## RS Aggarwal Class 10 Solutions Chapter 2 Polynomials Test Yourself

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

**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

**MCQ**

**Question 1.**

**Solution:**

**(b)** Polynomial is x^{2} – 2x – 3

=> x^{2} – 3x + x – 3

= x(x – 3) + 1(x – 3)

= (x – 3) (x + 1)

Either x – 3 = 0, then x = 3

or x + 1 = 0, then x = -1

Zeros are 3, -1

**Question 2.**

**Solution:**

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

Here, a = 1, b = -6, c = -1, d = 30

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

**Question 3.**

**Solution:**

**(c)**

**Question 4.**

**Solution:**

**(c)** Let α and β be the zeros of the polynomial 4x^{2} – 8kx + 9

**Short-Answer Questions**

**Question 5.**

**Solution:**

Either x + 15 = 0, then x = -15

or x – 13 = 0, then x = 13

Zeros are 13, -15

**Question 6.**

**Solution:**

The polynomial is (a^{2} + 9) x^{2} + 13x + 6a

**Question 7.**

**Solution:**

Zeros are 2 and -5

Sum of zeros = 2 + (-5) = 2 – 5 = -3

and product of zeros = 2 x (-5) = -10

Now polynomial will be

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

= x^{2} – (-3)x + (-10)

= x^{2} + 3x – 10

**Question 8.**

**Solution:**

(a – b), a, (a + b) are the zeros of the polynomial x^{3} – 3x^{2} + x + 1

Here, a = 1, b = -3, c = 1, d = 1

Now, sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -(-3) }{ 1 }\) = 3

a – b + a + a + b = 3

**Question 9.**

**Solution:**

Let f(x) = x^{3} + 4x^{2} – 3x – 18

If 2 is its zero, then it will satisfy it

Now, (x – 2) is a factor Dividing by (x – 2)

Hence, x = 2 is a zero of f(x)

**Question 10.**

**Solution:**

Sum of zeros = -5

and product of zeros = 6

Quadratic polynomial will be

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

= x^{2} – (-5) x + 6

= x^{2} + 5x + 6

**Question 11.**

**Solution:**

**Question 12.**

**Solution:**

p(x) = x^{3} + 3x^{2} – 5x + 4

g(x) = x – 2

Let x – 2 = 0, then x = 2

Remainder = p(2) = (2)^{3} + 3(2)^{2} – 5(2) + 4 = 8 + 12 – 10 + 4 = 14

**Question 13.**

**Solution:**

f(x) = x^{3} + 4x^{2} + x – 6

and g(x) = x + 2

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

f(-2) = (-2)^{3} + 4(-2)^{2} + (-2) – 6 = -8 + 16 – 2 – 6 = 0

Remainder is zero, x + 2 is a factor of f(x)

**Question 14.**

**Solution:**

**Question 15.**

**Solution:**

**Question 16.**

**Solution:**

f(x) = x^{4} + 4x^{2} + 6

=> (x^{2})^{2} + 4(x^{2}) + 6 = y^{2} + 4y + 6 (Let x^{2} = y)

Let α, β be the zeros of y^{2} + 4y + 6

Sum of zeros = -4

and product of zeros = 6

But there are no factors of 6 whose sum is -4 {Factors of 6 = 1 x 6 and 2 x 3}

Hence, f(x) Has no zero (real).

**Long-Answer Questions**

**Question 17.**

**Solution:**

3 is one zero of p(x) = x^{3} – 6x^{2} + 11x – 6

(x – 3) is a factor of p(x)

Dividing, we get

**Question 18.**

**Solution:**

**Question 19.**

**Solution:**

p(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Dividing by x^{2} + 3x + 1,

we get,

Quotient = 3x^{2} – 4x + 2

**Question 20.**

**Solution:**

Let p(x) = x^{3} + 2x^{2} + kx + 3

g(x) = x – 3

and r(x) = 21

Dividing p(x) by g(x), we get

But remainder = 21

3 + 3k + 45 = 21

3k = 21 – 45 – 3

=> 3k = 21 – 48 = -27

k = -9

Second method:

x – 3 is a factor of p(x) : x = 3

Substituting the value of x in p(x)

p(3) = 3^{3} + 2 x 3^{2} + k x 3 + 3

= 27 + 18 + 3k + 3

48 + 3k = 21

=> 3k = -48 + 21 = -27

k = -9

Hence, k = -9

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

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