# RS Aggarwal Solutions Chapter 2 Polynomials Exercise - 2B Class 10 Maths

 Chapter Name RS Aggarwal Chapter 2 Polynomials Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 2AExercise 2C Related Study NCERT Solutions for Class 10 Maths

### Exercise 2B Solutions

1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x3– 2x2– 5x + 6) and verify the relation between it zeros and coefficients.

Solution

The given polynomial is p(x) = (x3 – 2x2 – 5x + 6)

∴ p(3) = (33 – 2×32 - 5×3 + 6) = (27–18–15 + 6) = 0

p(-2) = [(–23) – 2×(–2)2 – 5×(–2) + 6] = (–8 –8 +10+6) = 0

p(1) = (13 – 2×12 – 5×1 +6) = (1-2–5+6) = 0

∴ 3, –2 and 1are the zeroes of p(x),

Let α = 3, β = –2 and γ = 1. Then we have:

(α + β + γ) = (3 – 2 + 1) = 2 = (-coefficient of x2)/(coefficient of x3)

(αβ + βγ + γα) = (-6 – 2 + 3) = -5/1 = (coefficient of x)/(coefficient of x3)

αβγ = {(-3 × (-2) × 1} = -6/1 = (-coefficient term)/(coefficient of x3)

2. Verify that 5, -2 and 13are the zeroes of the cubic polynomial p(x) = (3x3 – 10x2 – 27x + 10) and verify the relation between its zeroes and coefficients.

Solution

p(x) = (3x3 – 10x2 – 27x + 10)

p(5) = (3×53 – 10×52 – 27×5 + 10) = (375–250–135+10) = 0

p(–2) = [3×(–23) – 10×(–22) – 27×(–2) + 10] = (–24 – 40 + 54 + 10) = 0

p(1/3) = {3×(1/3)3 – 10×(1/3)2 – 27× 1/3 + 10 } = (3× 1/27 – 10× 1/9 – 9 + 10)

= (1/9 – 10/9 + 1) = (1 – 10 – 9)/9 = (0/9) = 0

∴ 5, -2 and 1/3 are zeros of p(x).

Let α = 5, β = –2 and γ = 1/3. Then we have:

(α + β + γ) = (5 – 2 + 1/3) = 10/3 = (-coefficient of x2)/(coefficient of x3)

(αβ + βγ + γα) = (-10 – 2/3 + 5/3) = -27/3 = (coefficient of x)/(coefficient of x3)

αβγ = {5 × (-2) × 1/3} = -10/3 = (-coefficient term)/(coefficient of x3)

3. Find a cubic polynomial whose zeroes are 2, -3 and 4.

Solution

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as

x3 – (a + b + c)x2 + (ab + bc + ca)x – abc …(1)

Let a = 2, b = –3 and c = 4

Substituting the values in 1, we get

x3 – (2 – 3 + 4)x2 + (– 6 – 12 + 8)x – (–24)

⇒ x3 – 3x2 – 10x + 24

4. Find a cubic polynomial whose zeroes are 1/2, 1 and –3.

Solution

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as

x3 – (a + b + c)x2 + (ab + bc + ca)x – abc …(1)

Let a = 1/2, b = 1 and c = –3

Substituting the values in (1), we get

x3 – (1/2 + 1 − 3)x2 + (1/2 – 3 – 3/2)x – (-3/2)

⇒ x3 – (-3/2)x2 – 4x + 3/2

⇒ 2x3 + 3x2 – 8x + 3

5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively.

Solution

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as

x3 – (sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x – product of zeroes

Therefore, the required polynomial is

x3– 5x2 – 2x + 24

6. If f(x) = x3 – 3x + 5x – 3 divided by g(x) = x2 – 2

Solution

Quotient q(x) = x – 3

Remainder r(x) = 7x – 9

7. If f(x) = x4 – 3x2 + 4x + 5 is divided by g(x)= x2 – x + 1

Solution

Quotient q(x) = x2 + x – 3

Remainder r(x) = 8

8. If f(x) = x4 – 5x + 6 is divided by g(x) = 2 – x2

Solution

We can write

f(x) as x4 + 0x3 + 0x2 – 5x + 6 and g(x) as – x2 + 2

Quotient q(x) = – x2– 2

Remainder r(x) = –5x + 10

9. By actual division, show that x2 – 3 is a factor of 2x4 + 3x3 – 2x2 – 9x – 12.

Solution

Let f(x) = 2x4 + 3x3– 2x2– 9x – 12 and g(x) as x2 – 3

Quotient q(x) = 2x2 + 3x + 4

Remainder r(x) = 0

Since, the remainder is 0.

Hence, x2– 3 is a factor of 2x4 + 3x3– 2x2– 9x – 12

10. On dividing 3x3 + x2 + 2x + 5 is divided by a polynomial g(x), the quotient and remainder are (3x – 5) and (9x + 10) respectively. Find g(x).

Solution

By using division rule, we have

Dividend = Quotient × Divisor + Remainder

∴ 3x3 + x2 + 2x + 5 = (3x – 5)g(x) + 9x + 10

⇒ 3x3 + x2 + 2x + 5 – 9x – 10 = (3x – 5)g(x)

⇒ 3x3 + x2 – 7x – 5 = (3x – 5)g(x)

⇒ g(x) = (3x3 + x2 – 7x – 5)/(3x – 5)

∴ g(x) = x2 + 2x + 1

11. Verify division algorithm for the polynomial f(x) = (8 + 20x + x2 – 6x3) by g(x) = (2 + 5x – 3x2).

Solution

We can write f(x) as – 6x3 + x2 + 20x + 8 and g(x) as –3x2 + 5x + 2

Quotient = 2x + 3

Remainder = x + 2

By using division rule, we have

Dividend = Quotient × Divisor + Remainder

∴ – 6x3 + x2 + 20x + 8 = (– 3x2 + 5x + 2) (2x + 3) + x + 2

⇒ – 6x3 + x2 + 20x + 8 = –6x3 + 10x2 + 4x – 9x2 + 15x + 6 + x + 2

⇒ – 6x3 + x2 + 20x + 8 = – 6x3 + x2 + 20x + 8

12. It is given that –1 is one of the zeroes of the polynomial x3 + 2x2– 11x – 12. Find all the zeroes of the given polynomial.

Solution

Let f(x) = x3 + 2x2 – 11x – 12

Since – 1 is a zero of f(x), (x + 1) is a factor of f(x).

On dividing f(x) by (x + 1), we get

f(x) = x3 + 2x2– 11x – 12

= (x + 1)(x2 + x – 12)

= (x + 1){x2 + 4x – 3x – 12}

= (x + 1){x(x + 4) – 3(x + 4)}

= (x + 1)(x – 3)(x + 4)

∴ f(x) = 0 ⇒ (x + 1)(x – 3)(x + 4) = 0

⇒ (x + 1) = 0 or (x – 3) = 0 or (x + 4) = 0

⇒ x = – 1 or x = 3 or x = – 4

Thus, all the zeroes are – 1, 3 and – 4.

13. If 1 and –2 are two zeroes of the polynomial (x3– 4x2– 7x + 10), find its third zero.

Solution

Let f(x) = x3– 4x2– 7x + 10

Since 1 and –2 are the zeroes of f(x), it follows that each one of (x – 1) and (x + 2) is a factor of f(x).

Consequently, (x – 1)(x + 2) = (x2 + x – 2) is a factor of f(x).

On dividing f(x) by (x2 + x – 2), we get:

f(x) = 0 ⇒ (x2 + x – 2)(x – 5) = 0

⇒ (x – 1)(x + 2)(x – 5) = 0

⇒ x = 1 or x = – 2 or x = 5

Hence, the third zero is 5.

14. If 3 and –3 are two zeroes of the polynomial (x4 + x3– 11x2– 9x + 18), find all the zeroes of the given polynomial.

Solution

Let x4 + x3 – 11x2 – 9x + 18

Since 3 and – 3 are the zeroes of f(x), it follows that each one of (x + 3) and (x – 3) is a factor of f(x).

Consequently, (x – 3) (x + 3) = (x2– 9) is a factor of f(x).

On dividing f(x) by (x2– 9), we get:

f(x) = 0 ⇒ (x2 + x – 2) (x2 – 9) = 0

⇒ (x2 + 2x – x – 2) (x – 3) (x + 3)

⇒ (x – 1) (x + 2) (x – 3) (x + 3) = 0

⇒ x = 1 or x = -2 or x = 3 or x = -3

Hence, all the zeroes are 1, -2, 3 and -3.

15. If 2 and -2 are two zeroes of the polynomial (x4 + x3– 34x2– 4x + 120), find all the zeroes of the given polynomial.

Solution

Let f(x) = x4 + x3– 34x2– 4x + 120

Since 2 and -2 are the zeroes of f(x), it follows that each one of (x – 2) and (x + 2) is a factor of f(x).

Consequently, (x – 2) (x + 2) = (x2– 4) is a factor of f(x).

On dividing f(x) by (x2– 4), we get:

f(x) = 0

⇒ (x2 + x – 30) (x2 – 4) = 0

⇒ (x2 + 6x – 5x – 30) (x – 2) (x + 2)

⇒ [x(x + 6) – 5(x + 6)] (x – 2) (x + 2)

⇒ (x – 5) (x + 6) (x – 2) (x + 2) = 0

⇒ x = 5 or x = -6 or x = 2 or x = -2

Hence, all the zeroes are 2, -2, 5 and -6.

16. Find all the zeroes of (x4 + x3 – 23x2 – 3x + 60), if it is given that two of its zeroes are √3 and –√3.

Solution

Let f(x) = x4 + x3 – 23x2 – 3x + 60

Since √3 and –√3 are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).

Consequently, (x – √3) (x + √3) = (x2 – 3) is a factor of f(x).

On dividing f(x) by (x2 – 3), we get:

f(x) = 0

⇒ (x2 + x – 20) (x2 – 3) = 0

⇒ (x2 + 5x – 4x – 20)(x2 – 3)

⇒ [x(x + 5) – 4(x + 5)](x2 – 3)

⇒ (x – 4) (x + 5) (x – √3)(x + √3) = 0

⇒ x = 4 or x = -5 or x = √3 or x = -√3

Hence, all the zeroes are √3, -√3, 4 and -5.

17. Find all the zeroes of (2x4 – 3x3 – 5x2 + 9x – 3), it is being given that two of its zeroes are √3 and –√3.

Solution

The given polynomial is f(x) = 2x4 – 3x3 – 5x2 + 9x – 3

Since √3 and –√3 are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).

Consequently, (x – √3) (x + √3) = (x2 – 3) is a factor of f(x).

On dividing f(x) by (x2 – 3), we get:

f(x) = 0

⇒ 2x4 – 3x3 – 5x2 + 9x – 3 = 0

⇒ (x2 – 3) (2x2 – 3x + 1) = 0

⇒ (x2 – 3) (2x2 – 2x – x + 1) = 0

⇒ (x – √3) (x + √3) (2x – 1) (x – 1) = 0

⇒ x = √3 or x = -√3 or x = 1/2 or x = 1

Hence, all the zeroes are √3, -√3, 1/2 and 1.

18. Obtain all other zeroes of (x4 + 4x3 – 2x2– 20x – 15) if two of its zeroes are √5 and –√5.

Solution

The given polynomial is f(x) = x4 + 4x3– 2x2– 20x – 15.

Since (x – √5) and (x + √5) are the zeroes of f(x) it follows that each one of (x – √5) and

(x + √5) is a factor of f(x).

Consequently, (x – √5) (x + √5) = (x2 – 5) is a factor of f(x).

On dividing f(x) by (x2 – 5), we get:

f(x) = 0

⇒ x4 + 4x3 – 7x2 – 20x – 15 = 0

⇒ (x2 – 5) (x2 + 4x + 3) = 0

⇒ (x – √5) (x + √5) (x + 1) (x + 3) = 0

⇒ x = √5 or x = -√5 or x = -1 or x = -3

Hence, all the zeroes are √5, -√5, -1 and -3.

19. Find all the zeroes of polynomial (2x4– 11x3 + 7x2 + 13x – 7), it being given that two of its zeroes are (3 + √2) and (3 – √2).

Solution

The given polynomial is f(x) = 2x4 – 11x3 + 7x2 + 13x – 7.

Since (3 + √2) and (3 – √2) are the zeroes of f(x) it follows that each one of (x + 3 + √2) and (x + 3 – √2) is a factor of f(x).

Consequently, [(x – (3 + √2)] [(x – (3 – √2)] = [(x – 3) - √2] [(x – 3) + √2] = [(x – 3)2 – 2] = x2 – 6x + 7, which is a factor of f(x).

On dividing f(x) by (x2 – 6x + 7), we get:

f(x) = 0

⇒ 2x4 – 11x3 + 7x2 + 13x – 7 = 0

⇒ (x2 – 6x + 7) (2x2 + x – 7) = 0

⇒ (x + 3 + √2) (x + 3 – √2) (2x – 1) (x + 1) = 0

⇒ x = –3 – √2 or x = –3 + √2 or x = 12or x = -1

Hence, all the zeroes are (–3 – √2), (–3 + √2), 1/2 and -1.