Question 110316
For factoring quadratics with the leading coefficent greater than 1 is easy. I'll do a harder one that the one you listed (because leading coefficent prime it's actually the same as if it was 1, it will be in the for (3w+something)(w+something) if it's factorable)

Let
{{{f(x)=4x^2+8x+3}}}. Here your answer must in the form (2x+something)(2x+something) or (4x+something)(x+something) because the only way to multiple two integers to get 4 is 4 times 1 or 2 times 2. Then it's a guess and check time game. Clearly, the two contants in the factors have to be 3 and 1. So the only possibilities are (4x+1)(x+3) and (4x+3)(x+1) and (2x+1)(2x+3). Which one works?