Important class 10 maths questions for chapter 2 polynomials are provided here to help the students practice and score well in the CBSE class 10 maths exam. These questions from the polynomials chapter of NCERT class 10 covers most concepts and will help the students to get acquainted with wide variations of questions and thus, develop problem-solving skills to a great extent.

**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

## Important Polynomials Questions for Class 10- Chapter 12 (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 equation 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 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:**

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

As (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)

Now, multiply equation (ii) by 2 and subtract from equation (i),

=> α = 20 – 12 = 8

Now, substitute this value in equation (ii),

=> β = 6-8 = -2

By substituting the value of α and β in equation (iii), the value of “y” can be obtained.

y = αβ = -16

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

### More Topics Related to Class 10 Polynomials

### Extra Questions For Class 10 Chapter 2: Polynomials (NCERT)

Very nice app with all solutions of any question .