CSIR NET MATHEMATICS| Abstract Algebra | Use Inspection Method to solve the problem (DEC 2018)

In this video, we are going to discuss example of abstract algebra from CSINET December 2018. Example number 82. Booklet C. Set C. It is a problem of 4.75 marks. Let fx is equal to x power 7 minus 100 and 5x plus 12. Then which of the following are correct? First, fx is divisible over q. Second, there exists an integer m such that fm equal to 105 Third option there exists an integer m such that f of m equal to 2 and the fourth option fm is not a prime number for any integer m. We begin with the inspection method. So we inspect the given polynomial fx is an even integer then m equal to 2k so fm equal to 2k power 7 minus 105 2k plus 12. So seventh power of 2k is even then even into odd is also a even number. plus 12 is even number so sum sum and the subtraction of even numbers will be the even number if m is an odd number or odd integer then m equal to 2k plus 1 so f of m equal to 2k plus 1 power 7 minus 100 and 5 2k plus 1 plus 12 so seventh power of odd is odd number odd into odd will also be odd number plus 12 is even number addition or subtraction of of odd numbers. will be the even number so even plus even equal to even therefore if m is an odd integer fm has even value and if we put m equal to 0 fm equal to 12 therefore for any integer m fm will be the even numbers so the second option where fm is equal to 105 which is an odd therefore We cannot find any integer m such that f of m is an odd number. Hence, the second option is wrong now. For the third option, there exists an integer m such that fm equal to 2. So, this 2 is an even number. So, there may be possibility of m such that fm will be 2. And the fourth option, fm is not a prime number for any integer m. So, prime numbers are odd except 2. So therefore fm equal to 2 is correct then the fourth option is wrong and fm equal to 2 is wrong then fourth option is correct and the first option is a reducible over q. So we have one theorem statement Einstein's reducible criteria. The criteria says that if fx is equal to an x power n plus an minus 1 x power n minus 1 plus an minus 2 x power n minus 2 a 0. Is an integer polynomial that means an, an-1, an-2, a0 are integers in qx then fx is irreducible over q if there exists some prime number p such that p divides an-1, p divides an-2, p divides a0 but p does not divide an and p square does not divide a0. For the given polynomial, fx is equal to x power 7 minus 105x plus 12. If we P equal to 3, then prime P equal to 3 divides 105, 3 divides 12, but 3 does not divide an is equal to 1 and P square, that means 3 square does not divide 12, that is a 0. Therefore, it satisfies this condition. So, f(x) is irreducible over Q. So, the first option that f(x) is reducible over Q, which is wrong. Now, for the third option, f(m) equal to 2. So m power 7 minus 105 m plus 12 equal to 2. So by simplifying m power 7 minus 105 m plus 10 equal to 0. Now take the assumption g(m) is equal to m power 7 minus 105 m plus If we take p equal to 5, then 5 divides 105, 5 divides 10, but 5 does not divide 1, that means a1, and 5 squared does not divide 10, that means a0. Therefore, it also satisfies the Eisenstein criteria, and therefore g(m) is irreducible over Q, therefore it is irreducible over Z. That means we are not able to find any integer such that If g(m) has value 0, ultimately f(m) equal to 2 will not be possible for any integer and therefore the third option is wrong. That means on the fourth option is correct. That is fm is not a prime number for any integer m because f(m) is always an even number and f(m) equal to 2 is not possible. We already discussed in the third option. So fourth option is only correct option. Thanks for watching. For more videos, like, subscribe, press the bell icon of NextGenMaths.