Let’s find G.C.D of the following algebraic expression:

x² + 3x + 2, x² + 4x + 3, x² + 5x + 6

Factors of x² + 3x + 2

= x² + 2x + x + 2

=x(x + 2) + 1(x + 2)

=(x + 1)(x + 2)
Factors of x² + 4x + 3

= x² + 3x + x + 3

=x(x + 3) + 1(x + 3)

=(x + 1)(x + 3)
Factors of x² + 5x + 6

= x² + 2x + 3x + 6
=x(x + 2) + 3(x + 2)

=(x + 2)(x + 3)

GCD is the greatest common divisor, which is equal to the product of all the common divisors.

∴ the GCD of x² + 3x + 2, x² + 4x + 3, x² + 5x + 6 is 1 because no other factor is common to all the three and only 1 which is a universal factor of every number is the answer.