## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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**Class 10 NCERT Solutions**will help you in gaining more marks in the examinations and prepare for the board exams in better manner.**Exercises 2.1**

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

**Answer**

*x*-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the

*x*-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the

*x*-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the

*x*-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the

*x*-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the

*x*-axis at 3 points.

Page No: 33

**Exercise 2.2**

(i)

*x*

^{2}â€“ 2

*x*â€“ 8

(ii) 4

*s*

^{2}â€“ 4

*s*+ 1

(iii) 6

*x*

^{2}â€“ 3 â€“ 7

*x*

(iv) 4

*u*

^{2}+ 8

*u*

(v)

*t*^{2}â€“ 15
(vi) 3

*x*^{2}â€“*x*â€“ 4**Answer**

(i)

*x*

^{2}â€“ 2

*x*â€“ 8

= (

*x*- 4) (

*x*+ 2)

The value of

*x*

^{2}â€“ 2

*x*â€“ 8 is zero when

*x*- 4 = 0 or

*x*+ 2 = 0, i.e., when

*x*= 4 or

*x*= -2

Therefore, the zeroes of

*x*

^{2}â€“ 2

*x*â€“ 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of

*x*)/Coefficient of*x*^{2}
Product of zeroes = 4 Ã— (-2) = -8 = -8/1 = Constant term/Coefficient of

*x*^{2}(ii) 4

*s*

^{2}â€“ 4

*s*+ 1

= (2

*s*-1)

^{2}

The value of 4

*s*

^{2}- 4

*s*+ 1 is zero when 2

*s*- 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s

^{2}- 4s + 1 are 1/2 and 1/2.Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of

*s)*/Coefficient of

*s*

^{2}

Product of zeroes = 1/2 Ã— 1/2 = 1/4 = Constant term/Coefficient of

*s*

^{2}.

(iii) 6

*x*

^{2}â€“ 3 â€“ 7

*x*

*=*6

*x*

^{2 }â€“ 7

*x*â€“ 3

= (3

*x*Â + 1) (2

*x*- 3)

The value of 6

*x*

^{2}- 3 - 7

*x*is zero when 3

*x*+ 1 = 0 or 2

*x*- 3 = 0, i.e.,

*x*= -1/3 or

*x*= 3/2

Therefore, the zeroes of 6

*x*^{2}- 3 - 7*x*are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of

*x*)/Coefficient ofÂ

*x*

^{2}

Product of zeroes = -1/3 Ã— 3/2 = -1/2 = -3/6Â = Constant term/Coefficient ofÂ

*x*

^{2}.

(iv) 4

*u*

^{2}Â + 8

*u*

*=Â*4

*u*

^{2}Â + 8

*uÂ +*0

= 4

*u*(

*u*+ 2)

The value of 4

*u*

^{2}+ 8

*u*is zero when 4

*u*= 0 or

*u*+ 2 = 0, i.e.,

*u*= 0 or

*u*= - 2

Therefore, the zeroes of 4

*u*

^{2}+ 8

*u*are 0 and - 2.

Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of

*u*)/Coefficient of

*u*

^{2}

Product of zeroes = 0 Ã—Â (-2)Â = 0 = 0/4Â = Constant term/Coefficient of

*u*

^{2}.

(v)Â

*t*

^{2}Â â€“ 15

=Â

*t*

^{2Â }- 0.

*t*- 15

= (

*t*- âˆš15) (

*t*Â + âˆš15)

The value of

*t*

^{2}- 15 is zero when

*t*- âˆš15Â = 0 or

*t*+Â âˆš15Â = 0, i.e., when

*t*= âˆš15 or

*tÂ*= -âˆš15

Therefore, the zeroes of

*t*

^{2}- 15 are âˆš15 and -âˆš15.Sum of zeroes =Â âˆš15Â +Â -âˆš15Â = 0 = -0/1Â = -(Coefficient ofÂ

*t*)/Coefficient ofÂ

*t*

^{2}

Product of zeroes = (âˆš15) (-âˆš15)Â = -15 = -15/1Â = Constant term/Coefficient ofÂ

*u*

^{2}.

(vi) 3

*x*

^{2}Â â€“

*Â x*Â â€“ 4

= (3

*x*- 4) (

*x*Â + 1)

The value of 3

*x*

^{2}Â â€“

*Â x*Â â€“ 4 is zero when 3

*x*Â - 4 = 0 andÂ

*x*Â + 1 = 0,i.e., when

*x*= 4/3 or

*x*= -1

Therefore, the zeroes of 3

*x*

^{2}Â â€“

*Â x*Â â€“ 4 are 4/3 and -1.

Sum of zeroes = 4/3 +Â (-1) = 1/3 = -(-1)/3Â = -(Coefficient ofÂ

*x*)/Coefficient ofÂ*x*^{2}
Product of zeroes = 4/3 Ã— (-1) = -4/3Â = Constant term/Coefficient ofÂ

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

(i) 1/4 , -1*x*^{2}.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(ii) âˆš2 , 1/3Â

(iii) 0, âˆš5

(iv) 1,1Â

(v) -1/4 ,1/4Â

(vi) 4,1

**Answer**

(i) 1/4 , -1

Let the polynomial be

*ax*

^{2}Â +

*bx*Â +

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ = 1/4 = -

*b*/

*a*

Î±ÃŸ = -1 = -4/4 =

*c*/

*a*

If

*a*= 4, then

*b*= -1,

*c*= -4

Therefore, the quadratic polynomial is 4

*x*

^{2}-

*x*-4.

(ii) âˆš2Â , 1/3

Let the polynomial beÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ =Â âˆš2 = 3âˆš2/3Â = -

*b*/

*a*

Î±ÃŸ = 1/3 =Â

*c*/

*a*

If

*a*= 3, thenÂ

*b*Â = -3âˆš2,Â

*c*Â = 1

Therefore, the quadratic polynomial is 3

*x*

^{2}Â -3âˆš2

*x*Â +1.

(iii) 0, âˆš5

Let the polynomial beÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ = 0 = 0/1Â = -

*b*/

*a*

Î±ÃŸ =Â âˆš5Â =Â âˆš5/1Â =Â

*c*/

*a*

If

*a*= 1, thenÂ

*b*Â = 0,Â

*c*Â = âˆš5

Therefore, the quadratic polynomial is

*x*

^{2}Â + âˆš5.

(iv) 1, 1

Let the polynomial beÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ = 1Â = 1/1Â = -

*b*/

*a*

Î±ÃŸ = 1Â = 1/1Â =Â

*c*/

*a*

IfÂ

*a*Â = 1, thenÂ

*b*Â = -1,Â

*c*Â = 1

Therefore, the quadratic polynomial is

*x*

^{2}Â -Â

*x*Â +1.

(v) -1/4 ,1/4

Let the polynomial beÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ = -1/4Â = -

*b*/

*a*

Î±ÃŸ = 1/4Â =Â

*c*/

*a*

IfÂ

*a*Â = 4, thenÂ

*b*Â = 1,Â

*c*Â = 1

Therefore, the quadratic polynomial isÂ 4

*x*

^{2}Â +Â

*x*Â

(vi) 4,1

Let the polynomial beÂ

*ax*

^{2}Â +Â

*bx*Â +Â

*c*, and its zeroes be Î± and ÃŸ

Î±Â + ÃŸ = 4 = 4/1Â = -

*b*/

*a*

Î±ÃŸÂ = 1Â = 1/1Â =Â

*c*/

*a*

IfÂ

*a*Â = 1, thenÂ

*b*Â = -4,Â

*c*Â = 1

Therefore, the quadratic polynomial isÂ

*x*

^{2}Â - 4

*x*Â

Page No: 36

**Exercise 2.3**

1. Divide the polynomial

*p*(

*x*) by the polynomial

*g*(

*x*) and find the quotient and remainder in each of the following:

^{}

**Answer**

(i)Â

*p*(

*x*) =Â

*x*

^{3}Â â€“ 3

*x*

^{2}Â + 5

*x*Â â€“ 3,Â

*g*(

*x*) =Â

*x*

^{2}Â â€“ 2

*x*-3 and remainder 7

*x*- 9

(ii)Â

*p*(*x*) =Â*x*^{4}Â â€“ 3*x*^{2}Â + 4x + 5,Â*g*(*x*) =Â*x*^{2}Â + 1 â€“Â*x**x*

^{2}+Â

*xÂ*- 3 and remainder 8

(iii)Â

*p*(*x*) =Â*x*^{4}Â â€“ 5*x*Â + 6,Â*g*(*x*) = 2 â€“Â*x*^{2}*x*

^{2}-2 and remainder -5

*x*+10

^{}

second polynomial by the first polynomial:

**Answer**

(i)

*t*^{2}â€“ 3, Â 2*t*^{4}+ 3*t*^{3}â€“ 2*t*^{2}â€“ 9*t*â€“ 12*t*^{2}Â â€“ 3 exactly divides Â 2*t*^{4}Â + 3*t*^{3}Â â€“ 2*t*^{2}Â â€“ 9*t*Â â€“ 12 leaving no remainder. Hence, it is a factor ofÂ Â 2*t*^{4}Â + 3*t*^{3}Â â€“ 2*t*^{2}Â â€“ 9*t*Â â€“ 12.*x*

^{2}+ 3

*x*+ 1, 3

*x*

^{4}+ 5

*x*

^{3}â€“ 7

*x*

^{2}+ 2

*x*+ 2

*x*

^{2}Â + 3

*x*Â + 1 exactly divides 3

*x*

^{4}Â + 5

*x*

^{3}Â â€“ 7

*x*

^{2}Â + 2

*x*Â + 2 leaving no remainder. Hence, it is factor of 3

*x*

^{4}Â + 5

*x*

^{3}Â â€“ 7

*x*

^{2}Â + 2

*x*Â + 2.

*x*

^{3}â€“ 3

*x*+ 1,

*x*

^{5}â€“ 4

*x*

^{3}+

*x*

^{2}+ 3

*x*+ 1

*x*

^{3}Â â€“ 3

*x*Â + 1 didn't divides exactlyÂ

*x*

^{5}Â â€“ 4

*x*

^{3}Â +Â

*x*

^{2}Â + 3

*x*Â + 1 and leaves 2 as remainder. Hence, it not a factor ofÂ

*x*

^{5}Â â€“ 4

*x*

^{3}Â +Â

*x*

^{2}Â + 3

*x*Â + 1.

3. Obtain all other zeroes of 3

*x*^{4}+ 6*x*^{3}â€“ 2*x*^{2}â€“ 10*x*â€“ 5, if two of its zeroes are âˆš(5/3)**Answer**

*p*(

*x*) = 3

*x*

^{4}Â + 6

*x*

^{3}Â â€“ 2

*x*

^{2}Â â€“ 10

*x*Â â€“ 5

Since the two zeroes are âˆš(5/3)Â and - âˆš(5/3).

We factorize

*x*^{2}+ 2*xÂ*+ 1
= (

*xÂ*+ 1)^{2}
Therefore, its zero is given by

*x*Â + 1 = 0*x*= -1

*xÂ*+ 1)

^{2}Â , therefore, there will be 2 zeroes at

*x*= - 1.

Hence, the zeroes of the given polynomial areÂ âˆš(5/3)Â and - âˆš(5/3), - 1 and - 1.

4. Â On dividingÂ

*x*

^{3}Â - 3

*x*

^{2}Â +

*x*+ 2 by a polynomial

*g*(

*x*), the quotient and remainder were

*x*- 2 andÂ

-2

*x*+Â 4, respectively. Find

*g*(

*x*).

**Answer**

Here in the given question,

Dividend =Â

*x*^{3}Â - 3*x*^{2}Â +*x*+ 2
Quotient =Â

*x*- 2
Remainder =Â -2

*x*+Â 4
Divisor =

*g*(*x*)
We know that,

Dividend = Quotient Ã— DivisorÂ + Remainder

â‡’Â

â‡’Â

â‡’

âˆ´Â

5.Give examples of polynomial

(i) deg

(ii) deg

(iii) deg

(ii) Let us assume the division of

Here,

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

(iii) Let us assume the division of

Here,

g(x) = x

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

Page No: 36

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2

(ii)

(i)Â

Now for zeroes, putting the given value in x.

=Â (2Ã—1/8)Â + 1/4 - 5/2Â + 2

= 1/4Â + 1/4 - 5/2Â + 2

= 1/2 - 5/2Â + 2 = 0

=Â (2Ã—1)Â + 1 - 5 + 2

= 2Â + 1 - 5 + 2 = 0

=Â (2 Ã— -8)Â + 4 + 10Â + 2

= -16 + 16Â = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also,Â Î±=1/2, Î²=1 and Î³=-2

Now,

-b/a = Î±+Î²+Î³

â‡’ 1/2 = 1/2Â+ 1 - 2

â‡’ 1/2 = 1/2

c/a =Â Î±Î²+Î²Î³+Î³Î±

â‡’ -5/2 = (1/2 Ã— 1)Â + (1 Ã— -2)Â + (-2 Ã— 1/2)

â‡’ -5/2 = 1/2 - 2 - 1

â‡’ -5/2 = -5/2

-d/a = Î±Î²Î³

â‡’ -2/2 = (1/2 Ã— 1 Ã— -2)

â‡’ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii) Â

Now for zeroes, putting the given value in x.

= 8 - 16 +Â 10 - 2

= 0

= 1 - 4Â +Â 5 - 2

= 0

= 1 - 4Â +Â 5 - 2

= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also,Â Î±=2, Î²=1 and Î³=1

Now,

-b/a = Î±+Î²+Î³

â‡’ 4/1 = 2Â+ 1 + 1

â‡’ 4 = 4

c/a =Â Î±Î²+Î²Î³+Î³Î±

â‡’ 5/1 = (2 Ã— 1)Â + (1 Ã— 1)Â + (1 Ã— 2)

â‡’ 5 = 2 + 1Â + 2

â‡’ 5 = 5

-d/a = Î±Î²Î³

â‡’ 2/1 = (2 Ã— 1 Ã— 1)

â‡’ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, â€“7, â€“14 respectively.

Let the polynomial be

Then, Î± +Â Î² +Â Î³ = -(-2)/1 = 2 = -b/a

Î±Î² +Â Î²Î³Â + Î³Î± = -7 = -7/1 = c/a

Î±Î²Î³ = -14 = -14/1 = -d/a

âˆ´ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will beÂ

3. If the zeroes of the polynomialÂ

Since, (a - b), a, (a + b) are the zeroes of the polynomialÂ

Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

â‡’ 3a = 3 â‡’ a =1

âˆ´ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a

â‡’ 3a

Putting the value of a,

â‡’ 3(1)

â‡’ 3 - b

â‡’ b

â‡’ b = Â±âˆš2

Hence, a = 1 and b = Â±âˆš2

4. If two zeroes of the polynomial x

2+âˆš3 and 2-âˆš3 are two zeroes of the polynomial p(x) = x

Let x = 2Â±âˆš3

So, x-2 = Â±âˆš3

On squaring, we get x

â‡’ x

Now, dividing p(x) by x

âˆ´ p(x) = x

= (x

= (x

= (x

= (x

âˆ´ (x + 5) and (x - 7) are other factors of p(x).

âˆ´ - 5 and 7 are other zeroes of the given polynomial.

Chapter 2 Polynomials is mainly concerned about finding zeroes of given polynomials. You will also get to know about division algorithm which is very important. We have already studied in Class 9 that an expression of the form p(x) = a

This chapter is divided into four sections:

â€¢ First section is about introduction in which you will get to know about degrees of polynomials.

â€¢ The second section is Geometrical Meaning of the zeroes of a Polynomial on which one exercise has been provided which only has one question.

â€¢ The third section is namedÂ Relationship between zeroes and coefficient of a Polynomial. There are two questions in this exercise, in the first you have to find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients and in the another one you have to find a quadratic polynomial with the given numbers.

â€¢ The fourth section is division algorithm for Polynomials in which you have to divide the polynomials and find remainder and quotient.

Other than NCERT Solutions for Class 10 Maths Chapter 2 Polynomials, you can also check Notes of Chapter 2 Polynomials Class 10, MCQ of Class 10 Chapter 2 Poynomials.

There are total four exercises in this chapter in which last one is optional. We have provided Class 10 Maths Polynomials Solutions of every exercises step by step. For finding the exercise wise solutions you can click the links below.

*x*^{3}Â - 3*x*^{2}Â +*x*+ 2 = (*x*- 2) Ã—*g*(*x*)Â + (-2*x*Â + 4)â‡’Â*x*^{3}Â - 3*x*^{2}Â +*x*+ 2 - (-2*x*Â + 4) = (*x*- 2)Â Ã—Â*g*(*x*)â‡’Â

*x*^{3}Â - 3*x*^{2}Â + 3*x*- 2 = (*x*- 2) Ã—*g*(*x*)â‡’

*g*(*x*) = Â (*x*^{3}Â - 3*x*^{2}Â + 3*x*- 2)/(*x*- 2)*g*(*x*) = (*x*^{2}Â -*x*Â + 1)5.Give examples of polynomial

*p*(*x*),*g*(*x*),*q*(*x*) and*r*(*x*), which satisfy the division algorithm and(i) deg

*p*(*x*) = deg*q*(*x*)(ii) deg

*q*(*x*) = deg*r*(*x*)(iii) deg

*r*(*x*) = 0**Answer**
(i) Let us assume the division of 6

Here,

Degree of

Checking for division algorithm,

Or, 6

Hence, division algorithm is satisfied.

*x*^{2}+ 2*x*+ 2 by 2Here,

*p*(*x*) = 6*x*^{2}+ 2*x*+ 2*g*(*x*) = 2*q*(*x*) = 3*x*^{2}+*x*+ 1*r*(*x*) = 0Degree of

*p*(*x*) and*q*(*x*) is same i.e. 2.Checking for division algorithm,

*p*(*x*) =*g*(*x*)Â Ã—Â*q*(*x*) +*r*(*x*)Or, 6

*x*^{2}+ 2*x*+ 2 = 2*x*(3*x*^{2}+*x*+ 1)Hence, division algorithm is satisfied.

(ii) Let us assume the division of

*x*^{3}+*x*by*x*^{2},Here,

*p*(*x*) =*x*^{3}+*x**g*(*x*) =*x*^{2}*q*(*x*) =*x*and*r*(*x*) =*x*Clearly, the degree of

*q*(*x*) and*r*(*x*) is the same i.e., 1.Checking for division algorithm,

*p*(*x*) =*g*(*x*) Ã—*q*(*x*) +*r*(*x*)*x*^{3}+*x*= (*x*^{2}) Ã—*x*+*x**x*^{3}+*x*=*x*^{3}+*x*Thus, the division algorithm is satisfied.

(iii) Let us assume the division of

*x*^{3}+ 1 by*x*^{2}.Here,

*p*(*x*) =*x*^{3}+ 1g(x) = x

^{2}*q*(*x*) =*x*and*r*(*x*) = 1Clearly, the degree of

*r*(*x*) is 0.Checking for division algorithm,

*p*(*x*) =*g*(*x*) Ã—*q*(*x*) +*r*(*x*)*x*^{3}+ 1 = (*x*^{2}) Ã—*x*+ 1*x*^{3}+ 1 =*x*^{3}+ 1Thus, the division algorithm is satisfied.

Page No: 36

**Exercise 2.4 (Optional)**1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2

*x*^{3}Â +*x*^{2}Â -Â 5*xÂ*+ 2; 1/2, 1, -2(ii)

*x*^{3}Â - 4*x*^{2}Â +Â 5*x -*Â 2; 2, 1, 1**Answer**(i)Â

*p*(*x*) = 2*x*^{3}Â +Â*x*^{2}Â -Â 5*xÂ*+ 2Now for zeroes, putting the given value in x.

*p*(*1/2*) = 2(1/2)^{3}Â +Â*(1/2)*^{2}Â - 5(1/2)*Â*+ 2=Â (2Ã—1/8)Â + 1/4 - 5/2Â + 2

= 1/4Â + 1/4 - 5/2Â + 2

= 1/2 - 5/2Â + 2 = 0

*p*(*1*) = 2(1)^{3}Â +Â*(1)*^{2}Â - 5(1)*Â*+ 2=Â (2Ã—1)Â + 1 - 5 + 2

= 2Â + 1 - 5 + 2 = 0

*p*(*-2*) = 2(-2)^{3}Â +Â (-2)^{2}Â - 5(-2)*Â*+ 2=Â (2 Ã— -8)Â + 4 + 10Â + 2

= -16 + 16Â = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with a

*x*^{3}Â +Â b*x*^{2}Â +Â c*xÂ*+ d, we get a=2, b=1, c=-5, d=2Also,Â Î±=1/2, Î²=1 and Î³=-2

Now,

-b/a = Î±+Î²+Î³

â‡’ 1/2 = 1/2Â

â‡’ -5/2 = (1/2 Ã— 1)Â + (1 Ã— -2)Â + (-2 Ã— 1/2)

â‡’ -5/2 = 1/2 - 2 - 1

â‡’ -5/2 = -5/2

-d/a = Î±Î²Î³

â‡’ -2/2 = (1/2 Ã— 1 Ã— -2)

â‡’ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii) Â

*p*(*x*) =Â*x*^{3}Â - 4*x*^{2}Â +Â 5*x -*Â 2Now for zeroes, putting the given value in x.

*p*(*2*) = 2^{3}Â - 4*(2)*^{2}Â + 5(2)*Â*- 2= 8 - 16 +Â 10 - 2

= 0

*p*(*1*) = 1^{3}Â - 4*(1)*^{2}Â +Â 5(1)*Â*- 2= 1 - 4Â +Â 5 - 2

= 0

*p*(*1*) = 1^{3}Â - 4*(1)*^{2}Â +Â 5(1)*Â*- 2= 1 - 4Â +Â 5 - 2

= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a

*x*^{3}Â +Â b*x*^{2}Â +Â c*xÂ*+ d, we get a=1, b=-4, c=5, d=-2Also,Â Î±=2, Î²=1 and Î³=1

Now,

-b/a = Î±+Î²+Î³

â‡’ 4/1 = 2Â

â‡’ 5/1 = (2 Ã— 1)Â + (1 Ã— 1)Â + (1 Ã— 2)

â‡’ 5 = 2 + 1Â + 2

â‡’ 5 = 5

-d/a = Î±Î²Î³

â‡’ 2/1 = (2 Ã— 1 Ã— 1)

â‡’ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, â€“7, â€“14 respectively.

**Answer**Let the polynomial be

*ax*^{3}Â +Â*bx*^{2Â }+*cxÂ + d*and the zeroes be Î±, Î² and Î³Then, Î± +Â Î² +Â Î³ = -(-2)/1 = 2 = -b/a

Î±Î² +Â Î²Î³Â + Î³Î± = -7 = -7/1 = c/a

Î±Î²Î³ = -14 = -14/1 = -d/a

âˆ´ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will beÂ

*x*^{3}Â - 2*x*^{2 Â }-*Â 7xÂ + 14*3. If the zeroes of the polynomialÂ

*x*^{3}Â â€“ 3*x*^{2}Â +Â*x*Â + 1 are aâ€“b, a, a+b, find a and b.**Answer**Since, (a - b), a, (a + b) are the zeroes of the polynomialÂ

*x*^{3}Â â€“ 3*x*^{2}Â +Â*x*Â + 1.Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

â‡’ 3a = 3 â‡’ a =1

âˆ´ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a

^{2}Â - ab + a^{2}Â + ab + a^{2}Â - b^{2Â }= 1â‡’ 3a

^{2}Â - b^{2}Â =1Putting the value of a,

â‡’ 3(1)

^{2}Â - b^{2}Â = 1â‡’ 3 - b

^{2}Â = 1â‡’ b

^{2}Â = 2â‡’ b = Â±âˆš2

Hence, a = 1 and b = Â±âˆš2

4. If two zeroes of the polynomial x

^{4}Â â€“ 6x^{3}Â â€“ 26x^{2}Â + 138x â€“ 35 are Â 2Â±âˆš3,Â Â find other zeroes.**Answer**2+âˆš3 and 2-âˆš3 are two zeroes of the polynomial p(x) = x

^{4}Â â€“ 6x^{3}Â â€“ 26x^{2}Â + 138x â€“ 35.Let x = 2Â±âˆš3

So, x-2 = Â±âˆš3

On squaring, we get x

^{2}Â - 4x + 4 = 3,â‡’ x

^{2}Â - 4x + 1= 0^{2}Â - 4x + 1^{4}Â - 6x^{3}Â - 26x^{2}Â + 138x - 35= (x

^{2}Â - 4x + 1) (x^{2}Â - 2x - 35)= (x

^{2}Â - 4x + 1) (x^{2}Â - 7x + 5x - 35)= (x

^{2}Â - 4x + 1) [x(x - 7) + 5 (x - 7)]= (x

^{2}Â - 4x + 1) (x + 5) (x - 7)âˆ´ (x + 5) and (x - 7) are other factors of p(x).

âˆ´ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial x

^{4}Â â€“ 6x^{3}Â + 16x^{2}Â â€“ 25x + 10 is divided by another polynomial x^{2}Â â€“ 2x + k, the remainder comes out to be x + a, find k and a.**Answer**

On dividing x

^{4}Â â€“ 6x^{3}Â + 16x^{2}Â â€“ 25x + 10 by x^{2}Â â€“ 2x + k
âˆ´ Remainder = (2k - 9)x - (8 - k)k + 10Â

But the remainder is given as x+ a.Â

On comparing their coefficients,

2k - 9 = 1

â‡’ k = 10Â

â‡’ k = 5 and,

-(8-k)k +Â 10 = a

â‡’ a = -(8 - 5)5 + 10 =- 15 + 10 = -5Â

Hence, k = 5 and a = -5Â

**Go Back To NCERT Solutions for Class 10th Maths****NCERT Solutions for Class 10 Maths Chapter 2 Polynomials**Chapter 2 Polynomials is mainly concerned about finding zeroes of given polynomials. You will also get to know about division algorithm which is very important. We have already studied in Class 9 that an expression of the form p(x) = a

_{0}Â +Â a_{1}x + a_{2}x^{2}Â + .... +Â a_{n}x^{n}Â whereÂ a_{n}Â â‰ 0 is called a polynomialÂ inÂ variable x of degree n. where;Â a_{0},Â a_{2 ....}Â a_{n}Â are real numbers and each power of xÂ isÂ aÂ non negative integer.Â A polynomial which contains only constant term, is called a zero polynomial.This chapter is divided into four sections:

â€¢ First section is about introduction in which you will get to know about degrees of polynomials.

â€¢ The second section is Geometrical Meaning of the zeroes of a Polynomial on which one exercise has been provided which only has one question.

â€¢ The third section is namedÂ Relationship between zeroes and coefficient of a Polynomial. There are two questions in this exercise, in the first you have to find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients and in the another one you have to find a quadratic polynomial with the given numbers.

â€¢ The fourth section is division algorithm for Polynomials in which you have to divide the polynomials and find remainder and quotient.

Other than NCERT Solutions for Class 10 Maths Chapter 2 Polynomials, you can also check Notes of Chapter 2 Polynomials Class 10, MCQ of Class 10 Chapter 2 Poynomials.

There are total four exercises in this chapter in which last one is optional. We have provided Class 10 Maths Polynomials Solutions of every exercises step by step. For finding the exercise wise solutions you can click the links below.

### NCERT Solutions for Class 10 Maths Chapters:

**FAQ onÂ Chapter 2 Polynomials**

#### How many exercises in Chapter 2 Polynomials

There are only 4 exercises in the chapter. You will get detailed solutions of every questions and grasp the concepts embedded in it. These NCERT Solutions for Chapter 2 Polynomials Class 10 Maths will help you a lot in board exams.

#### What is Division Algorithm?

If we divide a polynomial p(x) by a polynomial g(x), then their exists polynomials q(x) and r(x) such that, p(x) = g(x) x q(x) + r(x), where r(x) = 0 or deg r(x) < deg g(x). It is known as Division Algorithm.

#### What is Linear Polynomial.

A polynomial of degree 1 is called a linear polynomial. A linear polynomial is generally written as ax + b (a â‰ 0), where a, b are real coefficients.

####
Find the zeroes of 2x^{2} - 8x + 6.

We have:

2x

= 2x (x - 3) - 2 (x - 3)

= (2x - 2) (x - 3)

= 2 (x - 1) (x - 3)

For 2x

Either, x - 1 = 0 â‡’ x = 1

or x - 3 = 0 â‡’ x = 3

âˆ´ The zeroes of 2x

2x

^{2}- 8x + 6 = 2x^{2}- 6x - 2x + 6= 2x (x - 3) - 2 (x - 3)

= (2x - 2) (x - 3)

= 2 (x - 1) (x - 3)

For 2x

^{2}- 8x + 6 to be zero,Either, x - 1 = 0 â‡’ x = 1

or x - 3 = 0 â‡’ x = 3

âˆ´ The zeroes of 2x

^{2}- 8x + 6 are 1 and 3.