NCERT SOLUTIONS FOR CLASS 10 MATHS

NCERT SOLUTIONS FOR CLASS 10 MATHS

NCERT SOLUTIONS FOR CLASS FOR CLASS 10 MATHS CHAPTER – 2

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-: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.

NCERT MATHS CLASS 10
                                                                       NCERT MATHS CLASS 10

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

10th maths solution
10th maths solution

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

 

 

10th maths
                                                                            10th maths

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

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

ncert 10 maths
ncert 10 maths

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

ncert chapter 2 solution

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

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

class 10 math

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

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

class 10 math solution

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,

class 10 maths chapter 2 solution
class 10 maths chapter 2 solution

 

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 ).

maths solution
maths solution

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).

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