In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 2 Polynomials Ex 2C Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 2 Polynomials Ex 2C Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C Maths book pdf download. Now you will get step by step solution to each question.

**RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Ex 2C**

**Question 1.****Solution:**

Let other zero of x^{2} – 4x + 1 be a, then

Sum of zeros = −ba = −(−4)1 = 4

But one zero is 2 + √3

Second zero = 4 – (2 + √3) =4 – 2 – √3 = 2 – √3

**Question 2.****Solution:**

Let f(x) = x^{2} + x – p(p + 1)

= x^{2} + (p + 1) x – px – p(p + 1)

= x(x + p + 1) – p(x + p + 1)

= (x + p + 1) (x – p)

Either x + p + 1 = 0, then x = -(p + 1)

or x – p = 0, then x = p

Hence, zeros are p and -(p + 1)

**Question 3.****Solution:**

p(x) = x^{2} – 3x – m(m + 3)

= x^{2} – (m + 3)x + mx – m(m + 3)

= x(x – m – 3) + m(x – m – 3)

= (x – m – 3)(x + m)

Either x – m – 3 = 0, then x = m + 3

or x + m = 0, then x = -m

Zeros are (m + 3), -m

**Question 4.****Solution:**

a and p are the zeros of a polynomial

and α + β = 6, αβ = 4

Polynomial = x^{2} – (α + β)x + αβ = x^{2} – (6)x + 4 = x^{2} – 6x + 4

**Question 5.****Solution:**

One zero of kx^{2} + 3x + k is 2

x = 2 will satisfy it

⇒ k(2)^{2} + 3 x 2 + k = 0

⇒ 4k + 6 + k= 0

⇒5k + 6 = 0

⇒ 5k = -6

⇒ k = −65

Hence, k = −65

**Question 6.****Solution:**

3 is a zero of the polynomial 2x^{2} + x + k

Then 3 will satisfy it

2x^{2} + x + k = 0

⇒ 2(3)^{2} + 3 + k = 0

⇒ 18 + 3+ k = 0

⇒ 21 + k = 0

⇒ k = -21

Hence, k = -21

**Question 7.****Solution:**

-4 is a zero of polynomial x^{2} – x – (2k + 2)

Then it will satisfy the equation

x^{2} – x – (2k + 2) = 0

⇒ (-4)2 – (-4) – 2k – 2 = 0

⇒ 16 + 4 – 2k – 2 = 0

⇒ -2k + 18 = 0

⇒ 2k = 18

k = 9

**Question 8.****Solution:**

1 is a zero of the polynomial ax^{2} – 3(a – 1)x – 1

Then 1 will satisfy the equation ax^{2} – 3(a – 1) x – 1 = 0

a(1)^{2} – 3(a – 1) x 1 – 1 = 0

⇒ a x 1 – 3a + 3 – 1 = 0

⇒ a – 3a + 2 = 0

⇒ -2a + 2 = 0

⇒ 2a = 2

⇒ a = 1

**Question 9.****Solution:**

-2 is a zero of 3x^{2} + 4x + 2k

It will satisfy the equation 3x^{2} + 4x + 2k = 5

3(-2)^{2} + 4(-2) + 2k = 0

⇒ 3 x 4 + 4(-2) + 2k = 0

⇒ 12 – 8 + 2k = 0

⇒ 4 + 2k=0

⇒ 2k = -4

⇒ k = -2

k = -2

**Question 10.****Solution:**

Let f(x) = x^{2} – x – 6

= x^{2} – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3)(x + 2)

(x – 3)(x + 2) = 0

Either x – 3 = 0, then x = 3

or x + 2 = 0, then x = -2

Zeros are 3, -2

**Question 11.****Solution:**

Sum of zeros = 1

and polynomial is kx^{2} – 3x + 5

Sum of zeros = −ba = −(−3)k = 3k

3k = 1

⇒ k = 3

Hence, k = 3

**Question 12.****Solution:**

Product of zeros of polynomial x^{2} – 4x + k is 3

Product of zeros = ca

⇒ k1 = 3

⇒ k = 3

**Question 13.****Solution:**

x + a is a factor of

f(x) = 2x^{2} + (2a + 5) x + 10

Let x + a = 0, then

Zero of f(x) = -a

Now f(-a) = 2 (-a)^{2} + (2a + 5)(-a) + 10 = 0

2a^{2} – 2a^{2} – 5a + 10 = 0

⇒ 5a = 10

⇒ a = 2

**Question 14.****Solution:**

(a – b), a, (a + b) are the zeros of 2x^{3} – 6x^{2} + 5x – 7

Sum of zeros = −ba

⇒ a – b + a + a + b = −(−6)2

⇒ 3a = 62

⇒ 3a = 3

⇒ a = 1

**Question 15.****Solution:**

f(x) = x^{3} + x^{2} – ax + 6 is divisible by x^{2} – x

Remainder will be zero

Now dividing f(x) by x^{2} – x

Remainder = (2 – a) x + b

(2 – a) x + b = 0

2 – a = 0

⇒ a = 2 and b = 0

Hence, a = 2, b = 0

**Question 16.****Solution:**

α and β are the zeros of polynomial f(x) = 2x^{2} + 7x + 5

**Question 17.****Solution:****Division algorithm for polynomials:**

If f(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomial q(x) and r(x).

f(x) = q(x) x g(x) + r(x)

where r (x) = 0

or [degree of r(x) < degree of g(x)]

or Dividend=Quotient x Division + Remainder

**Question 18.****Solution:**

Sum of zeros = −12

Product of zeros = -3

Polynomial: x^{2} – (Sum of zeros) x + product of zeros

**Short-Answer Questions****Question 19.****Solution:**

**Question 20.****Solution:**

**Question 21.****Solution:**

α and β are the zeros of polynomial f(x) = x^{2} – 5x + k

(1)^{2} = (5)^{2} – 4 k

1 ⇒ 25 – 4k

⇒ 4k = 25 – 1 = 24

Hence, k = 6

**Question 22.****Solution:**

**Question 23.****Solution:**

α and β are the zeros of polynomial

f(x) = 5x^{2} – 7x + 1

**Question 24.****Solution:**

**Question 25.****Solution:**

(a – b), a and (a + b) are the zeros of the polynomial

f(x) = x^{3} – 3x^{2} + x + 1

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