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Complete 10th Class Maths Solution
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.2 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.3 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.4 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 2 Exercise 2.1 Polynomials Real Number
EXERCISE:-2.3
Question: 1 DIVED THE POLYNOMIAL p(x) BY THE POLYNOMIAL g(x) AND FIND THE QUOTIENT AND REMAINDER IN EACH OF THE FOLLOWING:
(i) p(x) = x³-3x+5x-3 g(x)=x²-2
(ii) p(x)=x power4-3x²+4x+5 g(x)= x²+1-x
(iii) p(x)= x power4 -5x +6 g(x) =2-x²
SOL:-
(i) p(x) = x³-3x+5x-3 g(x)=x²-2
x²-2 ) x³-3x+5x-3 ( x-3
x³ -2x
– +
-3x² +7x -3
– 3x² +6
+ –
7x-9
quotient =x-3
remainder =7x-
(ii) p(x)=x power4-3x²+4x+5
g(x)= x²+1-x
x²+1-x ) x power 4-3x²+4x+5 (x²+ x+1
x power 4 – x³ +x²
– + –
x³ -4x² +4x+ 5
x³ -x² +x
– + –
-3x² +3x+5
-3x² +3x -3
+ – +
8
quitient =x²+x-3
remainder =8
(iii) p(x)= x power4 -5x +6 g(x) =2-x²
2-x²) x power4 -5x +6 (-x²-2
x power 4 -2x²
– +
2x² -5x+ 6
2x² -4
– +
-5x+10
quitient =x²-2
remainder =-5x+10
QUE:-2 CHECK WHETHER THE FIRST POLYNOMIAL IS A FACTORE OF THE SECOND POLYNOMIAL BY DIVIDING THE SECOND POLYNOMIAL BY THE FIRST POLYNOMIAL :
(i) t²-3, 2tpower(4)+3t³-2t²-9t-12
(ii) x²+3x +1 , 3xpower(4)+5x³-7x²+2x+2
(iii) x³-3x+1, xpower(5) -4x³+x²+3x+1
SOL:-
(i) t²-3, 2tpower(4)+3t³-2t²-9t-12
t²-3=t²+0t-3
t²+0t-3 ) 2tpower(4)+3t³-2t²-9t-12 ( 2t²+3t+4
2tpower(4) -6t²
– +
3t³ +4t²-9t-12
3t³ +ot² -9t
– – +
4t²-12
4t²-12
– +
0
since the remainder =0,
hence , t²-3 is a factor of 2tpower(4)+3t³-2t²-9t-12 .
(ii) x²+3x +1 , 3xpower(4)+5x³-7x²+2x+2
x²+3x +1 ) 3xpower(4)+5x³-7x²+2x+2 ( 2t²+3t+4
3xpower(4) +9x³ +3x²
– – –
-4x³ -10x²-2x-2
-4x³ -12x² -4x
+ + +
2x²+6x+2
2x²+6x+2
0
since the remainder =0,
hence , x²+3x +1 is a factor of 3xpower(4)+5x³-7x²+2x+2 .
(iii) x³-3x+1, xpower(5) -4x³+x²+3x+1
x³-3x+1 ) xpower(5) -4x³+x²+3x+1 ( x²-1
xpower(5) -3x³ +x²
– + –
-x³ +3x +1
-x³ +3x -1
+ – +
2
since the remainder ≠0,
hence , x²+3x +1 is not a factor of xpower(5) -4x³+x²+3x+1 .
QUE:-3OBTAIN ALL OTHER ZEROES OF 3x(POWER4)+6x³-2x²-10x-5,IF TWO OF ITS ZEROES ARE root5/3 AND -root5/3.
SOL:- p(x) = 3x(POWER4)+6x³-2x²-10x-5
since the two zeroes are root5/3 and-root5/3
∴ {x-root5/3}{x+root5/3} = {x²-5/3} is a factor of 3x(POWER4)+6x³-2x²-10x-5
therefore , we divide the given polynomial by x²-5/3
x²+0x-5/3 )3x(POWER4)+6x³-2x²-10x-5(3x² +6x +3
3x(POWER4)+0x³-5x²
– – +
6x³ +3x² -10x -5
6x³ -10x
– +
3x² -5
3x² -5
– +
0
we factorize x²+2x+1
= (x+1)²
therefore , its zero is given by x+1=0 or x=-1
as it has the term (x+1)² ,therefore, there will be 2 zeroes at x=-1
hence , the zeroes of the given polynomial are root5/3 ,-root5/3 -1 and -1
QUE:-4ON DIVIDING x³-3x²+x+2 POLYNOMIAL g(x) ,THE QUOTIENT AND REMAINDER WERE x-2 AND REMAINDER WERE x-2 AND -2x+4 , RESPECTIVELY FIND g(x).
SOL:-
p(x)= x³-3x²+x+2 (DIVIDEND)
g(x) = ? (DIVISOR)
QUTIENT=(x- 2)
REMAINDER =(-2x+4)
DIVIDEND=DIVISOR×QUOTIENT +REMAINDER
x³-3x²+x+2=g(x) ×(x- 2) +(-2x+4)
x³-3x²+x+2 +2x-4=g(x) ×(x- 2)
x³-3x²+3x-2 =g(x) ×(x- 2)
g(x) IS QUOTIENT WHEN WE DIVIDE x³-3x²+3x-2 BY(x- 2)
x-2 )x³-3x²+3x-2(x²-x+1
x³ -2x²
– +
-x² +3x -2
-x² +2x
+ –
x-2
x-2
– +
∴ g(x) =x²-x+1
QUE :-5 GIVE EXAMPLES OF POLYNOMIAL p(x),q(x),g(x)AND r(x) ,WHICH DIVISION ALGORITAM AND
(i) deg p(x) =deg q(x)
(ii) deg q(x)= deg r(x)
(iii) deg r(x) =0
SOL:-
(i) Accordindg to the division algoritham , if p(x) and g(x) are two polynomials with g(x)≠0,then we can find polynomials q(x) and r(x)such that
p(x)=g(x) × q(x) × r(x),
where r(x) = 0 or degree of r(x) <degree of g(x)
degree of a polynomial is the highest power of the variable in the polynomial.
deg p(x)=deg q(x)
degree of quotient will be equal to degree of dividend when divisor is constant .
let us assume the division of 8x² +6x +2 by 2 .
here ,p(x) = 8x² +6x +2
g(x) = 2
q(x) = 4x² +3x +1 and r(x) =0
degree of p (x) and q(x) is the same i.e.,2.
checking for division algorithm ,p(x)=g(x) × q(x) × r(x)
8x² +6x +2=(2)(4x² +3x +1)+0
thus , the division algorithm is satisfied .
(ii) deg q(x)= deg r(x)
According to the division algoritham , if p(x) and g(x) are two polynomials with g(x)≠0,then we can find polynomials q(x) and r(x)such that
p(x)=g(x) × q(x) × r(x),
where r(x) = 0 or degree of r(x) <degree of g(x)
degree of a polynomial is the highest power of the variable in the polynomial.
deg p(x)=deg q(x)
degree of quotient will be equal to degree of dividend when divisor is constant .
let us assume the division of t³ +t by t²
here , p(x) =t³ +t g(x) =t² q(x) =t and r(x) =t
degree of q (x) and r(x) is the same
checking for division algorithm ,p(x)=g(x) × q(x) × r(x)
t³ +t =( t²)tt+t t³+t=t³ +t
thus , the division algorithm is satisfied .
(iii) deg r(x) =0
Accordindg to the division algoritham , if p(x) and g(x) are two polynomials with g(x)≠0,then we can find polynomials q(x) and r(x)such that
p(x)=g(x) × q(x) × r(x),
where r(x) = 0 or degree of r(x) <degree of g(x)
degree of a polynomial is the highest power of the variable in the polynomial.
deg p(x)=deg q(x)
degree of quotient will be equal to degree of dividend when divisor is constant .
degree of remainder will be 0 when remainder comes to a constant.
let us assume the division of x³ +1 by x²
clearly , the degree of r(x) is 0. checking for division algorithm,
p(x)=g(x) × q(x) × r(x)
x³ +1 =(x²)xx+1x³+1=x³ +1
thus , the division algorithm is satisfied.
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