If our purpose is factorizing any polinomial of degree n, we must rewrite it as a product of a polynomial of degree n-1 and another one of degree 1:
P_n (x_1, x_2, ..., x_{n-1}) = P_{n-1} (x_1, x_2, ..., x_{n-1}) cdot P_1 (x_1, x_2, ..., x_{n-1})
Although it is not necessary to know this, there must be n-1 variables, with a maximum degree of n-1 (i.e. x^{n-1} must be the maximum power).
In our exercise, we want to write 14x^2+23xy+3y^2 as a product like this one:
(alpha x + beta y) cdot (gamma x + delta y)
where alpha, beta, gamma, delta are the coefficients we want to determine.
If we solve the product:
(alpha x + beta y) cdot (gamma x + delta y) =
= (alpha gamma cdot x^2) + (alpha delta cdot xy) + (beta gamma cdot yx) + (beta delta cdot y^2)
If we group all terms according to the power of variables x and y, we obtain:
(alpha gamma) x^2 + (alpha delta + beta gamma) xy + (beta delta)y^2
And we can relate it with our trinomial:
14 x^2 + 23 xy + 3 y^2 = (alpha gamma) x^2 + (alpha delta + beta gamma) xy + (beta delta) y^2
Every coefficient at the left-hand side must be equal to the other one at the right-hand side, so:
14 = alpha cdot gamma
23 = alpha cdot delta + beta cdot gamma
3 = beta cdot delta
We have 3 equations and 4 unknowns, so we can give a random value to one of them (for instance, beta = 1). This is completely OK, because factorization is not unique (there are infinite factorizations, each one of them with a random value of beta).
Solving sistem, we obtain:
{alpha = 7, beta = 1, gamma = 2, delta = 3}
This way, factorization is:
14 x^2 + 23 xy + 3 y^2 = (7x + y) cdot (2x + 3y)
*Tip: you can factorize this polynomial (and any other one) by writing it on WolframAlpha *
