## NCERT SOLUTIONS FOR CLASS 10 MATHS

### NCERT SOLUTIONS FOR CLASS FOR CLASS 10 MATHS CHAPTER – 2

Hello students I am Bijoy and welcome to my educational forum/portal. Today we will discuss about * NCERT SOLUTIONS FOR CLASS 10 MATHS* , so let’s start.

-:*POLYNOMIALS:-*

What is polynomial ?

A mathematical expression on various types of function like algebraic, exponential, trigonometric etc. which maintain the any n-th degrees in variable, that expression is called as **polynomial** on that function. Where ‘n’ is an integer.

Note – but in the class of ten (X) we only discuss the algebraic polynomial on 1th, 2^{nd}, 3^{rd} degree polynomials. Ex – x^3+ 2x^2-3

**EXERCISE – 2.1**

Using formula –** Graphical representation**

A given polynomial p(x) with degree of n, then the number of zero or zeroes is/are represented by the no of intersecting point / points on x – axis of the graph of y = p(x).

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.

i) Ans. **no of zeroes = no of intersected point on x – axis, graph of y = p(x)**

Hence according to question, the first polynomial has no zero, because in this case has no intersecting point on x -axis or here the graph does not intersect anywhere on the x – axis.

ii) No of zero is equal to ‘1’ (one)

Note – Students, in this case you can see have no intersected point of the graph of y = p(x) at x – axis, but remember just one thing that no of zeroes is equal to no of points of intersecting only on x – axis, does not count from y – axis.

iii) In the no of question 3^{rd} the value of zeroes are 3 (three). Because here the graph of y = p (x) cut x – axis at the three points.

iv) In this case no of zeroes are equal to ‘2’ (two).

Note – here the graph of y = p(x) intersects the x- axis at the two points.

v) Students you just observe that in the question no (v) the graph of y = p(x) intersects the x -axis at the four points.

Hence no of zeroes are equal to ‘4’ (four).

vi) In the question (vi) the graph of y = p(x) touches the x – axis at the three points, hence the no of zeroes are ‘3’ (three).

#### CLASS 10 MATHS SOLUTION

**EXERCISE – 2.2**

Using formula :-

i) find the value of zeroes

y = P(x) = 0

ii) Required polynomial by the given sum and product of zeroes = x ^{2}– (α+β) x + α.β

solution :

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

i) Given, x^{2} -x-8

Now let y = p(x) = x^{2} -x-8

hence p(x) = 0

or, x^{2} -x-8 = 0

Now factorize the polynomial and splitting 2x we get,

Or, x^{2} -(4-2)x-8=0

note :- first divide the ‘8’ in two numbers that after minus this two numbers just get 2 (i.e 4-2=2).

Or, x^{2} -4x+2x-8=0

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

Now taking common (x-4), we get

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

Hence the required zeroes are

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

∴ X=4 or ∴ x=-2

∴ The required zeroes of given polynomial are 4 and -2.

How to verify the relationship between the zeroes and the coefficients ?

Let, the zeroes a = 4 and b = -2

∴ a+b = 4+(-2 )= 2 = – coefficient of x/coefficient of x^{2}

∴ a.b = 4.(-2) = -8 = constant term/coefficient of x^{2}

ii) Given, polynomial 4s^{2} – 4s + 1

hence, 4s^{2} – 4s + 1= 0

now split the 4 as (2+2) we get,

or, 4s^{2}– (2+2) s + 1= 0

or, 4s^{2 }-2s -2s +1 = 0

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

now taking common (2s-1)

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

hence the required zeroes are

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

∴ S =1/2 or, ∴ s=1/2

∴ The required zeroes of given polynomial are 1/2.

Now check the relationship between the coefficients and zeroes

a=1/2 and b=1/2

Hence, a+b=1/2+1/2=1= -(coefficient of x)/coefficient of s^{2}

a.b= \( \frac{1}{2}*\frac{1}{2}=\frac{1}{4} \)= (constant term)/coefficient of s^{2}

iii) Given, p(x)=6x^{2}-3-7x

hence 6x^{2}-3-7x=0

or, 6x^{2}-7x-3=0

now split the 7 as (9-2=7)

or, 6x^{2}-(9-2)x-3=0

or, 6x^{2}-9x+2x-3=0

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

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

hence the required zeroes are

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

Therefore x=3/2 or, x=-1/3

∴ The required zeroes of given polynomial are 3/2 and -1/3.

Now checking the relation between coefficients and zeroes

a=3/2 and b= -1/3

Hence a+b=3/2+(-1/3)=7/6= -(coefficient of x )/coefficient of x^{2}

a.b= \( \frac{3}{2}*\frac {-1}{3}=\frac{-1}{2} \)= constant term/coefficient of x^{2}

iv) given, p(x) = 4u^{2}+8u

hence, 4u^{2}+8u+0=0

now taking common 4u and we get

4u(u+2)=0

Therefore the required zeroes are

4u=0 or, (u+2)=0

∴ U=0 or, ∴ u=-2

∴ The required zeroes of given polynomial are 0 and -2.

Now check the relationship between the coefficients and zeroes

a = 0 and b=-2

a+b=0-2=-2=-(coefficient of u)/coefficient of u^{2})

a.b=0.(-2)=0= constant term/coefficient of u^{2}

v) given, p(t)=t^{2}-15

hence t^{2}+0-15=0

or, t^{2} – (√15)^{2} = 0

or, ( t – √15 ) (t – √15 ) = 0

Hence the zeroes of given polynomial are

( t + √15 ) = 0 or ( t – √15 ) = 0

∴ t = – √15 or, t = √15

∴ The required zeroes of given polynomial are – √15 and √15.

Now check the relationship between the coefficients and zeroes

therefore a = – √15 and b = √15

∴ a + b = – √15 + √15 = 0 = -(coefficient of t)/coefficient of t^{2})

vi) given, p(x) = 3x^{2}-x-4

hence, 3x^{2}-x-4=0

now splitting the 1.x as (4-3=1)

we get, 3x^{2} – (4-3)x -4=0

or, 3x^{2} -4x+3x-4=0

or, x ( 3x – 4) + 1 (3x-4) = 0

now taking common by ( 3x – 4 ) we get,

or, ( 3x – 4 ) ( x + 1 ) = 0

hence the required zeroes are

( 3x – 4 ) = 0 or ( x + 1 ) = 0

∴ x = 4/3 or ∴ x = -1

∴ The zeroes of given polynomial are 4/3 and -1.

Now check the relationship between the coefficients and zeroes

Therefore a = 4/3 and b = -1

A+b = 4/3 + (-1) = 1/3 = – ( coefficient of x ) / ( coefficient of x^{2} )

And a.b = \( \frac{4}{3}*-1= -\frac{4}{3} \) = constant term / coefficient of x^{2}

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

Using formula : –

The required polynomial = x^{2} – ( sum of zeroes )·x + ( product of zeroes )

Solution –

(i) Given, sum of zeroes = \( \frac{1}{4} \) and product of zeroes = -1

Hence the required polynomial

= x^{2} – \( \frac{1}{4}x \) + (-1)

= 4x^{2} – x + 4

(ii) Given, sum of zeroes = √2 and product of zeroes = \( \frac{1}{3} \)

Hence the required polynomial is

= x^{2} – ( √2 ) x + \( \frac{1}{3} \)

= 3x^{2} – 3 √2 x + 1

(iii) Given, sum of zeroes = 0 and product of zeroes = √5

Hence the required polynomial is

= x^{2} – (0).x + √5

=x^{2} + √5

(iv) Given, sum of zeroes = 1 and product of zeroes = 1

Hence the required polynomial is

= x^{2} – (1) x + 1

= x^{2} – x + 1

(v) Given, sum of zeroes = -1/4 and product of zeroes = 1/4

Hence the required polynomial is

= x^{2} – (\( \frac{-1}{4} \)) x + \( \frac{1}{4} \)

= 4x^{2} + x + 1

(vi) Given, sum of zeroes = 4 and product of zeroes = 1

Hence the required polynomial is

= x^{2} – 4x + 1

**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 :

(i) p(x) = x^{3} – 3x^{2} + 5x – 3, and g(x) = x^{2} – 2

Hence the quotient is (x-3) and remainder is ( 7x – 9 )

(ii) Given, p(x) = x^{4} – 3x^{2} + 4x + 5 and g(x) = x^{2} + 1 – x

∴ The quotient is x^{2} + x – 3 and remainder is 8

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

Hence the quotient is – x^{2} – 2 and remainder is – 5x + 10.

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :

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

so, the remainder is zero (0), hence the first is the factor of second polynomial.

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

Since the remainder is zero, first polynomial is the factor of second poynomial.

(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

In this case remainder is 2 not zero, the the first polynomial is not factor of second polynomial.

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

Solution : Given the zeroes are √5/3 and – √5/3. Hence the two factors of given polynomial are ( x -√5/3 ) and ( x + √5/3 ).

Therefore , ( x + √5/3 ) ( x -√5/3 ) = ( x^{2} – \( \frac{5}{3} \) ) is also the factor of 3x^{4} + 6x^{3} – 2x^{2} 10x – 5.

Now divide the polynomial 3x^{4} + 6x^{3} – 2x^{2} 10x – 5 by ( x^{2} – \( \frac{5}{3} \) ) we get,

therefore the quotient is 3x^{2} + 6x + 3, hence 3x^{2} + 6x + 3 also the factor of 3x^{4} + 6x^{3} – 2x^{2} 10x – 5.

So, if we want to find another zeroes of the polynomial we need to do 3x^{2} + 6x + 3 = 0.

Now by solving the quadratic we get,

Or, 3x^{2} + 6x + 3 = 0

Or, 3x^{2} + 3x + 3x + 3 = 0

Or, 3x ( x + 1 ) +3 ( x + 1) = 0

Or, ( x + 1 ) ( 3x + 3 ) = 0

Hence the required zeroes are,

( x + 1) = 0 or, ( 3x + 3) = 0

∴ x = -1 or, ∴ x = – 1

Therefore the zeroes of 3x^{2} + 6x + 3, hence 3x^{2} + 6x + 3 are √5/3, -√5/3, -1 and -1

4. On dividing x^{3} – 3x^{2} +x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).

Solution :

We know, polynomial or dividend = divisor × quotient + remainder

Given, polynomial = x^{3} – 3x^{2} +x + 2, quotient = x – 2 and remainder = – 2x + 4, and now putting the value we get,

here divisor = g(x)

x^{3} – 3x^{2} +x + 2 = g(x) × (x – 2) + 4 – 2x

∴ g(x) = (x^{3} – 3x^{2} +3x – 2)/(x – 2), hence g(x) must be quotient when we divide x^{3} – 3x^{2} +x + 2 by ( x – 2 ).

Therefore the g(x) = x^{2} – x + 1

5. Give an examples of polynomials 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

Solution :

(i) p(x) = 2x^{5} – 6x^{3} + 4x^{2} + 10

∴ q(x) = ( x^{5} – 3x^{3} + 2x^{2} + 3 ), divisor = 2 and remainder [ r(x) ] is 4.

Therefore the degree of p(x) and q(x) are equal.

(ii) Assume p(x) = x^{3} – 2x +1 and g(x) = x^{2}, now if we divide p(x) by g(x) then quotient must be = x and remainder will be -2x + 1. means q(x) = x and r(x) = -2x + 1.

Hence the deg of q(x) = deg of r(x) .

(iii) deg r(x) = 0, means remainder must be a constant term and that example already in example no (i).