Find The G.C.D :-
6a²-13ab+6b² , 6a²+11ab-10b² and 6a²+2ab+4b²
Answers
Answer:
Just like we find gcd of numbers using prime factorisation, in same way we will find the gcd of the expressions using factorisation method. A change of sign in last expression has been made because otherwise the last expression would not break into factor and gcd would be 1. that is rather easy, however, feeling the intention of problem, a change has been made to have a plausible solution, please check it with original question)
6a^2 - 13ab + 6b^2
= 6a^2 - 9ab - 4ab + 6b^2
= 3a(2a - 3b) - 2b(2a - 3b)
= (2a - 3b) (3a - 2b)
6a^2 + 11ab - 10b^2
= 6a^2 + 15ab - 4ab - 10b^2
= 3a(2a +5b) - 2b(2a + 5b)
= (2a + 5b) (3a - 2b)
6a^2 + 2ab - 4b^2 (in question the term 4b^2 is +ve, however to have the problem a solution, 4b^2 has to be -ve, so I am solving by taking it -ve)
= 6a^2 + 6ab - 4ab - 4b^2
= 6a (a + b) - 4b(a + b)
= (a + b) (6a - 4b)
= 2(a + b) (3a - 2b)
As we can see that the term (3a - 2b) is a common factor of all the given expressions. Hence the required gcd is (3a - 2b).