Ncert Solutions for class 10 subject Maths Chapter 2 POLYNOMIALSin pdf Best Free NCERT Solutions for class 1 to 12 in pdf NCERT Solutions, cbse board, Maths, ncert Solutions for Class 10 Maths, class 10 Maths ncert solutions, POLYNOMIALS, Class 10, ncert solutions chapter 2 POLYNOMIALS, class 10 Maths, class 10 Maths ncert solutions, Maths ncert solutions class 10, Ncert Solutions Class 10 Mathematics Chapter 2 POLYNOMIALS

If p(x) is a polynomial in terms of x, the highest power of x in p(x) is called the degree of the polynomial p(x).

**For example:**

3x + 2 is a polynomial of x. Degree of expression is **1**

4z^**2** –z + 2 is a polynomial of z. Degree of expression is **2**,

3x^**3** – 2x – 2 is a polynomial of x. Degree of expression is **3**

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**Type of polynomials**

A polynomial of degree 1 is called a linear polynomial.

**Quadratic polynomial**** :-**

A polynomial of degree 2 is called a quadratic polynomial.

A polynomial of degree 3 is called a cubic polynomial

A real number t is called a zero of a polynomial if the value of f(t) = 0

**For example**

f(x) = x^2 – 6x +8

zeros of this equation are 2 and 4 because

f(2)= 2^2 -6*2 + 8 = 0

f(4)= 4^2 – 6*4 + 8 =0

Sum and product of root of quadratic equation :-

For a equation ax^2 + bx + c = 0 , if root are α and β ,

Roots for cubic equation :-

For a equation ax^3 + bx^2 + cx + d = 0

If p(x) and g(x) are any two polynomials with g(x) is not equal to 0, then we can find polynomials q(x) and r(x) such that

If r(x) = 0 or degree of r(x) < degree of g(x).

Dividend = Divisor × Quotient + Remainder

p(x) = g(x) × q(x) + r(x),

**Question 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. **

Answer

(i)the graph does not intersect the x axis so that there are no zero in this graph

(ii)the graph intersects the x axis at one places so that here are one zeros in this graph

(iii)the graph intersects the x axis at three places so that there are three zeros in this graph

(iv)the graph intersects the x axis at two places so there are two zeros in this graph

(v) the graph intersect the x axis at four places so there are four zeros in this graph

(vi) the graph intersects the x axis at three places so that there are three zeros in this graph

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (iii) 4u² + 8u**

(iv) 4u² + 8u

Factorize the equation

Compare the equation with au² + bu + c = 0

We get

a = 4 ,b=-8 c= 0

To factorize the value we can take 4u common there

4u(u+2) = 0

First zero

4u = 0

U = 0

second zero

u+2 = 0

u = - 2

sum of zero 0 - 2 = - 2

product of zero 0 x ( - 2 ) = 0

Sum of zero -b/a = -(8/4) = -2

Product of zero c/a = 0/4 = 0

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (vi)3x² – x – 4**

(vi) 3x² – x – 4

Factorize the equation

3x² – x – 4

Compare the equation with ax² + bx + c = 0

We get

a = 3 ,b=-1 c= -4

to factorize the value we have to find two value which

sum is equal to b =-1

product is a*c = 3*-4 = -12

3 and -4 are such number which

Sum is 3 – 4 = - 1

Product is 3 *- 4 = - 12

So we can write middle term -x = 3x – 4x

We get

3x² + 3x - 4x - 4 = 0

3x(x +1) -4(x + 1) = 0

(x + 1) (3x - 4) = 0

x + 1 = 0 , 3x - 4 = 0

Solve for first zero

x = - 1

solve for second zero

3x-4 = 0

3x = 4

x = 4/3

Sum of zeros -b/a = -(-1/3) = 1/3

Product of zero c/a = -4/3 = 4/3

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (v) t² – 15**

(v) t² – 15

Factorize the equation

t² – 15 = 0

Add 15 both side we get

t² = 15

Take square root both side

t= ± √15

First zero is √15

second zerois - √15

Sum of zero √15 - √15 = 0

Product of zero √15 x -√15 = -15

Compare the equation with at² + bt +c = 0

We get

a = 1, b = 0, c = -15

Sum of zero -b/a = -(0/1) = 0

Product of zero c/a = -15/1 = -15

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (iii) 6x² – 3 – 7x **

(iii) 6x² – 3 – 7x

Factorize the equation

6x² – 7x – 3

Compare the equation with ax² + bx + c = 0

We get

a = 6 ,b=-7 c= -3

To factorize the value we have to find two value which

sum is equal to b =-7

product is a x c = 6 x -3 = -18

2 and -9 are such number which

sum is 2 – 9 = - 7

product is 2 x - 9 = - 18

So we can write middle term -7x = 2x – 9x

We get

2x(3x + 1) -3(3x + 1) = 0

(3x + 1) (2x - 3)= 0

3x + 1 = 0 , 2x - 3 = 0

Solve for first zero

3x = - 1

x = -1/3

Solve for second zero

2x-3 = 0

2x = 3

X = 3/2

Plug the values of a , b and c we get

Sum of zeros -b/a = -(-7/6) = 7/6

Product of zero c/a = -3/6 = -1/2

**1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x² – 2x – 8 **

** (i)**

Factorize the equation

We get

a = 1 ,b=-2 c= -8

To factorize the value we have to find two value which

sum is equal to b =-2

and product is a x c = 1 x -8 = -8

2 and -4 are such number which

product is 2*-4 = - 8

So we can write middle term -2 x = 2 x – 4 x

We get

x²+2x-4x-8=0

x (x + 2)-4(x + 2) = 0

(x+2) (x-4) = 0

First zero

x + 2 = 0

x = -2

Second zero

x-4 = 0

x = 4

sum of zero -2 + 4 = 2

product of zero 2 x -4 = -8

For a equation ax² + bx + c = 0 , if zeroare α and β ,

Plug the values of a , b and c we get

Sum of zeros -b/a = -(-2/1) = 2

Product of zeros c/a =-8/1 = -8

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

**(ii) 4s² – 4s + 1**

Factorize the equation

Compare the equation with as² + bs + c = 0

We get

a = 4 , b = -4, c = 1

To factorize the value we have to find two value which

sum is equal to b = -4

product is a x c = 4

-2 and -2 are such number which

sum is (– 2)+( – 2) = - 4

product is (- 2 )* (- 2) = 4

So we can write middle term - 4s = -2s – 2s

We get

4s² - 2s - 2s + 1 = 0

2s(2s -1) -1(2s -1) = 0

(2s - 1) (2s - 1) = 0

2s - 1 =0

2s = 1

s = ½

2s - 1 =0

2s = 1

S = ½

product of zeros ½ x ½ = 1/4

Sum of zero = -b/a = -(-4/4) = 1

Product of zero= c/a = 1/4 = 1/4

**Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i)1/4 , -1 (ii) √2 , 1/3 (iii) 0, √5 (iv) 1,1 (v) -1/4 ,1/4 (vi) 4,1**

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

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(1/4)x -1 = 0

Multiply by 4 to remove denominator we get

4x² - x -4 = 0

**(ii) √2 , 1/3 **

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(√2)x + 1/3 = 0

Multiply by 3 to remove denominator we get

3x² - 3√2 x + 1 = 0

**(iii) 0, √5**

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(0)x + √5 = 0

simplify it we get

x² + √5 = 0

**(iv) 1,1 **

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(1)x + 1 = 0

simplify it we get

x² - x + 1 = 0

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

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(-1/4)x + 1/4 = 0

multiply by 4 we get

4x² + x + 1 = 0

**(vi) 4,1**

Now formula of quadratic equation is

x²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

x² –(4)x + 1 = 0

x^2 –4x + 1 = 0

**1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :**

**(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2**

So quotient = x-3 and remainder 7x - 9

**(ii)p(x) = x ^{4} – 3x² + 4x + 5, g(x) = x² + 1 – x**

So quotient = x² + x-3 and remainder 8

**(iii) p(x) = x ^{4} – 5x + 6, g(x) = 2 – x²**

**Question 2. Check whether the first polynomial is a factor of the second polynomial by iding the second polynomial by the first polynomial:**

**(i)t² – 3, 2t ^{4} + 3t³ – 2t² – 9t – 12**

Remainder is 0 hence t² -3 is a factor of 2t

**(ii)x² + 3x + 1, 3x ^{4} + 5x³ – 7x² + 2x + 2**

**(iii)x³ – 3x + 1, x ^{5} – 4x³ + x² + 3x + 1**

Remainder is 2 so x³ – 3x + 1 is not a factor

**Question 3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3)**

Quotient is x² + 2x + 1 = 0

Compare the equation with ax² + bx + c = 0

We get

a = 1 ,b = 2, c = 1

To factorize the value we have to find two value which

sum is equal to b = 2

product is a x c = 1 x 1 = 1

1 and 1 are such number which

sum is 1 + 1 = 2

product is 1 x 1 = 1

So we can write middle term 2x = x + x

We get

x² + x + x + 1 = 0

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

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

x +1 = 0, x + 1= 0

x = -1 , x = - 1

**so our zeroes are - 1 -1 , √(5/3) and - √(5/3)**

**Question 4. On iding x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).**

According to ision algorithm

Dividen= Divisor × Quotient + Remainder

p(x) = g(x) × q(x) + r(x),

plug the value in formula we get

x³ – 3x² + x + 2 = g(x) *(x-2) - 2x + 4

Add 2x and subtract 4 both side we get

x³ – 3x² + x + 2 + 2x – 4 = g(x) *(x-2)

simplify and ide by x/2 we get

(x³ – 3x² + 3x– 2)/(x-2) = g(x)

**So we get g(x) = x^2 – x +1 **

**Question5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the ision algorithm**

We can write many such examples

deg p(x) = deg q(x)

We can write many such examples

P(x) = x² , q(x) = x² g(x) = 1 R(x) = 0

(ii) deg q(x) = deg r(x)

Q(x) = x , R(x) = x , p(x) = x^3 + x g(x) = x²

(iii) deg r(x) = 0

Q(x) = 1 , R(x) = 1 , p(x) = x+1, g(x) = x

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