An algebra problem by Akshat Sharda

I'm going to be a bugger and use calculus! Taking $P'(x) = 0$ , or:

$P'(x) = \frac{(2x+45)(x+3) - (1)(x^2+45x+450)}{(x+3)^2} = 0 \Rightarrow \frac{(2x^2+51x+135) - (x^2+45x+450)}{(x+3)^2} = 0 \Rightarrow \frac{x^2+6x-315}{(x+3)^2} = 0$ .

which the numerator factors into $(x+21)(x-15) = 0 \Rightarrow x = -21, 15.$

Now the second derivative of $P(x)$ computes to:

$P''(x) = \frac{(2x+6)(x+3)^2 - 2(x+3)(x^2+6x-315)}{(x+3)^4}$ ;

and taking our two earlier critical points gives $P''(15) > 0$ and $P''(-21) < 0$ , which are respectively the local minimum and maximum of $P(x)$ . Hence $B = 15$ and $A = P(15) = 75$ , and $A + B = 75 + 15 = \boxed{90}$ .