# Class 10 RD SHARMA Solutions Maths Chapter 2 - Polynomials

## Polynomials Exercise Ex. 2.1

### Solution 1(i)

x^{2} - 2x - 8 = x^{2} - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)

The zeroes of the quadratic equation are 4 and -2.

Let ∝ = 4 and β = -2 Consider f(x) = x^{2} - 2x - 8

Sum of the zeroes = …(i)

Also, ∝ + β = 4 - 2 = 2 …(ii)

Product of the zeroes = …(iii)

Also, ∝ β = -8 …(iv)

Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(ii)

4s^{2}
- 4s + 1
= 4s^{2} - 2s - 2s + 1
= 2s(2s - 1) - (2s - 1)
= (2s - 1)(2s - 1)
The zeroes of the quadratic equation are and .
Let ∝ = and β =
Consider4s^{2} - 4s + 1
Sum of the zeroes = …(i)
Also, ∝ + β = …(ii)
Product of the zeroes = …(iii)
Also, ∝ β = …(iv)
Hence, from (i), (ii), (iii) and (iv),the relationship
between the zeroes and their coefficients is verified.

### Solution 1(iii)

h (t) = t^{2} - 15 = (t + √15)(t - √15)
The zeroes of the quadratic equation areand .
Let ∝ = and β =
Considert^{2} - 15 = t^{2} - 0t - 15
Sum of the zeroes = …(i)
Also, ∝ + β = …(ii)
Product of the zeroes = …(iii)
Also, ∝ β = …(iv)
Hence, from (i), (ii), (iii) and (iv),the relationship
between the zeroes and their coefficients is verified.

### Solution 1(iv)

f(s) = 0
6x^{2} - 3 - 7x =0 6x^{2} - 9x + 2x - 3 = 0
3x (2x - 3) + (2x - 3) = 0
(3x + 1) (2x - 3) = 0
The zeroes of a quadratic equation are and
.
Let ∝ = and β =
Consider6x^{2} - 7x - 3 = 0
Sum of the zeroes = …(i)
Also, ∝ + β = …(ii)
Product of the zeroes = …(iii)
Also, ∝ β = …(iv)
Hence, from (i), (ii), (iii) and (iv),the relationship
between the zeroes and their coefficients is verified.

### Solution 1(v)

The zeroes of a quadratic equation are and . Let ∝ = and β = Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(vi)

The zeroes of a quadratic equation are and . Let ∝ = and β = Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(vii)

The zeroes of a quadratic equation are and 1. Let ∝ = and β = 1 Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(viii)

The zeroes of a quadratic equation are a and. Let ∝ = a and β = Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(ix)

The zeroes of a quadratic equation are and . Let ∝ = and β = Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(x)

The zeroes of a quadratic equation are and . Let ∝ = and β = Consider=0 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(xi)

The zeroes of a quadratic equation are and. Let ∝ = and β = Consider=0 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 1(xii)

The zeroes of a quadratic equation are and. Let ∝ = and β = Consider=0 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 2(iv)

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

## Polynomials Exercise Ex. 2.2

### Solution 1

On comparing the given polynomial with the polynomial ax

^{3}+ bx

^{2}+ cx + d, we obtain a = 2, b = 1, c = -5, d = 2

Thus, the relationship between the zeroes and the coefficients is verified.

On comparing the given polynomial with the polynomial ax

^{3}+ bx

_{2}+ cx + d, we obtain a = 1, b = -4, c = 5, d = -2.

Thus, the relationship between the zeroes and the coefficients is verified.

**Concept insight:**The zero of a polynomial is that value of the variable which makes the polynomial 0. Remember that there are three relationships between the zeroes of a cubic polynomial and its coefficients which involve the sum of zeroes, product of all zeroes and the product of zeroes taken two at a time.

### Solution 2

### Solution 3

Let f(x) = 3x^{3} + 10x^{2} - 9x - 4

As 1 is one of the zeroes of the polynomial, so (x - 1) becomes the factor of f(x).

Dividing f(x) by (x - 1), we have

Hence, the zeroes are

### Solution 4

Let f(x) = x^{3} - 3x^{2} - 10x + 24

As 4 is one of the zeroes of the polynomial, so (x - 4) becomes the factor of f(x).

Dividing f(x) by (x - 4), we have

Hence, the zeroes are 4, -3 and 2.

### Solution 5

### Solution 6

## Polynomials Exercise Ex. 2.3

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

Let f(x) = 2x^{4} - 9x^{3} + 5x^{2} + 3x - 1

As are two of the zeroes of the polynomial, so becomes the factor of f(x).

Dividing f(x) by we have

Hence, the other zeros

### Solution 12

Let f(x) = 3x^{4} - 9x^{3} + x^{2} + 15x + k

As f(x) is
completely divisible by 3x^{2} - 5, it becomes one of the factors of f(x).

Dividing f(x) by 3x^{2}
- 5, we have

As (3x^{2}
- 5) is one of the factors, the remainder will be 0.

Therefore, k + 10 = 0

Thus, k = -10.

### Solution 13

### Solution 14

## Polynomials Exercise 2.61

### Solution 1

### Solution 2

So, the correct option is (d).

### Solution 3

## Polynomials Exercise 2.62

### Solution 4

So, the correct option is (b).

### Solution 5

### Solution 6

### Solution 7

Given that (α + 1) (β + 1) = 0

So, the correct option is (a).

### Solution 8

We know that, if the quadratic equation ax^{2} + bx + c = 0 has no real zeros

then

Case 1:

a > 0, the graph of quadratic equation should not intersect x - axis

It must be of the type

Case 2 :

a < 0, the graph will not intersect x - axis and it must be of type

According to the question,

a + b + c < 0

This means,

f(1) = a + b + c

f(1) < 0

Hence, f(0) < 0 [as Case 2 will be applicable]

So, the correct option is (c).

### Solution 9

### Solution 10

### Solution 11

### Solution 12

## Polynomials Exercise 2.63

### Solution 13

So, the correct option is (b).

### Solution 14

### Solution 15

So, the correct option is (c).

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

## Polynomials Exercise 2.64

### Solution 24

We know that, if α and β are roots of ax^{2 }+ bx + c = 0 then they must satisfy the equation.

According to the question, the equation is

x^{2 }- 5x + 4 = 0

If 3 is the root of equation it must satisfy equation.

x^{2 }- 5x + 4 = 0

but f(3) = 3^{2 }- 5(3) + 4 = -2

so, 2 has to be added in the equation.

So, the correct option is (b).

### Solution 25

We know that, if α and β are roots of ax^{2 } + bx + c = 0, then α and β must satisfy the equation.

According to the question, the equation is

x^{2 }- 16x + 30 = 0

If 15 is a root, then it must satisfy the equation x^{2 }- 16x + 30 = 0,

But f(15) = 15^{2 }- 16(15) + 30 = 225 - 240 + 30 = 15

and so 15 should be subtracted from the equation.

So, the correct option is (c).

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

We know that

Dividend = Divisor × quotient + remainder

Then according to question,

Required polynomial

= (-x^{2} + x - 1) (x - 2) + 3

= -x^{3} + 2x^{2} + x^{2} -2x - x + 2 + 3

= -x^{3} + 3x^{2} - 3x + 5

So, the correct option is (c).

### Solution 31

Correct option: (d)

The polynomials having -2 and 5 as the zeroes can be written in the form

k(x + 2)(x - 5), where k is a constant.

Thus, number of polynomials with roots -2 and 5 are infinitely many, since k can take infinitely many values.

### Solution 32

### Solution 33

The zeroes of the quadratic polynomial x^{2}
+ 99x + 127 are both negative since all terms are positive.

Hence, correct option is (b).

### Solution 34

### Solution 35

### Solution 36

### Solution 37

## Polynomials Exercise 2.65

### Solution 38

### Solution 39

### Solution 40

It is given that the zeros of the quadratic
polynomial ax^{2} + bx + c, c ≠ 0 are
equal.

⇒ Discriminant = 0

⇒
b^{2} - 4ac = 0

⇒
b^{2} = 4ac

Now, b^{2} can never be negative,

Hence, 4ac also can never be negative.

⇒ a and c should have same sign.

Hence, correct option is (c).

### Solution 41

### Solution 42

The graph of a quadratic polynomial crosses X-axis at atmost two points.

Hence, correct option is (d).