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If that's + 42q2, the answer is (p - 7q)(p - 6q)
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What are the factors and composite factors of 15?
A number will have 12 factors if it satisfies any of the following prime factor configurations: P11 (P1)5*(P2) (P1)3*(P2)2 (P1)2*(P2)*(P3)
What are the factors of p2-2p plus 2?
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
Is Composite numbers have only 2 factors?
It depends on whether 1 is considered a factor. If yes, then the answer to the question is no, since 1 and the number itself are the two factors implying that it is a prime. If 1 is not considered as a factor, then the square of any prime meets the requirements. If p is a prime, then p and p2 are the only two factors of p2.
What is the pmf of trinomial distribution?
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
What is the mathematical proof to show that the number of prime numbers is infinite?
The proof is by contradiction: assume there is a finite number of prime numbers and get a contradiction by requiring a prime that is not one of the finite number of primes. Suppose there are only a finite number of prime numbers. Then there are n of them.; and they can all be listed as: p1, p2, ..., pn in order with there being no possible primes between p(r) and p(r+1) for all 0 < r < n. Consider the number m = p1 × p2 × ... × pn + 1 It is not divisible by any prime p1, p2, ..., pn as there is a remainder of 1. Thus either m is a prime number itself or there is some other prime p (greater than pn) which divides into m. Thus there is a prime which is not in the list p1, p2, ..., pn. But the list p1, p2, ..., pn is supposed to contain all the prime numbers. Thus the assumption that there is a finite number of primes is false; ie there are an infinite number of primes. QED.
For any prime number p 5 prove that p square 2 is composite?
By definition, prime number p has two factors, 1 and p. P2 has three factors, 1, p and p2 Therefore, p2 is composite.
Can a multiple of a prime number be prime?
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
What are the factors and composite factors of 15?
A number will have 12 factors if it satisfies any of the following prime factor configurations: P11 (P1)5*(P2) (P1)3*(P2)2 (P1)2*(P2)*(P3)
What are the factors of p2-2p plus 2?
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
What is the largest three digit number that has four factors?
If factors include 1 and the number itself then the any number (n) with only 4 factors has to be the product of two prime numbers, p1 and p2. The factors are then n, p1, p2, and 1. The largest 3-digit number with only 4 factors is 998. The factors are 998, 499, 2 and 1
How do you find the factor of p2 plus 3p-4?
p2 + 3p - 4What two factors of - 4 add up to 3 ??(p - 1)(p + 4)===============so,p = 1p = - 4
How do you factorize 3p4 - 12p2?
3 and p2 are both factors of each part of the binomial, so divide it from each part and see what you have left. (3p4 / 3p2) - (12p2 / 3p2) (3p2)(p2-4)◄
How do you caluculate the GCD of two prime numbers?
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
What is buffer P2?
The composition of Buffer P2 is:200 mM NaOH1% SDS (w/v)Buffer P2 is the lysis buffer
What are the factors of p2 plus 4pq plus 4q2-b2-2bc-c2?
-(b + c - p - 2q)(b + c + p + 2q)
How much does a P2 cost?
A P2 costs $9.00 per user/month.
How do you Factorise p2 plus 10d plus 7?
p2+10d+7
