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

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

### Exercise 2A Solutions

1. Find the zeros of the polynomial f(x) = x2 + 7x + 12  and verify the relation between its zeroes and coefficients.

Solution

x2 + 7x + 12 = 0

⇒ x2 + 4x + 3x + 12 = 0

⇒ x(x + 4) + 3(x + 4) = 0

⇒ (x + 4) (x + 3) = 0

⇒ (x + 4) = 0 or (x + 3) = 0

⇒ x = −4 or x = −3

Sum of zeroes = −4 + (−3) = -7/1 = -coefficient of x/coefficient of x2

Product of zeroes = (−4) (−3) = 12/1 = constant term/coefficient of x2

2. Find the zeroes of the polynomial f(x) = x2 ˗ 2x ˗ 8 and verify the relation between its zeroes and coefficients.

Solution

x2˗ 2x ˗ 8 = 0

⇒ x2˗ 4x + 2x ˗ 8 = 0

⇒ x(x ˗ 4) + 2(x ˗ 4) = 0

⇒ (x ˗ 4) (x + 2) = 0

⇒ (x ˗ 4) = 0 or (x + 2) = 0

⇒ x = 4 or x = −2

Sum of zeroes = 4 + (−2) = 2 = 2/1 = -(coefficient of x)/(coefficient of x2)

Product of zeros = (4)(-2) = -8/1 = (constant term)/(coefficient of x2)

3. Find the zeroes of the quadratic polynomial f(x) = x2 + 3x ˗ 10 and verify the relation between its zeroes and coefficients.

Solution

We have:

f(x) = x2 + 3x ˗ 10

= x2 + 5x ˗ 2x ˗ 10

= x(x + 5) ˗ 2(x + 5)

= (x ˗ 2) (x + 5)

∴ f(x) = 0

⇒ (x ˗ 2) (x + 5) = 0

⇒ x ˗ 2 = 0 or x + 5 = 0

⇒ x = 2 or x = −5.

So, the zeroes of f(x) are 2 and −5.

Sum of zeroes = 2 + (−5) = −3 = -3/1 = -(coefficient of x)/(coefficient of x2)

Product of zeroes = 2 × (−5) = −10 = -10/1 = (constant term)/(coefficient of x2)

4. Find the zeroes of the quadratic polynomial f(x) = 4x2˗ 4x ˗ 3 and verify the relation between its zeroes and coefficients.

Solution

We have:

f(x) = 4x2˗ 4x ˗ 3

= 4x2˗ (6x ˗ 2x) ˗ 3

= 4x2˗ 6x + 2x ˗ 3

= 2x (2x ˗ 3) + 1(2x ˗ 3)

= (2x + 1) (2x ˗ 3)

∴ f(x) = 0

⇒ (2x + 1) (2x ˗ 3) = 0

⇒ 2x + 1 = 0 or 2x ˗ 3 = 0

⇒ x = -1/2 or x = 3/2

So, the zeroes of f(x) are -1/2 and 3/2.

Sum of zeroes = (-1/2) + (3/2)

= (-1 + 3)/2

= 2/2

= 1 = -(coefficient of x)/(coefficient of x2)

Product of zeros = (-1/2) × (3/2)

= -3/4 = (constant term)/(coefficient of x2)

5. Find the zeroes of the quadratic polynomial f(x) = 5x2˗ 4 ˗ 8x and verify the relationship between the zeroes and coefficients of the given polynomial.

Solution

We have,

f(x) = 5x2˗ 4 ˗ 8x

= 5x2 ˗ 8x ˗ 4

= 5x2˗ (10x ˗ 2x) ˗ 4

= 5x2˗ 10x + 2x ˗ 4

= 5x (x ˗ 2) + 2(x ˗ 2)

= (5x + 2)(x ˗ 2)

∴ f(x) = 0

⇒ (5x + 2) (x ˗ 2) = 0

⇒ 5x + 2 = 0 or x ˗ 2 = 0

⇒ x = -2/5 or x = 2

So, the zeroes of f(x) are -2/5 and 2.

Sum of zeroes = (-2/5) + 2

= (-2 + 10)/5

= 8/5 = -(coefficient of x)/(coefficient of x2)

Product of zeros = (-2/5 × 2)

= -4/5 = (constant term)/(coefficient of x2)

6. Find the zeroes of the polynomial f(x) = 2√3x2 -5x + √3 and verify the relation between its zeroes and coefficients.

Solution

2√3x2 ˗ 5x + √3

⇒ 2√3x2 ˗ 2x ˗ 3x + √3

⇒ 2x (√3x ˗ 1) ˗ √3 (√3x ˗ 1) = 0

⇒ (√3x ˗ 1) or (2x− √3) = 0

⇒ (√3x ˗ 1) = 0 or (2x − √3) = 0

⇒ x = 1/√3 or x = √3/2

⇒ x = 1/√3 × 1/√3 = √3/3 or x = √3/2

Sum of zeros = √3/3 + √3/2 + (5√3)/6 = -(coefficient of x)/(coefficient of x2)

Product of zeros = √3/3 × √3/2 = √3/6 = (constant term)/(coefficient of x2)

7. Find the zeroes of the quadratic polynomial 2x2 - 11x + 15 and verify the relation between the zeroes and the coefficients.

Solution

f(x) = 2x2˗ 11x + 15

= 2x2˗ (6x + 5x) + 15

= 2x2˗ 6x ˗ 5x + 15

= 2x(x ˗ 3) ˗ 5(x ˗ 3)

= (2x ˗ 5)(x ˗ 3)

∴ f(x) = 0

⇒ (2x ˗ 5) (x ˗ 3) = 0

⇒ 2x ˗ 5 = 0 or x ˗ 3 = 0

⇒ x = 5/2 or x = 3

So, the zeroes of f(x) are 5/2 and 3.

Sum of zeros = 5/2 + 3

= (5 + 6)/2

= 11/2 = -(coefficient of x)/(coefficient of x2)

Product of zeros = 5/2 × 3

= -15/2 = (constant term)/(coefficient of x2)

8. Find the zeroes of the quadratic polynomial 4x2˗ 4x + 1 and verify the relation between the zeroes and the coefficients.

Solution

4x2 ˗ 4x + 1 = 0

⇒ (2x)2 ˗ 2(2x)(1) + (1)2 = 0

⇒ (2x ˗ 1)2 = 0  [∵ a2 – 2ab + b2 = (a–b)2

⇒ (2x ˗ 1)2 = 0

⇒ x = 1/2 or x = 1/2

Sum of zeroes = 1/2 + 1/2

= 1

= 1/1 = -(coefficient of x)/(coefficient of x2)

Product of zeroes = 1/2 × 1/2

= 1/4 = (constant term)/(coefficient of x2)

9. Find the zeroes of the quadratic polynomial (x2 ˗ 5) and verify the relation between the zeroes and the coefficients.

Solution

We have:

f(x) = x2˗ 5

It can be written as x2 + 0x ˗ 5.

= (x2 − (√5)2

= (x + √5) (x − √5)

∴ f(x) = 0 ⇒ (x + √5) (x − √5) = 0

⇒ x + √5 = 0 or x − √5 = 0

⇒ x = −√5 or x = √5

So, the zeroes of f(x) are −√5 and √5.

Here, the coefficient of x is 0 and the coefficient of x2 is 1.

Sum of zeroes = -√5 + √5

= 1

= 0/1 = -(coefficient of x)/(coefficient of x2)

Product of zeroes = -√5 × √5

= -5/1 = (constant term)/(coefficient of x2)

10. Find the zeroes of the quadratic polynomial (8x2˗ 4) and verify the relation between the zeroes and the coefficients.

Solution

We have:

f(x) = 8x2 ˗ 4

It can be written as 8x2 + 0x ˗ 4

= 4{(√2x)2˗ (1)2

= 4(√2x + 1)(√2x ˗ 1)

∴ f(x) = 0

⇒ (√2x + 1) (√2x ˗ 1) = 0

⇒ (√2x + 1) = 0 or √2x ˗ 1 = 0

⇒ x = -1/√2 or x = 1/√2

So, the zeroes of f(x) are -1/√2 and 1/√2

Here the coefficient of x is 0 and the coefficient of x2 is √2

Sum of zeroes = -1/√2 + 1/√2

= (-1 + 1)/√2

= 0/√2 = -(coefficient of x)/(coefficient of x2)

Product of zeros = -1/√2 × 1/√2

= (-1 × 4)/(2 × 4)

= -4/8 = (constant term/(coefficient of x2)

11. Find the zeroes of the quadratic polynomial (5y2 + 10y) and verify the relation between the zeroes and the coefficients.

Solution

We have,

f (u) = 5u2 + 10u

It can be written as 5u (u + 2)

∴ f (u) = 0

⇒ 5u = 0 or u + 2 = 0

⇒ u = 0 or u = −2

So, the zeroes of f (u) are −2 and 0.

Sum of the zeroes = −2 + 0

= −2 = (-2 × 5)/(1 × 5)

= -10/5 = (-coefficient of x)/(coefficient of u2)

Product of zeros = -2 × 0

= 0

= (0 × 5)/(1 × 5)

= -0/5 = (constant term)/(-coefficient of u2)

12. Find the zeroes of the quadratic polynomial (3x2˗ x ˗ 4) and verify the relation between the zeroes and the coefficients.

Solution

3x2˗ x ˗ 4 = 0

⇒ 3x2˗ 4x + 3x ˗ 4 = 0

⇒ x (3x ˗ 4) + 1 (3x ˗ 4) = 0

⇒ (3x ˗ 4) (x + 1) = 0

⇒ (3x ˗ 4) or (x + 1) = 0

⇒ x = 4/3 or x = ˗ 1

Sum of zeroes = 4/3 + (-1)

= 1/3 = -(coefficient of x)/(coefficient of x2)

Product of zeroes = 4/3 × (-1)

= -4/3 = (constant term)/(coefficient of x2)

13. Find the quadratic polynomial whose zeroes are 2 and -6. Verify the relation between the coefficients and the zeroes of the polynomial.

Solution

Let α = 2 and β = -6

Sum of the zeroes, (α + β) = 2 + (-6) = -4

Product of the zeros, αβ = 2 × (-6) = -12

∴ Required polynomial = x2 – (α + β)x + αβ

= x2 - (-4x) – 12

= x2 + 4x – 12

Sum of zeros = -4

= -4/1 (-coefficient of x)/(coefficient of x2)

Product of zeros = -12

= -12/1 = (constant term)/(coefficient of x2)

14. Find the quadratic polynomial whose zeroes are 2/3 and -1/4 . Verify the relation between the coefficients and the zeroes of the polynomial.

Solution

Let = α = 2/3 and β = -1/4

Sum of the zeroes = (α + β) = 2/3 + (-1/4)

= (8 – 3)/12

= 5/12

Product of zeroes = αβ = 2/3 × (-1/4)

= -2/12

= -1/6

∴ Required polynomial = x2 – (α + β)x + αβ

= x2 – 5/12x + (-1/6)

= x2 – 5/12x – 1/6

Sum of the zeros = 5/12 = -(coefficient of x)/(coefficient of x2)

Product of zeros = -1/6 = (constant term)/(coefficient of x2)

15. Find the quadratic polynomial, sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.

Solution

Let α and β be the zeroes of the required polynomial f(x).

Then (α + β) = 8 and αβ = 12

∴ f(x) = x2˗ (α + β)x + αβ

⇒ f(x) = x2˗ 8x + 12

Hence, required polynomial f(x) = x2 ˗ 8x + 12

∴ f(x) = 0

⇒ x2 ˗ 8x + 12 = 0

⇒ x2˗ (6x + 2x) + 12 = 0

⇒ x2˗ 6x ˗ 2x + 12 = 0

⇒ x (x – 6) – 2 (x – 6) = 0

⇒ (x – 2) (x – 6) = 0

⇒ (x – 2) = 0 or (x – 6) = 0

⇒ x = 2 or x = 6

So, the zeroes of f(x) are 2 and 6.

16. Find the quadratic polynomial, sum of whose zeroes is 0 and their product is -1. Hence, find the zeroes of the polynomial.

Solution

Let α and β be the zeroes of the required polynomial f(x).

Then (α + β) = 0 and αβ = -1

∴ f(x) = x2 ˗ (α + β)x + αβ

⇒ f(x) = x2˗ 0x + (-1)

⇒ f(x) = x2 ˗ 1

Hence, required polynomial f(x) = x2˗ 1.

∴ f(x) = 0

⇒ x2˗ 1 = 0

⇒ (x + 1) (x – 1) = 0

⇒ (x + 1) = 0 or (x – 1) = 0

⇒ x = -1 or x = 1

So, the zeroes of f(x) are -1 and 1.

17. Find the quadratic polynomial, sum of whose zeroes is (5/2) and their product is 1. Hence, find the zeroes of the polynomial.

Solution

Let α and β be the zeroes of the required polynomial f(x).

Then (α + β) = 5/2 and αβ = 1

∴ f(x) = x- (α + β) x + αβ

⇒ f(x) = x2 -5/2x + 1

⇒ f(x) = 2x– 5x + 2

Hence, the required polynomial is f(x) = 2x– 5x + 2

∴ f(x) = 0 ⇒ 2x– 5x + 2 = 0

⇒ 2x– (4x + x) + 2 = 0

⇒ 2x– 4x – x + 2 = 0

⇒ 2x (x – 2) – 1 (x – 2) = 0

⇒ (2x – 1) (x – 2) = 0

⇒ (2x – 1) = 0 or (x – 2) = 0

⇒ x = 1/2 or x = 2

So, the zeros of f(x) are 1/2 and 2.

18. Find the quadratic polynomial, sum of whose zeroes is √2 and their product is (1/3).

Solution

We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula

x– (Sum of the roots)x + Product of roots = 0

⇒ x– √2x + 1/3 = 0

⇒ 3x– 3√2x + 1 = 0

19. If x = 2/3 and x = -3 are the roots of the quadratic equation ax3 + 2ax + 5x + 10 then find the value of a and b.

Solution

Given: ax+ 7x + b = 0

Since, x = 2/3 is the root of the above quadratic equation

Hence, it will satisfy the above equation.

Therefore, we will get

a(2/3)2 + 7(2/3) + b = 0

⇒ 4/9a + 14/3 + b = 0

⇒ 4a + 42 + 9b = 0

⇒ 4a + 9b = – 42 …(1)

Since, x = –3 is the root of the above quadratic equation

Hence, It will satisfy the above equation.

Therefore, we will get

a (–3)+ 7 (–3) + b = 0

⇒ 9a – 21 + b = 0

⇒ 9a + b = 21 ...(2)

From (1) and (2), we get

a = 3, b = –6

20. If (x + a) is a factor of the polynomial 2x+ 2ax + 5x + 10, find the value of a.

Solution

Given: (x + a) is a factor of 2x+ 2ax + 5x + 10

So, we have

x + a = 0

⇒ x = – a

Now, it will satisfy the above polynomial.

Therefore, we will get

2 (–a)2 + 2a(–a) + 5(–a) + 10 = 0

⇒ 2a2 –2a2 – 5a + 10 = 0

⇒ – 5a = – 10

⇒ a = 2

21. One zero of the polynomial 3x3 + 16x2 + 15x – 18 is 2/3. Find the other zeros of the polynomial.

Solution

Given: x = 2/3 is one of the zero of 3x3 + 16x2 + 15x – 18

Now, we have

x = 2/3

⇒ x – 2/3 = 0

Now, we divide 3x3 + 16x2 + 15x – 18 by x – 2/3 to find the quotient

So, the quotient is 3x2 + 18x + 27

Now,

3x2 + 18x + 27 = 0

⇒ 3x2 + 9x + 9x + 27 = 0

⇒ 3x(x + 3) + 9(x + 3) = 0

⇒ (x + 3) (3x + 9) = 0

⇒ (x + 3) = 0 or (3x + 9) = 0

⇒ x = –3 or x = –3