RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

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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) = x2 – 2x – 8

Solution:

Given,

f(x) = x2 – 2x – 8

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

⇒ x2 – 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) = 4s2 – 4s + 1

Solution:

Given,

g(s) = 4s2 – 4s + 1

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

⇒ 4s2 – 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)=t2 – 15

Solution:

Given,

h(t) = t2 – 15 = t2 +(0)t – 15

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

⇒ t2 – 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) = 6x2 – 3 – 7x

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

⇒ 6x2 – 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) = x2 + 2√2x – 6

Solution:

Given,

p(x) = x2 + 2√2x – 6

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

⇒ x2 + 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)=√3x2 + 10x + 7√3

Solution:

Given,

q(x) = √3x2 + 10x + 7√3

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

⇒ √3x2 + 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) = x2 – (√3 + 1)x + √3

Solution:

Given,

f(x) = x2 – (√3 + 1)x + √3

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

⇒ x2 – (√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

⇒ ax2 − 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) = 2s2 – (1 + 2√2)s + √2

Solution:

Given,

h(s) = 2s2 – (1 + 2√2)s + √2

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

⇒ 2s2 – (1 + 2√2)s + √2 = 0

⇒  2s2 – 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) = v2 + 4√3v – 15

Solution:

Given,

f(v) = v2 + 4√3v – 15

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

⇒ v2 + 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) = y2 + (3√5/2)y – 5

Solution:

Given,

p(y) = y2 + (3√5/2)y – 5

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

⇒ y2 + (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) = 7y2 – (11/3)y – 2/3

Solution:

Given,

q(y) = 7y2 – (11/3)y – 2/3

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

⇒ 7y2 – (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 factorisation.

(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√3 and √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]

⇒

7. If one of the zeros 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 α4β3+α3β4.

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, α4β3+α3β4

⇒ α3β3 (α +β)

⇒ (αβ)3 (α +β)

⇒ (3)3 (4) = 27 x 4 = 108

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RD Sharma Solutions for Class 10 Maths Chapter 2 Exercise 2.2 Page No: 2.43

1. Verify that the numbers given alongside 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, the sum of the product of its zeroes taken two at a time, and the 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 a 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

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

Solution:

Given,

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

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

Thus,

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

(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

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

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

Since the remainder r(x) = 2 and is 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

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, by performing the division algorithm, we get

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 the division algorithm, we get

⇒ f(x)= (x2 + 11x + 10)( x + 2)

So, by 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 the 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

⇒ f(x)= (x2 – 3x + 2)( x2 – 3)

So, by 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 the 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

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

So, by 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 the 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

⇒ f(x)= (x2 + x – 30)( x2 – 4)

So, by 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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Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 2

Q1

What is the use of practising RD Sharma Solutions for Class 10 Maths Chapter 2?

Practising RD Sharma Solutions for Class 10 Maths Chapter 2 provides students with an idea about the sample of questions that will be asked in the board exam, which would help students prepare competently. These solutions are useful resources which can provide them with all the vital information in the most precise form. These solutions cover all the topics included in the RD Sharma syllabus, prescribed by the current CBSE board.

Q2

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

Polynomials are algebraic expressions that consist of variables and coefficients. Variables are also sometimes called indeterminates. We can perform arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variables. By referring to these solutions, students get to clear their doubts instantly and also can exercise additional questions.

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Can RD Sharma Solutions for Class 10 Maths Chapter 2 be viewed only online?

For effective learning of concepts, the solutions have also been provided in PDF format so that the students can download them for free and refer to the solutions offline as well. These RD Sharma Solutions for Class 10 Maths Chapter 2 can be viewed online.

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