## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Class 10 Solutions Chapter 2 Polynomials MCQS

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2A
- RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2B
- RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Ex 2C
- RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS
- RS Aggarwal Solutions Class 10 Chapter 2 Polynomials Test Yourself

**Choose the correct answer in each of the following questions.**

**Question 1.**

**Solution:**

**(d)** √2 x^{2} – 3√3 x + √6 is polynomial, others are not polynomial.

**Question 2.**

**Solution:**

**(d)** x + \(\frac { 3 }{ x }\) is not a polynomial, other are polynomial.

**Question 3.**

**Solution:**

**(c)** Let f(x) = x^{2} – 2x – 3

= x^{2} – 3x + x – 3

= x(x – 3) + 1(x – 3)

= (x – 3)(x + 1)

If x – 3 = 0, then x – 3

and if x + 1 = 0, then x = -1

Zeros are 3, -1

**Question 4.**

**Solution:**

**(b)**

**Question 5.**

**Solution:**

**(c)**

**Question 6.**

**Solution:**

**(b)** Polynomial is x^{2} + \(\frac { 1 }{ 6 }\) x – 2

**Question 7.**

**Solution:**

**(a)**

**Question 8.**

**Solution:**

**(c)** Sum of zeros = 3

Product of zeros = -10

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

= x^{2} – 3x – 10

**Question 9.**

**Solution:**

**(c)** Zeros are 5 and -3

Sum of zeros = 5 – 3 = 2

Product of zeros = 5 x (-3) = -15

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

= x^{2} – 2x – 15

**Question 10.**

**Solution:**

**(d)**

**Question 11.**

**Solution:**

**(b)** Let f(x) = x^{2} + 88x +125

Here, sum of roots = \(\frac { -b }{ a }\) = -88

and product = \(\frac { c }{ a }\) = 125

Product is positive,

Both zeros can be both positive or both negative.

Sum is negative.

Both zeros are negative.

**Question 12.**

**Solution:**

**(b)** α and β are the zeros of x^{2} + 5x + 8

Then sum of zeros (α + β) = \(\frac { -b }{ a }\) = \(\frac { -5 }{ 1 }\) = -5

**Question 13.**

**Solution:**

**(c)** α and β are the zeros of 2x^{2} + 5x – 9

Product of zeros (αβ) = \(\frac { c }{ a }\) = \(\frac { -9 }{ 2 }\)

**Question 14.**

**Solution:**

**(d)** 2 is a zero of kx^{2} + 3x + k

It will satisfy the quadratic equation kx^{2} + 3x + k = 0

k(2)^{2} + 3x^{2} + 1 = 0

4k + 6 + k = 0

=> 5k = -6

k = \(\frac { -6 }{ 5 }\)

**Question 15.**

**Solution:**

**(b)** -4 is a zero of (k – 1) x^{2} + 4x + 1

-4 will satisfy the equation (k – 1) x^{2} + kx + 1 = 0

=> (k – 1)(-4)^{2} + k(-4) + 1 =0

=> 16k – 16 – 4k + 1 = 0

=> 12k – 15 = 0

=> 12k = 15

=> k = \(\frac { 15 }{ 12 }\) = \(\frac { 5 }{ 4 }\)

**Question 16.**

**Solution:**

**(c)**

**Question 17.**

**Solution:**

**(a)**

**Question 18.**

**Solution:**

**(d)** Polynomial: kx^{2} + 2x + 3k

**Question 19.**

**Solution:**

**(b)** α, β are the zeros of the polynomial x^{2} + 6x + 2

**Question 20.**

**Solution:**

**(a)** α, β, γ are the zeros of x^{3} – 6x^{2} – x + 30

Then αβ + βγ + γα = \(\frac { c }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

**Question 21.**

**Solution:**

**(a)** α, β, γ are the zeros of 2x^{3} + x^{2} – 13x + 6, then

αβγ = \(\frac { -d }{ a }\) = \(\frac { -6 }{ 2 }\) = -3

**Question 22.**

**Solution:**

**(c)** α, β, γ are the zeros of p(x) such that

**Question 23.**

**Solution:**

**(a)**

**Question 24.**

**Solution:**

**(b)** If one zero of cubic polynomial ax^{3} + bx^{2} + cx + d = 0

Let a be zero, then

**Question 25.**

**Solution:**

**(c)**

**Question 26.**

**Solution:**

**(d)**

**Question 27.**

**Solution:**

**(c)** p(x) is divided by q(x), then

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

Either r(x) = 0

Degree of r(x) < deg of g(x)

**Question 28.**

**Solution:**

**(d)** (a) is not a linear polynomial.

(b) is trinomial not binomial.

(c) is not a monomial.

(d) 5x^{2} is monomial is true.

Hope given RS Aggarwal Solutions Class 10 Chapter 2 Polynomials MCQS are helpful to complete your math homework.

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