How do I factorise difficult quadratic equations?
A quadratic equation such as 6x2 + x - 35 = 0 may appear much harder to solve than a simpler equation which students might be more used to, such as x2 + 5x + 6 = 0. In reality, there are a couple of methods which make tackling these kinds of equations much easier.Let's look at the example I've given: 6x2 + x - 35 = 0. Here, our 'a' component is 6, our 'b' component is 1 and our 'c' component is -35. We need to start by finding factors which add to make 'b' (1) and multiply to make 'ac' (6 * -35 = -210). Since they multiply to make a negative number, one of these factors must be negative and one must be positive. Since they add to make 1, they must be quite similar in size! At this point, we can use some guesswork. We know that -20 x 10 = -200, which is not far off. We are probably looking for two numbers between 10 and 20. If I write out the factors of 210, I get 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210. An intimidating amount of options, but only two fit our description of close in size and between 10 and 20! We must use 14 and 15, and one will have to be negative and the other will have to be positive.At this point we must think about the factors of 'a', 6. These are 1, 2, 3 and 6. We may notice that 2 goes into 14 and 3 goes into 15. From this we can see that the x components in our two sets of brackets must be (2x ± ?) and (3x ± ?). Since 2 x 7 = 15 and 3 x 5 = 15, we can work out that the solution must be (2x + 5)(3x - 7) = 0.
