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Posted by Last Bencher 5 years, 8 months ago
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Posted by Jeevith Arumugam 5 years, 8 months ago
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Yogita Ingle 5 years, 8 months ago
Since there are 2 zeroes, the polynomial is of degree 2.
Let a and b represents the zeroes of the polynomial.
Let {tex}a=2+\sqrt{3} and b=2-\sqrt{3}{/tex}
Since, the polynomial is of degree 2
Polynomial would be of the form {tex}{x^{2}-(a+b) x+a b} {/tex}
Therefore computing the values,
{tex}{a+b=(2+\sqrt{3})+(2-\sqrt{3})=4}{/tex}
{tex}{a b=(2+\sqrt{3}) \times(2-\sqrt{3})=4-3=1}{/tex}
Therefore, the polynomial is {tex}{x^{2}-4 x+1} {/tex}
Posted by Bharani Ajith 5 years, 8 months ago
- 1 answers
Gaurav Seth 5 years, 8 months ago
Let a be any positive integer
by EDL a = bq +r
0 ≤ r < b
possible remainders are 0, 1, 2 , 3
this shows that a can be in the form of 4q, 4q+1, 4q+2, 4q+3 q is quotient
as a is odd a can't be the form of 4q or 4q+2 as they are even
so a ill be in the form of 4q + 1 or 4q+3
hence proved
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Krishn Joshi 5 years, 8 months ago
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Posted by Mansi Sachdeva 5 years, 8 months ago
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Yogita Ingle 5 years, 8 months ago
Let p be any positive integer.
On dividing p by 5 , Let M be Quotient and R be Remainder.
Then , By Euclid division lemma , we have
p = 5q+1 , where r = 0,1,2,3,4
p = 5q , where R = 0
p =5q + 1 , where R = 1
p = 5q + 2 , where R = 2
p = 5q + 3 , where R = 3
p = 5q + 4 , where R = 4
p = 5q , 5q+2 , 5q+4 are even values of N.
Thus
When p is odd , it is in the form of 5q , 5q+1 and 5q+3.
Posted by Anu Mishra 5 years, 8 months ago
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Posted by Kavitha Kombiah 5 years, 8 months ago
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Gaurav Seth 5 years, 8 months ago
Let n be a positive odd integer. We need to show that n can be written in any one of the form of 8q+1, 8q+3, 8q+5 or 8q+7
According to division algorithm,
we can write any number ‘a’ in the form
a = 8q + r
where q is any integer and 0 <= r <= 7. So r can be 0, 1, 2, 3, 4, 5, 6 or 7.
Thus, a can be written as
a = 8q
a = 8q+2
a = 8q+3
a = 8q+4
a = 8q+5
a = 8q+6
a = 8q+7
We need only odd numbers. Since 8q, 8q+2, 8q+4, and 8q+6 are divisible by 2, they are even numbers.
So any odd integer can be written as any one of the remaining forms which are (8q+1, 8q+3, 8q+5 or 8q+7.).
Posted by Gian Singh 5 years, 8 months ago
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Eshnoor ???? 5 years, 8 months ago
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Posted by Skm 2006 5 years, 8 months ago
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Yogita Ingle 5 years, 8 months ago
let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ <r<3
so the remainders may be 0,1 and 2
so n may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3
CASE 2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3
CASE 3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3
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