is a factor of f(x). Hence, performing division algorithm, we get
R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 9

⇒ f(x)= (x2 – 3x + 2)( x2 – 3)

So, putting x2 – 3x + 2 = 0 we can get the other 2 zeros.

⇒ (x – 2)(x – 1) = 0

∴ x = 2 or 1

Hence, all the zeros of the polynomial are −√3, 1, √3 and 2.

6.  Obtain all zeroes of the polynomial f(x)= 2x4 – 2x3 – 7x2 + 3x + 6, if the two of its zeroes are −√(3/2) and √(3/2).

Solution:

Given,

f(x)= 2x4 – 2x3 – 7x2 + 3x + 6

Since, two of the zeroes of polynomial are −√(3/2) and √(3/2) so, (x + √(3/2)) and (x –√(3/2)) are factors of f(x).

⇒ x2 – (3/2) is a factor of f(x). Hence, performing division algorithm, we get

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 10

⇒ f(x)= (2x2 – 2x – 4)( x2 – 3/2)= 2(x2 – x – 2)( x2 – 3/2)

So, putting x2 – x – 2 = 0 we can get the other 2 zeros.

⇒ (x – 2)(x + 1) = 0

∴ x = 2 or -1

Hence, all the zeros of the polynomial are −√(3/2), -1, √(3/2) and 2.

7.  Find all the zeroes of the polynomial x4 + x3 – 34x2 – 4x + 120, if the two of its zeros are 2 and -2.

Solution:

Let,

f(x) = x4 + x3 – 34x2 – 4x + 120

Since, two of the zeroes of polynomial are −2 and 2 so, (x + 2) and (x – 2) are factors of f(x).

⇒ x2 – 4 is a factor of f(x). Hence, performing division algorithm, we get

R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 11

⇒ f(x)= (x2 + x – 30)( x2 – 4)

So, putting x2 + x – 30 = 0 we can get the other 2 zeros.

⇒ (x – 6)(x + 5) = 0

∴ x = 6 or -5

Hence, all the zeros of the polynomial are -5, -2, 2 and 6.


 

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