NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Exercise 2.2 

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\]

Ans:

Given: \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\].

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(x}-\text{4)(x+2)}\]

The value of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] is zero.

when \[\text{x}-\text{4=0}\] or \[\text{x+2=0}\]. i.e., \[\text{x = 4}\] or \[\text{x = }-\text{2}\]

Therefore, the zeroes of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] are \[\text{4}\] and \[-2\].

Now, Sum of zeroes\[\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}} $

Product of zeroes \[\text{=4 }\!\!\times\!\!\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \Pr oduct\,of\,zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

(ii) \[4{{s}^{2}}-4s+1\]

Ans:

Given: \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}\]

The value of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] is zero.

when \[\text{2s}-\text{1=0}\], \[\text{2s}-\text{1=0}\]. i.e., \[\text{s =}\dfrac{\text{1}}{\text{2}}\] and \[\text{s =}\dfrac{\text{1}}{\text{2}}\]

Therefore, the zeroes of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] are \[\dfrac{\text{1}}{\text{2}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

\[\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

Product of zeroes\[=\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

$ \therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}} $ .

(iii) \[6{{x}^{2}}-3-7x\]

Ans:

Given: \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\]

\[\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(3x+1)(2x}-\text{3)}\]

The value of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] is zero.

when \[\text{3x+1=0}\] or \[\text{2x}-\text{3=0}\]. i.e., \[\text{x =}\dfrac{-\text{1}}{\text{3}}\] or \[\text{x =}\dfrac{\text{3}}{\text{2}}\].

Therefore, the zeroes of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] are \[\dfrac{\text{-1}}{\text{3}}\] and \[\dfrac{\text{3}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $

Product of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

(iv) \[4{{u}^{2}}+8u\]

Ans:

Given: \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\]

\[\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}\]

\[\Rightarrow \text{4u(u+2)}\]

The value of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] is zero.

when \[\text{4u=0}\] or \[\text{u+2=0}\]. i.e., \[\text{u = 0}\] or \[\text{u =}-\text{2}\]

Therefore, the zeroes of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] are \[\text{0}\] and \[\text{-2}\].

Now, Sum of zeroes\[\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $

Product of zeroes\[\text{=0 }\!\!\times\!\!\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $

(v) \[{{t}^{2}}-15\]

Ans:

Given: \[{{\text{t}}^{\text{2}}}-\text{15}\]

\[\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}\]

The value of \[{{\text{t}}^{\text{2}}}-\text{15}\] is zero. 

when \[\text{t}-\sqrt{\text{15}}\text{=0}\] or \[\text{t+}\sqrt{\text{15}}\text{=0}\], i.e., \[\text{t=}\sqrt{\text{15}}\] or \[\text{t=}-\sqrt{\text{15}}\]

Therefore, the zeroes of \[{{\text{t}}^{\text{2}}}-\text{15}\] are \[\sqrt{\text{15}}\] and \[-\sqrt{\text{15}}\].

Now, Sum of zeroes\[\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $

Product of zeroes\[\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ .

(vi) \[3{{x}^{2}}-x-4\]

Ans: 

Given: \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\]

Now factorize the given polynomial to get the roots.

$ \Rightarrow \left( 3x-4 \right)(x+1) $

The value of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] is zero.

when \[\text{3x}-\text{4=0}\] or \[\text{x+1=0}\], i.e., \[\text{x=}\dfrac{\text{4}}{\text{3}}\] or \[\text{x=}-\text{1}\]

Therefore, the zeroes of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] are \[\dfrac{\text{4}}{\text{3}}\] and \[\text{-1}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $

Product of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

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

(i) \[\dfrac{1}{4},-1\]

Ans:

Given: \[\dfrac{\text{1}}{\text{4}}\text{,-1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then, 

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}-\text{1}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

$ \Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1 $

$ \Rightarrow 4{{x}^{2}}-x-4 $

Therefore, the quadratic polynomial is \[\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\].

(ii) \[\sqrt{2},\dfrac{1}{3}\]

Ans:

Given: \[\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\sqrt{\text{2}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{3}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}\]

\[\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\]

Therefore, the quadratic polynomial is \[3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\].

(iii) \[0,\sqrt{5}\]

Ans:

Given: \[\text{0,}\sqrt{\text{5}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 0}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\sqrt{\text{5}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}\].

(iv) \[1,1\]

Ans:

Given: \[\text{1,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =1}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}-\text{x+1}\].

(v) \[-\dfrac{1}{4},\dfrac{1}{4}\]

Ans:

Given: \[-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}-\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

$ \Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4} $

$ \Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $

Therefore, the quadratic polynomial is  $ \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ .

(vi) \[4,1\]

Ans: 

Given: \[\text{4,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 4}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

$ \Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1} $ 

Therefore, the quadratic polynomial is  $ {{\text{x}}^{\text{2}}}-\text{4x+1} $.

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

Definition 

If x be a variable and x be a positive integer with a1, a2,  a3, ……., an be constants then, f(x) = anxn + an-1x-1 + ….. + a1x + a0   is called a polynomial in x.

In this polynomial anxn + an-1x-1, ………, a1x and a0 are called the terms of the polynomial, and an, an-1,…., a1, a0 are called their coefficients.

Example: p(x) = 4x -1  is a polynomial in variable x.

     q(y) = 2y2 – 3y + 5 is a polynomial in variable y.

Note: Expressions like 3x2 - 2√x + 7, 1/(3x-5), etc. are not polynomials as some powers of x are not integers. 

Degree of a Polynomial

The exponent of the highest degree term in a polynomial is called its degree. 

Example:

1. f(x) = 4x -1 is a polynomial in x of degree 1.

2. q(y) = 2y2 - 3y + 5 is a polynomial in y of degree 2.

Types of Polynomial

Let us see the different types of polynomials.

1. Constant Polynomial: It is a polynomial of degree zero.

Ex: f(x) = 5, g(x) = 2, etc.

The constant polynomial f(x) = 0 is called the zero polynomial. The degree of the zero polynomial is not defined as f(x) = 0, g(x) =0x, h(x) = 0x2, p(x) = 0x3, etc. are all equal to the zero polynomial.

2. Linear Polynomial: It is a polynomial of degrees.

Ex: f(x) = 4x -1.

A polynomial q(y) = 3y2 + 4 is not linear as it has degree 2.

In general, any linear polynomial in x with real coefficients is of the form f(x) = ax + b where a & b are real numbers and a 0.

Note: A linear polynomial may be a monomial (having only one term. Ex: q(x) = 5x) or a binomial (having 2 terms. Ex: f(x) = 4x-1)

3. Quadratic Polynomial: It is a polynomial of degree 2.

Ex: f(x) = 5x2 - 3x + 4.

In general, a quadratic polynomial is of the form f(x) = ax2 + bx + c, where a, b & c are real numbers and a 0.

4. Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial. 

Ex: f(x) = 2x3 – 5x2 + 7x – 9.

The general form of a cubic polynomial is f(x) = ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a 0.

Value of a Polynomial

If f(x) is a polynomial and ‘a’ is a real number then the real number obtained by replacing x by ‘a’ , is called the value f(x) at x = a and is denoted by f(a). For example: if f(x) = 2x2 - 3x -2, then its value

1. at x = 1 is given by f(1) = 2(1)2 - 3 1 -2 = 2-3-2 = -3

2. at x = -2, is given by f(-2) = 2(-2)2 -3(-2)-2 = 8 + 6 -2 = 12

Zero of a Polynomial

Consider the cubic polynomial f(x) = x3 – 6x2 + 11x - 6. The value of this polynomial at x =1, 2, 3 are respectively

f(1) =13 – 6 * 12 + 11 * 1 – 6 = 1-6 + 11 -6 =0

f(2) = 23 – 6 * 22 + 11 * 2 – 6 = 8 – 24 + 22 – 6 = 0

f(3) = 33 – 6* 32 + 11 * 3 -6 = 27  -54 + 33 -6 = 0

1, 2, 3 are called the zeroes of the cubic polynomial f(x) = x3 -6x2 + 11x - 6.

A real number is 0 of a polynomial f(x), if f() = 0.

To find this we solve the equation f(x) = 0.

For Example: if the polynomial is f(x) = x -3 then putting x -3 = 0, we get x = 3

Then the real number 3 is the zero of the polynomial f(x) = x -3.

There can be more than one zero of a polynomial as can be seen in the example f(x) = x3-6x2 + 11x - 6.

Graph of a Polynomial

The graph of a polynomial f(x) is the set of all points (x, y), where y = f(x). This graph is a smooth free hand curve, passing through the points (x1, y1), (x2, y2), (x3,y3), ……, etc. y1, y2, y3, …… are the estimated values of the polynomial f(x) at x1, x2, x3,..... respectively.

Geometrical meaning of the zeroes of a linear polynomial

We take an example f(x) = 4x - 2

Let y  4x - 2

We make a table with values of y corresponding to different values of x.

The points A(0, 1) and B (2, 6) are plotted on the graph paper on a suitable scale. A line drawn runs through these points to derive the graph of the given polynomial. It can be seen that the graph intersects the x-axis at C (½, 0), where y = f(x) = 0 so that the zero of the given polynomial is ½.

Meaning of the Zeroes of a Quadratic Polynomial

Graph of a quadratic polynomial f(x) = x2 - 2x - 8

Let y = x2 - 2x - 8

We construct a table containing the values of y corresponding to various values of x:

We now plot the points (-4, 16), ( -3, 7), (-2, 0) , (-1, -5), (0 -8), (1, 9), (2, 8), (3, -5), (4, 0) , (5, 7) and (6, 16) on a graph paper and draw a smooth free hand curve passing through these points. Hence, the curve derived demonstrates the graph of the polynomial f(x) = x2 -2x -8. Such a curve is called a parabola. The lowest point of the graph (1, 9) is called the vertex of the parabola. 

The Following Observations Can Be Made On Class 10 Maths ex 2.2

1. The graph is symmetrical about the line parallel to the y-axis through the vertex (1, -9). This line is called the axis of the parabola.

2. The concavity of the parabola is towards the positive direction of the y-axis (this happens when the coefficient of x2 is positive.

3. As  x2 -2x -8 = (x-4) (x + 2), the polynomial f(x) has two factors (x-4) & (x+2), so that the parabola cuts the x-axis at two distinct points (4,0) & (-2, 0). The x-coordinates of these two points: 4 & -2 are the two zeroes of f(x).

Note in Exercise 2.2 Class 10 Math:

1. The number of zeroes of a polynomial depends on its degree. For a polynomial of degree n, the number of zeroes can be utmost n or less. For example, for a quadratic, polynomial the no. of zeroes can be 2 or less.

2. For a quadratic polynomial, its graph (ex: a parabola)  may cut the x-axis at two points in fig (i), or at one point in fig (ii) or at no point at all in fig (iii).

Conclusion

Your one-stop resource for learning how to solve these expressions might be Vedanta's Ex 2.2 Class 10 NCERT Solutions. These solutions simplify difficult subjects like factorization and finding zeroes into simple steps with illustrations. Pay particular attention to comprehending how a polynomial's zeroes and coefficients relate to one another.You'll come across a number of polynomial problems throughout the chapter, and this is the key to solving them. By using these solutions for practice, you'll get more confident and ace the test!

NCERT Solutions Class 10 Maths Chapter 2 Exercises

Other Related Links for Chapter 2 Polynomials Of Class 10

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.