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# Class 10 RD SHARMA Solutions Maths Chapter 2 - Polynomials

## Polynomials Exercise Ex. 2.1

### Solution 1(i)

x2 - 2x - 8 = x2 - 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) = x2 - 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)

4s2 - 4s + 1 = 4s2 - 2s - 2s + 1 = 2s(2s - 1) - (2s - 1) = (2s - 1)(2s - 1) The zeroes of the quadratic equation are and . Let =  and β = Consider4s2 - 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) = t2 - 15 = (t + √15)(t - √15)  The zeroes of the quadratic equation areand . Let =  and β = Considert2 - 15 = t2 - 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 6x2 - 3 - 7x =0 6x2 - 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 = Consider6x2 - 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.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

## Polynomials Exercise Ex. 2.2

### Solution 1

On comparing the given polynomial with the polynomial ax3 + bx2 + 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 ax3 + bx2 + 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 3

Let f(x) = 3x3 + 10x2 - 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) = x3 - 3x2 - 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.

## Polynomials Exercise Ex. 2.3

### Solution 11

Let f(x) = 2x4 - 9x3 + 5x2 + 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) = 3x4 - 9x3 + x2 + 15x + k

As f(x) is completely divisible by 3x2 - 5, it becomes one of the factors of f(x).

Dividing f(x) by 3x2 - 5, we have

As (3x2 - 5) is one of the factors, the remainder will be 0.

Therefore, k + 10 = 0

Thus, k = -10.

## Polynomials Exercise 2.61

### Solution 2

So, the correct option is (d).

## Polynomials Exercise 2.62

### Solution 4

So, the correct option is (b).

### Solution 7

Given that (α + 1) (β + 1) = 0

So, the correct option is (a).

### Solution 8

We know that, if the quadratic equation ax2 + 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).

## Polynomials Exercise 2.63

### Solution 13

So, the correct option is (b).

### Solution 15

So, the correct option is (c).

## Polynomials Exercise 2.64

### Solution 24

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

According to the question, the equation is

x- 5x + 4 = 0

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

x- 5x + 4 = 0

but f(3) = 3- 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 + bx + c = 0, then α and β must satisfy the equation.

According to the question, the equation is

x- 16x + 30 = 0

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

But f(15) = 15- 16(15) + 30 = 225 - 240 + 30 = 15

and so 15 should be subtracted from the equation.

So, the correct option is (c).

### Solution 30

We know that

Dividend = Divisor × quotient  + remainder

Then according to question,

Required polynomial

= (-x2 + x - 1) (x - 2) + 3

= -x3 + 2x2 + x2 -2x - x + 2 + 3

= -x3 + 3x2 - 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 33

The zeroes of the quadratic polynomial x2 + 99x + 127 are both negative since all terms are positive.

Hence, correct option is (b).

## Polynomials Exercise 2.65

### Solution 40

It is given that the zeros of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal.

Discriminant = 0

b2 - 4ac = 0

b2 = 4ac

Now, b2 can never be negative,

Hence, 4ac also can never be negative.

a and c should have same sign.

Hence, correct option is (c).

### Solution 42

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

Hence, correct option is (d).