# RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

## RD Sharma Solutions Class 10 Maths Chapter 2 – Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 2 –Â PolynomialsÂ is provided here for students to study and excel in their board exams. Mathematics is one of the scoring subjects in Class 10. And thatâ€™s why we at BYJUâ€™S have created the RD Sharma Solutions. This is created by our expert faculty in order to provide a clear understanding of important concepts by giving detailed explanations for the problems. This is a very important study source for any student to get high marks in their Mathematics examinations.

Our specialist tutors formulate these RD Sharma Solutions for Class 10 Maths Chapter 2 to help you with your exam preparation to achieve good marks in Maths. This exercise can be used as a model of reference by the students to improve their conceptual knowledge and understand the different ways used to solve the problems. By practising RD Sharma Solutions for Class 10 Maths, students will be able to grasp the concepts perfectly. It also helps in boosting their confidence, which plays a crucial role in the examinations. Letâ€™s get an insight into this chapter to get a better idea of what itâ€™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

## Download RD Sharma Solutions For Class 10 Maths Chapter 2 Polynomials PDF

### RD Sharma Solutions for Class 10 Maths Chapter 2 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) = x2Â â€“ 2x â€“ 8

Solution:

Given,

f(x) = x2Â â€“ 2x â€“ 8

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

â‡’ x2Â â€“ 2x â€“ 8 = 0

â‡’Â  x2Â – 4x + 2x – 8 = 0

â‡’ x(x – 4) + 2(x – 4) = 0

â‡’ (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) = 4s2Â â€“ 4s + 1

Solution:

Given,

g(s) = 4s2Â â€“ 4s + 1

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

â‡’ 4s2Â â€“ 4s + 1 = 0

â‡’Â  4s2Â – 2s – 2s + 1= 0

â‡’ Â 2s(2s – 1) – (2s – 1)Â = 0

â‡’ (2s – 1)(2s â€“ 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)=t2Â â€“ 15

Solution:

Given,

h(t) = t2Â â€“ 15 = t2Â +(0)t â€“ 15

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

â‡’ t2Â â€“ 15 = 0

â‡’ (t + âˆš15)(t – âˆš15)= 0

This gives us 2 zeros, for

t = âˆš15 and t = -âˆš15

Hence, the zeros of the quadratic equation are âˆš15 and -âˆš15.

Now, for verification

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

âˆš15 + (-âˆš15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t2

âˆš15 x (-âˆš15) = -15/1

-15 = -15

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

(iv) f(x) = 6x2Â â€“ 3 â€“ 7x

Solution:

Given,

f(x) = 6x2Â â€“ 3 â€“ 7x

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

â‡’ 6x2Â â€“ 3 â€“ 7x = 0

â‡’Â  6x2Â – 9x + 2x – 3 = 0

â‡’ 3x(2x – 3) + 1(2x – 3) = 0

â‡’ (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) = x2Â + 2âˆš2x â€“ 6

Solution:

Given,

p(x) = x2Â + 2âˆš2x â€“ 6

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

â‡’ x2Â + 2âˆš2x â€“ 6 = 0

â‡’Â  x2Â + 3âˆš2x – âˆš2x – 6 = 0

â‡’ x(x + 3âˆš2) – âˆš2 (x + 3âˆš2) = 0

â‡’ (x – âˆš2)(x + 3âˆš2) = 0

This gives us 2 zeros, for

x = âˆš2 and x = -3âˆš2

Hence, the zeros of the quadratic equation are âˆš2 and -3âˆš2.

Now, for verification

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

âˆš2 + (-3âˆš2) = – (2âˆš2) / 1

-2âˆš2 = -2âˆš2

Product of roots = constant / coefficient of x2

âˆš2 x (-3âˆš2) = (-6) / 2âˆš2

-3 x 2 = -6/1

-6 = -6

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

(vi) q(x)=âˆš3x2Â + 10x + 7âˆš3

Solution:

Given,

q(x) = âˆš3x2Â + 10x + 7âˆš3

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

â‡’ âˆš3x2Â + 10x + 7âˆš3 = 0

â‡’Â  âˆš3x2Â + 3x +7x + 7âˆš3x = 0

â‡’ âˆš3x(x + âˆš3) + 7 (x + âˆš3) = 0

â‡’ (x + âˆš3)(âˆš3x + 7) = 0

This gives us 2 zeros, for

x = -âˆš3 and x = -7/âˆš3

Hence, the zeros of the quadratic equation are -âˆš3 and -7/âˆš3.

Now, for verification

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

-âˆš3 + (-7/âˆš3) = – (10) /âˆš3

(-3-7)/ âˆš3 = -10/âˆš3

-10/ âˆš3 = -10/âˆš3

Product of roots = constant / coefficient of x2

(-âˆš3) x (-7/âˆš3) = (7âˆš3) / âˆš3

7 = 7

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

(vii) f(x) = x2Â â€“ (âˆš3 + 1)x + âˆš3

Solution:

Given,

f(x) = x2Â â€“ (âˆš3 + 1)x + âˆš3

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

â‡’ x2Â â€“ (âˆš3 + 1)x + âˆš3 = 0

â‡’Â  x2Â – âˆš3x – x + âˆš3 = 0

â‡’ x(x – âˆš3) – 1 (x – âˆš3) = 0

â‡’ (x – âˆš3)(x – 1) = 0

This gives us 2 zeros, for

x = âˆš3 and x = 1

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, for verification

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

âˆš3 + 1 = – (-(âˆš3 +1)) / 1

âˆš3 + 1 = âˆš3 +1

Product of roots = constant / coefficient of x2

1 x âˆš3 = âˆš3 / 1

âˆš3 = âˆš3

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

(viii) g(x)=a(x2+1)â€“x(a2+1)

Solution:

Given,

g(x) = a(x2+1)â€“x(a2+1)

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

â‡’ a(x2+1)â€“x(a2+1) = 0

â‡’ ax2Â + a âˆ’ a2x â€“ x = 0

â‡’ ax2Â âˆ’ a2x â€“ x + a = 0

â‡’ ax(x âˆ’ a) âˆ’ 1(x â€“ a) = 0

â‡’ (x â€“ a)(ax â€“ 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) = 2s2Â â€“ (1 + 2âˆš2)s + âˆš2

Solution:

Given,

h(s) = 2s2Â â€“ (1 + 2âˆš2)s + âˆš2

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

â‡’ 2s2Â â€“ (1 + 2âˆš2)s + âˆš2 = 0

â‡’Â  2s2Â â€“ 2âˆš2s â€“ s + âˆš2 = 0

â‡’ 2s(sÂ â€“ âˆš2) -1(s – âˆš2) = 0

â‡’ (2s – 1)(s – âˆš2) = 0

This gives us 2 zeros, for

x = âˆš2 and x = 1/2

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, for verification

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

âˆš2 + 1/2 = – (-(1 + 2âˆš2)) / 2

(2âˆš2 + 1)/2 = (2âˆš2 +1)/2

Product of roots = constant / coefficient of s2

1/2 x âˆš2 = âˆš2 / 2

âˆš2 / 2 = âˆš2 / 2

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

(x) f(v) = v2Â + 4âˆš3v – 15

Solution:

Given,

f(v) = v2Â + 4âˆš3v â€“ 15

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

â‡’ v2Â + 4âˆš3v â€“ 15 = 0

â‡’Â  v2Â + 5âˆš3v – âˆš3v – 15 = 0

â‡’ v(v + 5âˆš3) – âˆš3 (v + 5âˆš3) = 0

â‡’ (v – âˆš3)(v + 5âˆš3) = 0

This gives us 2 zeros, for

v = âˆš3 and v = -5âˆš3

Hence, the zeros of the quadratic equation are âˆš3 and -5âˆš3.

Now, for verification

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

âˆš3 + (-5âˆš3) = – (4âˆš3) / 1

-4âˆš3 = -4âˆš3

Product of roots = constant / coefficient of v2

âˆš3 x (-5âˆš3) = (-15) / 1

-5 x 3 = -15

-15 = -15

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

(xi) p(y) = y2Â + (3âˆš5/2)y â€“ 5

Solution:

Given,

p(y) = y2Â + (3âˆš5/2)y â€“ 5

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

â‡’ y2Â + (3âˆš5/2)y â€“ 5 = 0

â‡’Â  y2Â – âˆš5/2 y + 2âˆš5y – 5 = 0

â‡’ y(y – âˆš5/2) + 2âˆš5 (y – âˆš5/2) = 0

â‡’ (y + 2âˆš5)(y – âˆš5/2) = 0

This gives us 2 zeros, for

y = âˆš5/2 and y = -2âˆš5

Hence, the zeros of the quadratic equation are âˆš5/2 and -2âˆš5.

Now, for verification

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

âˆš5/2 + (-2âˆš5) = – (3âˆš5/2) / 1

-3âˆš5/2 = -3âˆš5/2

Product of roots = constant / coefficient of y2

âˆš5/2 x (-2âˆš5) = (-5) / 1

– (âˆš5)2 = -5

-5 = -5

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

(xii) q(y) = 7y2Â – (11/3)y â€“ 2/3

Solution:

Given,

q(y) = 7y2Â – (11/3)y â€“ 2/3

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

â‡’ 7y2Â – (11/3)y â€“ 2/3 = 0

â‡’Â  (21y2 – 11y -2)/3 = 0

â‡’ 21y2 – 11y – 2 = 0

â‡’ 21y2 – 14y + 3y – 2 = 0

â‡’ 7y(3y – 2) â€“ 1(3y + 2) = 0

â‡’ (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,

â‡’ x2 – (-8/3)x + (4/3)

â‡’ x2 + 8/3x + (4/3)

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

â‡’ x2 + 8/3x + (4/3) = 0

â‡’ 3x2 + 8x + 4 = 0

â‡’ 3x2 + 6x + 2x + 4 = 0

â‡’ 3x(x + 2) + 2(x + 2) = 0

â‡’ (x + 2) (3x + 2) = 0

â‡’ (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,

â‡’ x2 – (21/8)x + (5/16)

â‡’ x2 – 21/8x + 5/16

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

â‡’ x2 – 21/8x + 5/16 = 0

â‡’ 16x2 – 42x + 5 = 0

â‡’ 16x2 – 40x – 2x + 5 = 0

â‡’ 8x(2x – 5) – 1(2x – 5) = 0

â‡’ (2x – 5) (8x – 1) = 0

â‡’ (2x – 5) = 0 and, or (8x – 1) = 0

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

(iii) -2âˆš3, -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 = -2âˆš3 and product of zero = -9

Thus,

The required polynomial f(x) is,

â‡’ x2 – (-2âˆš3)x + (-9)

â‡’ x2 + 2âˆš3x â€“ 9

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

â‡’ x2 + 2âˆš3x â€“ 9 = 0

â‡’ x2 + 3âˆš3x – âˆš3x â€“ 9 = 0

â‡’ x(x + 3âˆš3) – âˆš3(x + 3âˆš3) = 0

â‡’ (x + 3âˆš3) (x – âˆš3) = 0

â‡’ (x + 3âˆš3) = 0 and, or (x – âˆš3) = 0

Therefore, the two zeros are -3âˆš3and âˆš3.

(iv) -3/2âˆš5, -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 = -3/2âˆš5 and product of zero = -1/2

Thus,

The required polynomial f(x) is,

â‡’ x2 – (-3/2âˆš5)x + (-1/2)

â‡’ x2 + 3/2âˆš5x â€“ 1/2

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

â‡’ x2 + 3/2âˆš5x â€“ 1/2 = 0

â‡’ 2âˆš5x2 + 3x – âˆš5 = 0

â‡’ 2âˆš5x2 + 5x â€“ 2x – âˆš5 = 0

â‡’ âˆš5x(2x + âˆš5) – 1(2x + âˆš5) = 0

â‡’ (2x + âˆš5) (âˆš5x – 1) = 0

â‡’ (2x + âˆš5) = 0 and, or (âˆš5x – 1) = 0

Therefore, the two zeros are -âˆš5/2 and 1/âˆš5.

3. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(x) = x2 â€“ 5x + 4, find the value ofÂ 1/Î± + 1/Î² â€“ 2Î±Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (-5)/1 = 5

Product of the roots =Â Î±Î²Â = c/a = 4/1 = 4

To find, 1/Î± +1/Î² â€“ 2Î±Î²

â‡’ [(Î± +Î²)/ Î±Î²] â€“ 2Î±Î²

â‡’ (5)/ 4 â€“ 2(4) = 5/4 â€“ 8 = -27/ 4

4. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ p(y) = 5y2 â€“ 7y + 1, find the value ofÂ 1/Î±+1/Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (-7)/5 = 7/5

Product of the roots =Â Î±Î²Â = c/a = 1/5

To find, 1/Î± +1/Î²

â‡’ (Î± +Î²)/ Î±Î²

â‡’ (7/5)/ (1/5) = 7

5. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(x)=x2 â€“ x â€“ 4, find the value ofÂ 1/Î±+1/Î²â€“Î±Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (-1)/1 = 1

Product of the roots =Â Î±Î²Â = c/a = -4 /1 = – 4

To find, 1/Î± +1/Î² â€“ Î±Î²

â‡’ [(Î± +Î²)/ Î±Î²] â€“ Î±Î²

â‡’ [(1)/ (-4)] â€“ (-4) = -1/4 + 4 = 15/ 4

6. IfÂ Î± and Î²Â are the zeroes of the quadratic polynomialÂ f(x) = x2 + x â€“ 2, find the value ofÂ 1/Î± â€“ 1/Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (1)/1 = -1

Product of the roots =Â Î±Î²Â = c/a = -2 /1 = – 2

To find, 1/Î± – 1/Î²

â‡’ [(Î² – Î±)/ Î±Î²]

â‡’

7. If one of the zero of the quadratic polynomialÂ f(x) = 4x2 â€“ 8kx â€“ 9Â is negative of the other, then find the value of k.

Solution:

From the question, itâ€™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 Î± and â€“ Î±.

So, we can find

Sum of the roots =Â Î± – Î±Â = -b/a = – (-8k)/1 = 8k = 0 [âˆµ Î± – Î±Â = 0]

â‡’ k = 0

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

Solution:

Given,

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

And,

Sum of the roots = Product of the roots

â‡’ (-b/a) = (c/a)

â‡’ (-2/k) = (3k/k)

â‡’ (-2/k) = 3

âˆ´ k = -2/3

9. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ p(x) = 4x2 â€“ 5x â€“ 1, find the value ofÂ Î±2Î²+Î±Î²2.

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (-5)/4 = 5/4

Product of the roots =Â Î±Î²Â = c/a = -1/4

To find, Î±2Î²+Î±Î²2

â‡’ Î±Î²(Î± +Î²)

â‡’ (-1/4)(5/4) = -5/16

10. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(t)=t2â€“ 4t + 3, find the value ofÂ Î±4Î²3+Î±3Î²4.

Solution:

From the question, itâ€™s given that:

Î± and Î²Â 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 =Â Î±+Î²Â = -b/a = – (-4)/1 = 4

Product of the roots =Â Î±Î²Â = c/a = 3/1 = 3

To find, Î±4Î²3+Î±3Î²4

â‡’ Î±3Î²3 (Î± +Î²)

â‡’ (Î±Î²)3 (Î± +Î²)

â‡’ (3)3 (4) = 27 x 4 = 108

### RD Sharma Solutions for Class 10 Maths Chapter 2 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

â‡’ 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

â‡’ 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

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

Now,

Sum of zeros = -b/a

1/2 + 1 â€“ 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 (â€“ 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

â‡’ 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

â‡’ 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.

â‡’ f(x) = k [x3 â€“ (sum of roots)x2 + (sum of products of roots taken two at a time)x â€“ (product of roots)]

Where, k is any non-zero real number.

Here,

f(x) = k [x3 â€“ (3)x2 + (-1)x â€“ (-3)]

âˆ´ f(x) = k [x3 â€“ 3x2 – x + 3)]

where, k is any non-zero real number.

3. If the zeros of the polynomial f(x) = 2x3Â â€“ 15x2Â + 37x â€“ 30 are in A.P., find them.

Solution:

Let the zeros of the given polynomial be Î±, Î² and Î³. (3 zeros as itâ€™s a cubic polynomial)

And given, the zeros are in A.P.

So, letâ€™s consider the roots as

Î± = a â€“ d, Î² = a and Î³ = 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

â‡’ Sum of roots = Î± + Î² + Î³ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 â‡’ a = 5/2

â‡’ Product of roots = (a – d) x (a) x (a + d) = a(a2 â€“d2) = -d/a = -(30)/2 = 15

â‡’ a(a2 â€“d2) = 15

Substituting the value of a, we get

â‡’ (5/2)[(5/2)2 â€“d2] = 15

â‡’ 5[(25/4) â€“d2] = 30

â‡’ (25/4) â€“ d2 = 6

â‡’ 25 â€“ 4d2 = 24

â‡’ 1 = 4d2

âˆ´ 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

### RD Sharma Solutions for Class 10 Maths Chapter 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)Â f(x) = x3â€“ 6x2 + 11x â€“ 6, g(x) = x2 + x +1

Solution:

Given,

f(x) = x3â€“ 6x2 +11x â€“ 6, g(x) = x2 +x + 1

Thus,

q(x) = x â€“ 7 and r(x) = 17x +1

(ii) f(x) =Â 10x4 + 17x3 â€“ 62x2 + 30x â€“ 3, g(x) = 2x2Â + 7x + 1

Solution:

Given,

f(x) =Â 10x4 + 17x3 â€“ 62x2 + 30x â€“ 3 and g(x) = 2x2Â + 7x + 1

Thus,

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

(iii)Â f(x) = 4x3 + 8x2 + 8x + 7, g(x)= 2x2 â€“ x + 1

Solution:

Given,

f(x) = 4x3 + 8x2 + 8x + 7 and g(x)= 2x2 â€“ x + 1

Thus,

q(x) = 2x + 5 and r(x) = 11x + 2

(iv)Â f(x) = 15x3 â€“ 20x2 + 13x â€“ 12, g(x) = x2 â€“ 2x + 2

Solution:

Given,

f(x) = 15x3 â€“ 20x2 + 13x â€“ 12 and g(x) = x2 â€“ 2x + 2

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)Â g(t) = t2â€“3; f(t)=2t4 + 3t3 â€“ 2t2 â€“ 9t â€“ 12

Solution:

Given,

g(t) = t2 â€“ 3; f(t) =2t4 + 3t3 â€“ 2t2 â€“ 9t â€“ 12

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

(ii)Â g(x) = x3 â€“ 3x + 1; f(x) = x5 â€“ 4x3 + x2 + 3x + 1

Solution:

Given,

g(x) = x3 â€“ 3x + 1; f(x) = x5 â€“ 4x3 + x2 + 3x + 1

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) = 2x2â€“ x + 3; f(x) = 6x5 âˆ’ x4 + 4x3 â€“ 5x2 â€“ x â€“15

Solution:

Given,

g(x) = 2x2â€“ x + 3; f(x)=6x5 âˆ’ x4 + 4x3 â€“ 5x2 â€“ x â€“15

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 â€“ 14x2 â€“ 19xâ€“6, if two of its zeroes are -2 and -1.

Solution:

Given,

f(x)= 2x4 + x3 â€“ 14x2 â€“ 19x â€“ 6

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

â‡’ (x+2)(x+1) = x2 + x + 2x + 2 = x2 + 3x +2 â€¦â€¦ (i)

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

The quotient is 2x2 – 5x â€“ 3.

â‡’ f(x)= (2x2 – 5x â€“ 3)( x2 + 3x +2)

For obtaining the other 2 zeros of the polynomial

We put,

2x2 – 5x â€“ 3 = 0

â‡’ (2x + 1)(x â€“ 3) = 0

âˆ´ x = -1/2 or 3

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

4. Â 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

â‡’ f(x)= (x2 + 11x + 10)( x + 2)

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

â‡’ (x + 10)(x + 1) = 0

âˆ´ x = -10 or -1

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

5. Obtain all zeroes of the polynomialÂ f(x) = x4 â€“ 3x3 â€“ x2 + 9x â€“ 6, if the two of its zeroes areÂ âˆ’âˆš3 and âˆš3.

Solution:

Given,

f(x) = x4 â€“ 3x3 â€“ x2 + 9x â€“ 6

Since, two of the zeroes of polynomial areÂ âˆ’âˆš3 and âˆš3Â so,Â (x + âˆš3) and (xâ€“âˆš3)Â are factors of f(x).

â‡’ x2 â€“ 3 is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (x2 – 3x + 2)( x2 â€“ 3)

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

â‡’ (x – 2)(x – 1) = 0

âˆ´ x = 2 or 1

Hence, all the zeros of the polynomial are âˆ’âˆš3, 1, âˆš3 and 2.

6. Â Obtain all zeroes of the polynomialÂ f(x)= 2x4 â€“ 2x3 â€“ 7x2 + 3x + 6, if the two of its zeroes areÂ âˆ’âˆš(3/2) and âˆš(3/2).

Solution:

Given,

f(x)= 2x4 â€“ 2x3 â€“ 7x2 + 3x + 6

Since, two of the zeroes of polynomial areÂ âˆ’âˆš(3/2) and âˆš(3/2)Â so,Â (x + âˆš(3/2)) and (x â€“âˆš(3/2))Â are factors of f(x).

â‡’ x2 â€“ (3/2) is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (2x2 – 2x – 4)( x2 â€“ 3/2)= 2(x2 – x – 2)( x2 â€“ 3/2)

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

â‡’ (x – 2)(x + 1) = 0

âˆ´ x = 2 or -1

Hence, all the zeros of the polynomial are âˆ’âˆš(3/2), -1, âˆš(3/2) and 2.

7. Â Find all the zeroes of the polynomialÂ x4 + x3 â€“ 34x2 â€“ 4x + 120, if the two of its zeros are 2 and -2.

Solution:

Let,

f(x) = x4 + x3 â€“ 34x2 – 4x + 120

Since, two of the zeroes of polynomial areÂ âˆ’2 and 2Â so,Â (x + 2) and (x â€“ 2)Â are factors of f(x).

â‡’ x2 â€“ 4 is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (x2 + x – 30)( x2 â€“ 4)

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

â‡’ (x + 6)(x – 5) = 0

âˆ´ x = -6 or 5

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

## Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 2

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### What are Polynomials according to RD Sharma Solutions for Class 10 Maths Chapter 2?

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