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 *
