# NCERT SOLUTIONS FOR CLASS 10 MATHS

## 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, 2nd, 3rd  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 3rd  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, x2 -x-8

Now let y = p(x) = x2 -x-8

hence p(x) = 0

or, x2 -x-8 = 0

Now factorize the polynomial and splitting 2x we get,

Or, x2 -(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, x2 -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 x2

∴ a.b = 4.(-2) = -8 = constant term/coefficient of x2

ii) Given, polynomial 4s2 – 4s + 1

hence, 4s2 – 4s + 1= 0

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

or, 4s2– (2+2) s + 1= 0

or, 4s2 -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 s2

a.b= ​$$\frac{1}{2}*\frac{1}{2}=\frac{1}{4}$$​= (constant term)/coefficient of s2

iii) Given, p(x)=6x2-3-7x

hence 6x2-3-7x=0

or, 6x2-7x-3=0

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

or, 6x2-(9-2)x-3=0

or, 6x2-9x+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 x2

a.b= ​$$\frac{3}{2}*\frac {-1}{3}=\frac{-1}{2}$$​= constant term/coefficient of x2

iv) given, p(x) = 4u2+8u

hence, 4u2+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 u2)

a.b=0.(-2)=0= constant term/coefficient of u2

v) given, p(t)=t2-15

hence  t2+0-15=0

or, t2 – (√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 t2)

vi) given, p(x) = 3x2-x-4

hence, 3x2-x-4=0

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

we get,    3x2 – (4-3)x -4=0

or, 3x2 -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 x2 )

And   a.b = ​$$\frac{4}{3}*-1= -\frac{4}{3}$$​ = constant term / coefficient of x2

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

Using formula : –

The required polynomial = x2 – ( 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

= x2 – ​$$\frac{1}{4}x$$​ + (-1)

= 4x2 – x + 4

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

Hence the required polynomial is

= x2 – ( √2 ) x + ​$$\frac{1}{3}$$

= 3x2 – 3 √2 x + 1

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

Hence the required polynomial is

= x2 – (0).x + √5

=x2 + √5

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

Hence the required polynomial is

= x2 – (1) x + 1

= x2 – x + 1

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

Hence the required polynomial is

= x2 – (​$$\frac{-1}{4}$$​) x + ​$$\frac{1}{4}$$

= 4x2 + x + 1

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

Hence the required polynomial is

= x2 – 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) = x3 – 3x2 + 5x – 3, and g(x) = x2 – 2

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

(ii) Given, p(x) = x4 – 3x2 + 4x + 5  and g(x) = x2 + 1 – x

∴ The quotient is x2 + x – 3 and remainder is 8

(iii) Given, p(x) = x4 – 5x  + 6, and  g(x) = 2 – x2

Hence the quotient is – x2 – 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) t2 – 3, 2t4 + 3t3 – 2t2 – 9t 12

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

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

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

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 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 3x4 + 6x3 – 2x2 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 ) = ( x2 – ​$$\frac{5}{3}$$​ ) is also the factor of 3x4 + 6x3 – 2x2 10x – 5.

Now divide the polynomial 3x4 + 6x3 – 2x2 10x – 5 by ( x2 – ​$$\frac{5}{3}$$​ ) we get,

therefore the quotient is 3x2 + 6x + 3, hence 3x2 + 6x + 3 also the factor of 3x4 + 6x3 – 2x2 10x – 5.

So, if we want to find another zeroes of the polynomial we need to do 3x2 + 6x + 3 = 0.

Now by solving the quadratic we get,

Or, 3x2 + 6x + 3 = 0

Or, 3x2 + 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 3x2 + 6x + 3, hence 3x2 + 6x + 3 are √5/3, -√5/3, -1 and -1

4. On dividing x3 – 3x2 +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 = x3 – 3x2 +x + 2, quotient = x – 2 and remainder = – 2x + 4, and now putting the value we get,

here divisor = g(x)

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

∴ g(x) = (x3 – 3x2 +3x – 2)/(x – 2), hence g(x) must be quotient when we divide x3 – 3x2 +x + 2 by ( x – 2 ).

Therefore the g(x) = x2 – 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) = 2x5 – 6x3 + 4x2 + 10

∴ q(x) = ( x5 – 3x3 + 2x2 + 3 ), divisor = 2 and remainder [ r(x) ] is 4.

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

(ii) Assume p(x) = x3 – 2x +1 and g(x) = x2, 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).