Important questions of Chapter 2 polynomials for class 10 Maths are provided here as per NCERT book. These questions with answers will help the students prepare and score well in the CBSE class 10 maths exam. The questions here covers the latest syllabus as prescribed by the board. Also, get the important questions for all the chapters of 10th standard Maths subject here.

We are also providing here som extra questions so that students can revise the chapter by solving these problems in an instance. The chapter polynomials deal with expressions carrying unknown variables with varying degrees. The solutions of these expressions are termed as zeros of the polynomial. The polynomial with degree 1 is the linear polynomial, with degree 2 is a quadratic polynomial, with degree 3 is a cubic polynomial.

**Also Check:**

- Important 2 Marks Questions for CBSE 10th Maths
- Important 3 Marks Questions for CBSE 10th Maths
- Important 4 Marks Questions for CBSE 10th Maths

## Class 10 Maths Chapter 2 Important Questions With Solutions

A few important class 10 polynomials questions are provided below with solutions. These questions include both short and long answer questions and involve HOTS to let the students get completely acquainted with the in-depth concepts.

**1. Find the value of “p” from the polynomial x ^{2} + 3x + p, if one of the zeroes of the polynomial is 2.**

**Solution:**

As 2 is the zero of the polynomial,

x^{2} + 3x + p, for x = 2

Now, put x = 2

2^{2} + 3(2) + p = 0

=> 4 + 6 + p = 0

Or, p = -10

**2. Does the polynomial a ^{4} + 4a^{2} + 5 = 0 have real zeroes?**

**Solution:**

In the aforementioned equation, let a^{2} = x.

Now, the equation becomes,

x^{2} + 4x^{2} + 5 = 0

Here, b^{2} – 4ac will be = 4^{4 }– 4(1)(5) = -20

So, D = b^{2} – 4ac < 0

As the discriminant (D) is negative, the given polynomial does not have real roots or zeroes.

**3. Compute the zeroes of the polynomial 4x ^{2} – 4x – 8. Also, establish a relationship between the zeroes and coefficients.**

**Solution:**

Factorise the equation 4x^{2} – 4x – 8,

4x^{2} – 4x – 8 = 4x^{2} – 2x – 2x + 1

= 2x(2x – 1) – 1(2x -1) = (2x – 1) (2x – 1)

So, the roots of 4x^{2} – 4x – 8 are (½ and ½)

Relation between the sum of zeroes and coefficients:

½ + ½ = 1 = -4/4 i.e. (- coefficient of x/ coefficient of x^{2})

Relation between the product of zeroes and coefficients:

½ × ½ = ¼ i.e (constant/coefficient of x^{2})

**4. Find the quadratic polynomial if its zeroes are 0, √5.**

**Solution:**

A quadratic polynomial can be written using the sum and product of its zeroes as:

x^{2} +(α + β)x + αβ = 0

Where α and β are the roots of the polynomial equation.

Here, α = 0 and β = √5

So, the equation will be:

x^{2} +(0 + √5)x + 0(√5) = 0

Or, x^{2} + √5x = 0

**5. Find the value of “x” in the equation 2a ^{2} + 2xa + 5x + 10 if (a + x) is one of its factors.**

**Solution:**

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

Since, (a + x) is a factor of 2a^{2} + 2xa + 5x + 10, f(-x) = 0

So, f(-x) = 2x^{2} – 2x^{2} – 5x + 10 = 0

Or, -5x + 10 = 0

Thus, x = 2

**6. How many zeros does the polynomial (x – 3) ^{2} – 4 can have? Also, find its zeroes.**

**Solution:**

Given equation is (x – 3)^{2} – 4

Now, expand this equation.

=> x^{2} + 9 – 6x – 4

= x^{2} – 6x + 5

As the equation has a degree of 2, the number of zeroes it will have is 2.

Now, solve x^{2} – 6x + 5 = 0 to get the roots.

So, x^{2} – x – 5x + 5 = 0

=> x(x-1) -5(x-1) = 0

=> (x-1)(x-5)

So, the roots are 1 and 5.

**7. α and β are zeroes of the quadratic polynomial x ^{2} – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.**

**Solution:**

Let, f(x) = x² – 6 x + a

From the question,

3α + 2β = 20

From f(x),

α + β = 6———————(ii)

And,

αβ = y———————(iii)

Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),

=> α = 20 – 12 = 8

Now, substitute this value in equation (ii),

=> β = 6-8 = -2

put the value of α and β in equation (iii) to get the value of y, such as;

y = αβ = -16

**8. If the zeroes of the polynomial x^{3 }– 3x^{2 }+ x + 1 are a – b, a, a + b, **

**then find the value of a and b.**

**Solution:**

p(x) = x^{3 }– 3x^{2 }+ x + 1

Here, zeroes are given are a – b, a, and a + b

Now, compare the given polynomial equation with general expression.

px^{3 }+ qx^{2} + rx + s = x^{3 }– 3x^{2} + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

-q/p = 3a

Substitute the values q and p.

-(-3)/1 = 3a

Or, a = 1

So, the zeroes are 1-b, 1, 1+b.

For product of zeroes:

Product of zeroes = 1(1-b)(1+b)

-s/p=1-𝑏^{2}

=> -1/1=1-𝑏^{2}

Or, 𝑏^{2} = 1 + 1 =2

So, b = √2

Thus, 1-√2, 1, 1+√2 are the zeroes of equation 𝑥^{3} − 3𝑥^{2} + 𝑥 + 1.

**9. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

**(i) 1/4, -1**

**(ii) 1,1**

**(iii) 4, 1**

Solution:

(i) From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α + β

Product of zeroes = α β

Given,

Sum of zeroes = 1/4

Product of zeroes = -1

Therefore, if α and β are zeroes of any quadratic polynomial, then the equation can be written as:-

x^{2} – (α+β)x +αβ = 0

x^{2} – (1/4)x +(-1) = 0

4x^{2} – x – 4 = 0

Thus, 4x^{2} – x – 4 is the quadratic polynomial.

(ii) Given,

Sum of zeroes = 1 = α+β

Product of zeroes = 1 = αβ

Therefore, if α and β are zeroes of any quadratic polynomial, then the equation can be written as:-

x^{2} – (α+β)x +αβ = 0

x^{2} – x + 1 = 0

Thus, x^{2} – x + 1 is the quadratic polynomial.

(iii) Given,

Sum of zeroes, α + β = 4

Product of zeroes, α β = 1

Therefore, if α and β are zeroes of any quadratic polynomial, then the equation can be written as:-

x^{2} – (α+β)x +αβ = 0

x^{2} – 4x +1 = 0

Thus, x^{2} – 4x +1 is the quadratic polynomial.

**10. Obtain all other zeroes of 3x ^{4}+6x^{3}-2x^{2}-10x-5, if two of its zeroes are √(5/3) and-√(5/3).**

Solutions: Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and-√(5/3) are zeroes of polynomial f(x).

∴ [x-√(5/3)] [x+√(5/3)] = x^{2}-(5/3) = 0

(3x^{2}−5)=0, is a factor of given polynomial f(x).

Now, if we divide f(x) by (3x^{2}−5) the quotient thus obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x^{4}+6x^{3}-2x^{2}-10x-5 = (3x^{2} – 5) (x^{2} + 2x +1)

Now, on further factorizing (x^{2} + 2x +1) we get,

x^{2} + 2x +1 = x^{2} + x + x +1 = 0

x(x + 1) + 1(x+1) = 0

(x+1) (x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3) and-√(5/3), −1 and −1.

Hence, is the answer.

Very nice app with all solutions of any question .