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 useful tool for any student to get top marks in their Mathematics examinations.

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

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. Letâ€™s get an insight to 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

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

### 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 ^{2Â }â€“ 2x â€“ 8**

**Solution: **

Given,

f(x) = x^{2Â }â€“ 2x â€“ 8

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

â‡’ x^{2Â }â€“ 2x â€“ 8 = 0

â‡’Â x^{2}Â – 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 x^{2}

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x^{2}

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

-8 = -8

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

**(ii) g(s) = 4s ^{2Â }â€“ 4s + 1**

**Solution: **

Given,

g(s) = 4s^{2Â }â€“ 4s + 1

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

â‡’ 4s^{2Â }â€“ 4s + 1 = 0

â‡’Â 4s^{2}Â – 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 s^{2}

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s^{2}

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 ^{2Â }â€“ 15**

**Solution: **

Given,

h(t) = t^{2Â }â€“ 15 = t^{2Â }+(0)t â€“ 15

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

â‡’ t^{2Â }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 t^{2}

âˆš15 + (-âˆš15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t^{2}

âˆš15 x (-âˆš15) = -15/1

-15 = -15

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

**(iv) f(x) = 6x ^{2}Â â€“ 3 â€“ 7x**

**Solution: **

Given,

f(x) = 6x^{2Â }â€“ 3 â€“ 7x

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

â‡’ 6x^{2Â }â€“ 3 â€“ 7x = 0

â‡’Â 6x^{2}Â – 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 x^{2}

3/2 + (-1/3) = – (-7) / 6

7/6 = 7/6

Product of roots = constant / coefficient of x^{2}

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 ^{2Â }+ 2âˆš2x â€“ 6**

**Solution: **

Given,

p(x) = x^{2Â }+ 2âˆš2x â€“ 6

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

â‡’ x^{2Â }+ 2âˆš2x â€“ 6 = 0

â‡’Â x^{2}Â + 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 x^{2}

âˆš2 + (-3âˆš2) = – (2âˆš2) / 1

-2âˆš2 = -2âˆš2

Product of roots = constant / coefficient of x^{2}

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)=âˆš3x ^{2Â }+ 10x + 7âˆš3**

**Solution: **

Given,

q(x) = âˆš3x^{2Â }+ 10x + 7âˆš3

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

â‡’ âˆš3x^{2Â }+ 10x + 7âˆš3 = 0

â‡’Â âˆš3x^{2}Â + 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 x^{2}

-âˆš3 + (-7/âˆš3) = – (10) /âˆš3

(-3-7)/ âˆš3 = -10/âˆš3

-10/ âˆš3 = -10/âˆš3

Product of roots = constant / coefficient of x^{2}

(-âˆš3) x (-7/âˆš3) = (7âˆš3) / âˆš3

7 = 7

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

**(vii) f(x) = x ^{2Â }â€“ (âˆš3 + 1)x + âˆš3**

**Solution: **

Given,

f(x) = x^{2Â }â€“ (âˆš3 + 1)x + âˆš3

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

â‡’ x^{2Â }â€“ (âˆš3 + 1)x + âˆš3 = 0

â‡’Â x^{2}Â – 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 x^{2}

âˆš3 + 1 = – (-(âˆš3 +1)) / 1

âˆš3 + 1 = âˆš3 +1

Product of roots = constant / coefficient of x^{2}

1 x âˆš3 = âˆš3 / 1

âˆš3 = âˆš3

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

** (viii) g(x)=a(x ^{2}+1)â€“x(a^{2}+1)**

**Solution: **

Given,

g(x) = a(x^{2}+1)â€“x(a^{2}+1)

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

â‡’ a(x^{2}+1)â€“x(a^{2}+1) = 0

â‡’ ax^{2}Â + a âˆ’ a^{2}x â€“ x = 0

â‡’ ax^{2Â }âˆ’ a^{2}x 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 x^{2}

a + 1/a = – (-(a^{2 }+ 1)) / a

(a^{2} + 1)/a = (a^{2} + 1)/a

Product of roots = constant / coefficient of x^{2}

a x 1/a = a / a

1 = 1

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

**(ix) h(s) = 2s ^{2Â }â€“ (1 + 2âˆš2)s + âˆš2**

**Solution: **

Given,

h(s) = 2s^{2Â }â€“ (1 + 2âˆš2)s + âˆš2

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

â‡’ 2s^{2Â }â€“ (1 + 2âˆš2)s + âˆš2 = 0

â‡’Â 2s^{2Â }â€“ 2âˆš2s â€“ s + âˆš2 = 0

â‡’ 2s(s^{Â }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 s^{2}

âˆš2 + 1/2 = – (-(1 + 2âˆš2)) / 2

(2âˆš2 + 1)/2 = (2âˆš2 +1)/2

Product of roots = constant / coefficient of s^{2}

1/2 x âˆš2 = âˆš2 / 2

âˆš2 / 2 = âˆš2 / 2

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

**(x) f(v) = v ^{2Â }+ 4âˆš3v – 15**

**Solution: **

Given,

f(v) = v^{2Â }+ 4âˆš3v â€“ 15

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

â‡’ v^{2Â }+ 4âˆš3v â€“ 15 = 0

â‡’Â v^{2}Â + 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 v^{2}

âˆš3 + (-5âˆš3) = – (4âˆš3) / 1

-4âˆš3 = -4âˆš3

Product of roots = constant / coefficient of v^{2}

âˆš3 x (-5âˆš3) = (-15) / 1

-5 x 3 = -15

-15 = -15

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

**(xi) p(y) = y ^{2Â }+ (3âˆš5/2)y â€“ 5**

**Solution: **

Given,

p(y) = y^{2Â }+ (3âˆš5/2)y â€“ 5

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

â‡’ y^{2Â }+ (3âˆš5/2)y â€“ 5 = 0

â‡’Â y^{2}Â – 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 y^{2}

âˆš5/2 + (-2âˆš5) = – (3âˆš5/2) / 1

-3âˆš5/2 = -3âˆš5/2

Product of roots = constant / coefficient of y^{2}

âˆš5/2 x (-2âˆš5) = (-5) / 1

– (âˆš5)^{2} = -5

-5 = -5

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

**(xii) q(y) = 7y ^{2Â }– (11/3)y â€“ 2/3**

**Solution: **

Given,

q(y) = 7y^{2Â }– (11/3)y â€“ 2/3

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

â‡’ 7y^{2Â }– (11/3)y â€“ 2/3 = 0

â‡’Â (21y^{2} – 11y -2)/3 = 0

â‡’ 21y^{2} – 11y – 2 = 0

â‡’ 21y^{2} – 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 y^{2}

2/3 + (-1/7) = – (-11/3) / 7

-11/21 = -11/21

Product of roots = constant / coefficient of y^{2}

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) = x^{2} + -(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‡’ x^{2} – (-8/3)x + (4/3)

â‡’ x^{2} + 8/3x + (4/3)

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

â‡’ x^{2} + 8/3x + (4/3) = 0

â‡’ 3x^{2} + 8x + 4 = 0

â‡’ 3x^{2} + 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) = x^{2} + -(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‡’ x^{2} – (21/8)x + (5/16)

â‡’ x^{2} – 21/8x + 5/16

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

â‡’ x^{2} – 21/8x + 5/16 = 0

â‡’ 16x^{2} – 42x + 5 = 0

â‡’ 16x^{2} – 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) = x^{2} + -(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‡’ x^{2} – (-2âˆš3)x + (-9)

â‡’ x^{2} + 2âˆš3x â€“ 9

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

â‡’ x^{2} + 2âˆš3x â€“ 9 = 0

â‡’ x^{2} + 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) = x^{2} + -(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‡’ x^{2} – (-3/2âˆš5)x + (-1/2)

â‡’ x^{2} + 3/2âˆš5x â€“ 1/2

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

â‡’ x^{2} + 3/2âˆš5x â€“ 1/2 = 0

â‡’ 2âˆš5x^{2} + 3x – âˆš5 = 0

â‡’ 2âˆš5x^{2} + 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) = x ^{2 }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) = 5y ^{2 }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)=x ^{2} 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) = x ^{2 }+ 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) = 4x ^{2 }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)=kt ^{2 }+ 2t + 3kÂ is equal to their product, then find the value of k.**

**Solution: **

Given,

The quadratic polynomial f(t)=kt^{2 }+ 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) = 4x ^{2 }â€“ 5x â€“ 1, find the value ofÂ Î±^{2}Î²+Î±Î²^{2}.**

**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±^{2}Î²+Î±Î²^{2}

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)=t ^{2}â€“ 4t + 3, find the value ofÂ Î±^{4}Î²^{3}+Î±^{3}Î²^{4}.**

**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±^{4}Î²^{3}+Î±^{3}Î²^{4}

â‡’ Î±^{3}Î²^{3 }(Î± +Î²)

â‡’ (Î±Î²)^{3 }(Î± +Î²)

â‡’ (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) = 2x ^{3 }+ x^{2 }– 5x + 2; 1/2, 1, -2**

**Solution: **

Given, f(x) = 2x^{3 }+ x^{2 }– 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) = x ^{3 }– 4x^{2 }+ 5x – 2; 2, 1, 1**

**Solution: **

Given, g(x) = x^{3 }– 4x^{2 }+ 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 ax^{3 }+ bx^{2 }+ cx + d.

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

â‡’ f(x) = k [x^{3} â€“ (sum of roots)x^{2} + (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 [x^{3} â€“ (3)x^{2} + (-1)x â€“ (-3)]

âˆ´ f(x) = k [x^{3} â€“ 3x^{2} – x + 3)]

where, k is any non-zero real number.

**3.** **If the zeros of the polynomial f(x) = 2x ^{3}Â â€“ 15x^{2}Â + 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(a^{2} â€“d^{2}) = -d/a = -(30)/2 = 15

â‡’ a(a^{2} â€“d^{2}) = 15

Substituting the value of a, we get

â‡’ (5/2)[(5/2)^{2} â€“d^{2}] = 15

â‡’ 5[(25/4) â€“d^{2}] = 30

â‡’ (25/4) â€“ d^{2} = 6

â‡’ 25 â€“ 4d^{2} = 24

â‡’ 1 = 4d^{2}

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

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) = x ^{3}â€“ 6x^{2 }+ 11x â€“ 6, g(x) = x^{2 }+ x +1**

**Solution: **

Given,

f(x) = x^{3}â€“ 6x^{2 }+11x â€“ 6, g(x) = x^{2 }+x + 1

Thus,

q(x) = x â€“ 7 and r(x) = 17x -1

**(ii) f(x) =Â 10x ^{4 }+ 17x^{3 }â€“ 62x^{2 }+ 30x â€“ 3, g(x) = 2x^{2}Â + 7x + 1**

**Solution:**

**Given,**

**f(x) =Â **10x^{4 }+ 17x^{3 }â€“ 62x^{2 }+ 30x â€“ 3 and g(x) = 2x^{2}**Â + 7x + 1**

**Thus,**

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

**(iii)Â f(x) = 4x ^{3} + 8x^{2} + 8x + 7, g(x)= 2x^{2 }â€“ x + 1**

**Solution:**

Given,

f(x) = 4x^{3} + 8x^{2} + 8x + 7 and g(x)= 2x^{2 }â€“ x + 1

**Thus,**

q(x) = 15x + 10 and r(x) = 3x – 32

**(iv)Â f(x) = 15x ^{3 }â€“ 20x^{2} + 13x â€“ 12, g(x) = x^{2 }â€“ 2x + 2**

**Solution:**

Given,

f(x) = 15x^{3 }â€“ 20x^{2} + 13x â€“ 12 and g(x) = x^{2 }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) = t ^{2}â€“3; f(t)=2t^{4} + 3t^{3} â€“ 2t^{2} â€“ 9t â€“ 12**

**Solution: **

Given,

g(t) = t^{2 }â€“ 3; f(t) =2t^{4} + 3t^{3} â€“ 2t^{2} 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) = x ^{3} â€“ 3x + 1; f(x) = x^{5} â€“ 4x^{3} + x^{2} + 3x + 1**

**Solution: **

Given,

g(x) = x^{3} â€“ 3x + 1; f(x) = x^{5} â€“ 4x^{3} + x^{2} + 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) = 2x ^{2}â€“ x + 3; f(x) = 6x^{5 }âˆ’ x^{4} + 4x^{3} â€“ 5x^{2} â€“ x â€“15**

**Solution: **

Given,

g(x) = 2x^{2}â€“ x + 3; f(x)=6x^{5 }âˆ’ x^{4} + 4x^{3} â€“ 5x^{2} 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)= 2x ^{4 }+ x^{3 }â€“ 14x^{2 }â€“ 19xâ€“6, if two of its zeroes are -2 and -1.**

**Solution:**

Given,

f(x)= 2x^{4 }+ x^{3 }â€“ 14x^{2 }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) = x^{2 }+ x + 2x + 2 = x^{2 }+ 3x +2 â€¦â€¦ (i)

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

The quotient is 2x^{2} – 5x â€“ 3.

â‡’ f(x)= (2x^{2} – 5x â€“ 3)( x^{2 }+ 3x +2)

For obtaining the other 2 zeros of the polynomial

We put,

2x^{2} – 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) = x ^{3 }+ 13x^{2 }+ 32x + 20, if one of its zeros is -2.**

**Solution: **

Given,

**f(x)= **x^{3 }+ 13x^{2 }+ 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)= (x^{2} + 11x + 10)( x + 2)

So, putting x^{2} + 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) = x ^{4} â€“ 3x^{3} â€“ x^{2} + 9x â€“ 6, if the two of its zeroes areÂ âˆ’âˆš3 and âˆš3.**

**Solution: **

**Given, **

f(x) = x^{4} â€“ 3x^{3} â€“ x^{2} + 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‡’ x^{2 }â€“ 3 is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (x^{2} – 3x + 2)( x^{2 }â€“ 3)

So, putting x^{2} – 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)= 2x ^{4 }â€“ 2x^{3 }â€“ 7x^{2 }+ 3x + 6, if the two of its zeroes areÂ âˆ’âˆš(3/2) and âˆš(3/2).**

**Solution: **

**Given, **

f(x)= 2x^{4 }â€“ 2x^{3 }â€“ 7x^{2 }+ 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‡’ x^{2 }â€“ (3/2) is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (2x^{2} – 2x – 4)( x^{2 }â€“ 3/2)= 2(x^{2} – x – 2)( x^{2 }â€“ 3/2)

So, putting x^{2} – 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 x ^{4 }+ x^{3 }â€“ 34x^{2 }â€“ 4x + 120, if the two of its zeros are 2 and -2.**

**Solution: **

**Let, **

f(x) = x^{4} + x^{3} â€“ 34x^{2} – 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‡’ x^{2 }â€“ 4 is a factor of f(x). Hence, performing division algorithm, we get

â‡’ f(x)= (x^{2} + x – 30)( x^{2 }â€“ 4)

So, putting x^{2} + 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.