Is it possible to use system of equations when factoring trinomials?

Question

I was wounder if it was possible to use systems of equations when factoring trinomials as the product of two binomials. For example, if $x^2+5x+6$ can be understood be $x^2+ax+bx+ab$ and is factored as $(x+a)(x+b)$. Is it possible to solve for the variables "a" and "b" with a system of equations like $a+b=5 , ab=6$ using the the substitution method? I've tried a couple of times, and I'm not sure if it's poor manipulation of the equations on my part or if it just doesn't work. I've tried searching for examples of it being done via google an this website, but, thus far, I haven't found anything addressing it. I would greatly appreciate it if anyone could help me out, and assuming that it isn't possible, I'd like to know why.

Answer 1

If you do that, then, in the end, you'll get the original equation once again. In fact, from $a+b=5$, you get that $b=5-a$. And then, from $ab=6$, you get that $a(5-a)=6$. But$$a(5-a)=6\iff a^2-5a+6=0.$$So, this approach is a waste of time.

Answer 2

Solving the system $ \ a + b \ = \ B \ \ , \ \ a·b \ = \ C \ \ $ for factoring the quadratic polynomial $ \ x^2 + Bx + C \ = \ (x + a)·(x + b) \ $ can be done; it is an alternative way of obtaining the quadratic formula:

$$ a \ + \ b \ \ = \ \ B \ \ \Rightarrow \ \ (a \ + \ b)^2 \ \ = \ \ a^2 \ + \ 2·a·b \ + \ b^2 \ \ = \ \ B^2 \ \ ; $$ $$ a·b \ \ = \ \ C \ \ \Rightarrow \ \ 4·a·b \ \ = \ \ 4·C \ \ ; $$ $$ \Rightarrow \ \ a^2 \ + \ 2·a·b \ + \ b^2 \ - \ 4·a·b \ \ = \ \ B^2 \ - \ 4·C \ \ = \ \ a^2 \ - \ 2·a·b \ + \ b^2 \ \ = \ \ (a \ - \ b)^2 $$ $$ \Rightarrow \ \ a \ + \ b \ \ = \ \ B \ \ , \ \ a \ - \ b \ \ = \ \ \sqrt{B^2 \ - \ 4·C} \ \ \ \ [ \ a \ \ge \ b \ ]$$ (solving this system of equations) $$ \Rightarrow \ \ a \ \ = \ \ \frac{B \ + \ \sqrt{B^2 \ - \ 4·C}}{2} \ \ , \ \ b \ \ = \ \ \frac{B \ - \ \sqrt{B^2 \ - \ 4·C}}{2} \ \ . $$ Of course, in the time it takes to carry out the calculations this way, you could have applied the quadratic formula already; that is probably why you didn't find anyone working out the factors by this approach.