We need to be careful with this function, because it's quadratic and therefore it could have two roots, two solutions for x when the function=0. These solutions determine the factors. If we plot the graph of the function roughly we can see that when x is large and positive or negative, the function is positive. If we put x=0 we can see where the graph crosses the vertical axis, -3 in this case. To get to -3 the graph must intersect the x-axis at two points. The graph has a U shape. Let's put a couple of values into the function. At x=1 the function is 4. At x=-1 it's 14. So we know it cuts the x-axis twice between x=1 and -1. That gives us a start. Using a calculator to speed things up, we can try values in these limits. Let's go for x=0.5 and -0.5. The two values are respectively are -2.5 and 2.5. What does that tell us? Between x=0 and 0.5 the function goes from -3 to -2.5, and between -0.5 and 0 it goes from -3 to 2.5. So in one case the graph doesn't cross the x-axis but in the other case it does because we've gone from negative to positive between x=0 and -0.5.
We need to look first for the solution between -0.5 and 0, so we can try values between -0.4 and -0.1. When we do this we find we go from positive to negative between -0.4 and -0.3. So using trial and error we continue to the next decimal place and try values between -0.39 and -0.31. This time the sign change takes place between -0.34 and -0.33. This is beginning to look like -1/3. Let's put this value into the function. Yes! x=-1/3 is a value that makes the function zero. Therefore one factor is (3x+1). The other factor can be found by inspection. It must start with 4x to make 12x^2 in the function, and the numerical part must be -3 because we have -3 in the original function as the last term and -3*1=-3. So factorisation gives us (3x+1)(4x-3).
