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

### Exercise 2C Solutions

1. If one zero of the polynomial x2 – 4x + 1. Is (2 + √3), write the other zero.

Solution

Let the other zeroes of x2 – 4x + 1 be a.

By using the relationship between the zeroes of the quadratic polynomial.

We have, sum of zeroes = (-coefficient of x)/(coefficient of x2)

∴ 2 + √3 + a = -(-4)/1

⇒ a = 2 – √3

Hence, the other zeroes of x2 – 4x + 1 is 2 – √3.

2. Find the zeroes of the polynomial x2 + x – p(p + 1)

Solution

f(x) = x2 + x – p (p + 1)

By adding and subtracting px, we get

f(x) = x2 + px + x – px – p(p + 1)

= x2 + (p + 1) x – px –p(p + 1)

= x[x + (p + 1)] –p[x + (p + 1)]

= [x + (p + 1)] (x – p)

f(x) = 0

⇒ [x + (p + 1)] (x – p) = 0

⇒ [x + (p + 1)] = 0 or (x – p) = 0

⇒ x = – (p + 1) or x = p

So, the zeroes of f(x) are – (p + 1) and p.

3. Find the zeroes of the polynomial x2 – 3x – m(m + 3)

Solution

f(x) = x2 – 3x – m (m + 3)

By adding and subtracting mx, we get

f(x) = x2 – mx – 3x + mx – m (m + 3)

= x[x – (m + 3)] + m[x – (m + 3)]

= [x – (m + 3)] (x + m)

f(x) = 0

⇒ [x – (m + 3)] (x + m) = 0

⇒ [x – (m + 3)] = 0 or (x + m) = 0

⇒ x = m + 3 or x = –m

So, the zeroes of f(x) are –m and +3.

4. Find α, β are the zeros of polynomial α + β = 6and αβ = 4 then write the polynomial.

Solution

If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as x2 – (α + β)x + αβ ...(1)

Substituting the values in (1), we get

x2 – 6x + 4

5. If one zero of the quadratic polynomial kx2 + 3x + k is 2, then find the value of k.

Solution

Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k

Therefore, it will satisfy the above polynomial.

Now, we have

k(2)2 + 3(2) + k = 0

⇒ 4k + 6 + k = 0

⇒ 5k + 6 = 0

⇒ k = – 6/5

6. If 3 is a zero of the polynomial 2x2 + x + k, find the value of k.

Solution

Given: x = 3 is one zero of the polynomial 2x2 + x + k

Therefore, it will satisfy the above polynomial.

Now, we have

2(3)2 + 3 + k = 0

⇒ 21 + k = 0

⇒ k = – 21

7. If -4 is a zero of the polynomial x2– x – (2k + 2) is –4, then find the value of k.

Solution

Given: x = –4 is one zero of the polynomial x2 – x – (2k + 2)

Therefore, it will satisfy the above polynomial.

Now, we have

(–4)2 – (–4) – (2k + 2) = 0

⇒ 16 + 4 – 2k – 2 = 0

⇒ 2k = – 18

⇒ k = 9

8. If 1is a zero of the quadratic polynomial ax2 – 3(a – 1)x – 1 is 1, then find the value of a.

Solution

Given: x = 1 is one zero of the polynomial ax2– 3(a – 1) x – 1

Therefore, it will satisfy the above polynomial.

Now, we have

a(1)2 – (a – 1)1 – 1 = 0

⇒ a – 3a + 3 – 1 = 0

⇒ –2a = – 2

⇒ a = 1

9. If -2 is a zero of the polynomial 3x2 + 4x + 2k then find the value of k.

Solution

Given: x = –2 is one zero of the polynomial 3x2 + 4x + 2k

Therefore, it will satisfy the above polynomial.

Now, we have

3(–2)2 + 4(–2)1 + 2k = 0

⇒ 12 – 8 + 2k = 0

⇒ k = – 2

10. Write the zeros of the polynomial f(x) = x2– x – 6.

Solution

f(x) = x2 – x – 6

= x2 – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3) (x + 2)

f(x) = 0 ⇒ (x – 3) (x + 2) = 0

⇒ (x – 3) = 0 or (x + 2) = 0

⇒ x = 3 or x = –2

So, the zeroes of f(x) are 3 and –2.

11. If the sum of the zeros of the quadratic polynomial kx2 - 3x + 5 is 1 write the value of k.

Solution

By using the relationship between the zeroes of the quadratic polynomial. We have

Sum of zeroes = (-coefficient of x)/(coefficient of x2)

⇒ 1 = -(-3)/k

⇒ k = 3

12. If the product of the zero of the polynomial (x2 - 4x + k) is 3. Find the value of k.

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have

Product of zeroes = (constant term)/(coefficient of x2)

⇒ 3 = k/1

⇒ k = 3

13. If (x + a) is a factor of (2x2 + 2ax + 5x + 10), then find the value of

Solution

Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10

We have

x + a = 0

⇒ x = –a

Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10

Hence, It will satisfy the above polynomial

∴ 2(–a)2 + 2a(–a) + 5(–a) + 10 = 0

⇒ –5a + 10 = 0

⇒ a = 2

14. If (a - b), a and (a + b) are zeros of the polynomial 2x3 - 6x2 + 5x - 7 write the value of a.

Solution

By using the relationship between the zeroes of the quadratic polynomial. We have

Sum of zeros = (-coefficient of x2)/(coefficient of x3)

⇒ a – b + a + a + b = (-6)/2

⇒ 3a = 3

⇒ a = 1

15. If x3+ x2- ax + b is divisible by (x2-x), write the value of a and b.

Solution

Equating x2– x to 0 to find the zeroes, we will get

x(x – 1) = 0

⇒ x = 0 or x – 1 = 0

⇒ x = 0 or x = 1

Since, x3 + x2 – ax + b is divisible by x2 – x.

Hence, the zeroes of x2– x will satisfy x3 + x2– ax + b

∴ (0)3 + 02 – a(0) + b = 0

⇒ b = 0

And

(1)3 + 12 – a(1) + 0 = 0   [∵ b = 0]

⇒ a = 2

16. If α and β be the zeroes of the polynomial 2x2- 7x + k write the value of (α + β + αβ).

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have

Sum of zeroes = -(coefficient of x)/(coefficient of x2) and Product of zeroes = (constant term)/(coefficient of x2)

∴ α + β = -7/2 and αβ = 5/2

Now, α + β + αβ = -7/2 + 5/2 = -1

17. State Division Algorithm for Polynomials.

Solution

“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, there exists unique polynomials q(x) and r(x) such that

f(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) ˂ degree of g(x).

18. Find the sum of the zeros and the product of zeros of a quadratic polynomial, are -1/2and -3 respectively. Write the polynomial.

Solution

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

x2 – (sum of the zeroes)x + product of zeroes

⇒ x2 – (-1/2)x + (–3)

⇒ x2 + 1/2x – 3

Hence, the required polynomial is x2 + 1/2x – 3.

19. Find the zeroes of the quadratic polynomial f(x) = 6x2 – 3.

Solution

To find the zeroes of the quadratic polynomial we will equate f(x) to 0

∴ f(x) = 0

⇒ 6x2 – 3 = 0

⇒ 3(2x2 – 1) = 0

⇒ 2x2 – 1 = 0

⇒ 2x2 = 1

⇒ x2 = 1/2

⇒ x = ± 1/√2

Hence, the zeroes of the quadratic polynomial f(x) = 6x2 – 3 are 1/√2, -(1/√2)

20. Find the zeroes of the quadratic polynomial f(x) = 4√3x2 + 5x – 2√3.

Solution

To find the zeroes of the quadratic polynomial we will equate f(x) to 0

∴ f(x) = 0

⇒ 4√3x2 + 5x – 2√3 = 0

⇒ 4√3x2 + 8x – 3x – 2√3 = 0

⇒ 4x(√3x + 2) – √3(√3x + 2) = 0

⇒ (√3x + 2) = 0 or (4x – √3) = 0

⇒ x = -2/√3or x = √3/4

Hence, the zeroes of the quadratic polynomial f(x) = 4√3x2 + 5x – 2√3 are –2√3 or √3/4

21. If α, β are the zeroes of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find the value of k = ?

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have

Sum of zeroes = (-coefficient of x)/(coefficient of x2) and Product of zeros = (constant term)/(coefficient of x2)

∴ α + β = -(-5)/1 and αβ = k/1

⇒ α + β = 5 and αβ = k/1

Solving α – β = 1 and α + β = 5, we will get

α = 3 and β = 2

Substituting these values in αβ = k/1, we will get

k = 6

22. If α and β are the zeros of the polynomial f(x) = 6x2 + x – 2 find the value of (α/β + α/β)

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have

Sum of zeroes = (-coefficient of x)/(coefficient of x2) and product of zeros = (constant term)/(coefficient of x2)

∴ α + β = (-1)/6 and αβ = 1/3

Now, α/β + α/β = (α2 + β2)/αβ

= (α2 + β2 + 2αβ -2αβ)/αβ

= {(α + β)2 - 2αβ}/αβ

= {(-1/6)2 – 2-(1/3)}/-(1/3)

= (1/36 + 2/3)/-(1/3)

= -(25/12)

23. If α, β are the zeroes of the polynomial f(x) = 5x- 7x + 1, then 1/α + 1/β = ?

Solution

By using the relationship between the zeroes of the quadratic polynomial. We have

Sum of zeroes = (-coefficient of x)/(coefficient of x) and Product of zeroes = (constant term)/(coefficient of x2)

∴ α + β = -(-7)/1 and αβ = 1/5

⇒ α + β = 7/5 and αβ = 1/5

Now, 1/α + 1/β = (α + β)/αβ

= (7/5)/(1/5)

= 7

24. If α, β are the zeroes of the polynomial f(x) = x2– x + 2, (α/β - α/β).

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have

Sum of zeroes = -(coefficient of x)/(coefficient of x2) and Product of zeroes = (constant term)/(coefficient of x2)

∴ α + β = -1/1 and αβ = -2/1

⇒ α + β = -1 and αβ = -2

Now, (1/α – 1/β)2 = (β – α)2/αβ

= (α + β)2 - 4αβ/(αβ)2 [∵ (β – α)2 = (α + β)2 - 4αβ]

= {(-1)2 – 4(-2)}/(-2)2 [∵ α + β = -1 and αβ = -2]

= [(-1)2 – 4(-2)}/4

= 9/4

∵ (1/α – 1/β)2 = 9/4

⇒ 1/α – 1/β = ± 3/2

25. If the zeroes of the polynomial f(x) = x3 – 3x2 + x + 1 are (a – b), a and (a + b), find the values of a and b.

Solution

By using the relationship between the zeroes of the quadratic polynomial.

We have, Sum of zeroes = -(coefficient of x2)/(coefficient of x2)

∴ a – b + a + a + b = -(-3)/1

⇒ 3a = 3

⇒ a = 1

Now, Product of zeroes = -(constant term)/(coefficient of x3)

∴ (a – b) (a) (a + b) = -1/1

⇒ (1 – b) (1) (1 + b) = –1 [∵ a = 1]

⇒ 1 – b2 = –1

⇒ b2 = 2

⇒ b = ± √2