The middle term method is a very good way to factorise quadratic equations. For example lets say you have the equation: ax^2 + bx + c = 0 (where a , b and c are the coefficients of x^2, x and x^0 repectively). Using the middle term method the first task would be to multiply a and c. Then find the LCM of the result which will give you possible combinations to (a*c). You must then determine which combination to use which will be helpful in replacing b. After choosing your combination you replace b with your values by either adding or subtracting. Seems a bit complicated but its not. Determining if you should add or subtract the 2 LCM terms of a*c is determined by the sign of c. If c is "-" then subtract if c is "+" then add. For example lets say we are faced with this equation: 4x^2 - 8x + 3 here a = 4, b = 8 and c = 3 First we multiply a and c so 4 * 3 = 12 Now we find the LCM of 12 = 2,2,3 the possbile combinations we get are: 12*1, 2*6, 3*4 Now we can see b = 8 and the sign on c is positive. So we must add either of these to get 8. We can see that adding 2 and 6 = 8. Therefore we use that combination and put it in the equation to equal b so it becomes (The 2 signs are both "-" because the sign of b is "-") 4x^2 - 2x - 6x + 3 Taking common 2x(x-1) - 3( x-1) Basically here you need to look out for the signs. Ive put both -2x and -6x as negative because the sing of 8x is negative. And i have added them becasue the signs are the same and because 3 is positive. So the result is: (x-1)(2x-3) Another example: 2x^2 + 5x + 2 a * c = 2 * 2 = 4 LCM of ac which is 4 = 1*4 , 2*2 Sign on c = postive so we must add the combination to get 5 we use 1 and 4 because if we add them we get 5. Equation becomes 2x^2 + x + 4x + 2 Taking common x(2x+1)+2(2x+1) (2x+1)(x+2) Final example: -8x^2 + 20x + 12 Ive chosen this because a lot of people get stuck thinkin what to do if the coefficient of x^2 is negative. What you simply do is take the negative out with any other common value. So it becomes -4(2x^2 - 5x - 3) Now just look at 2x^2 - 5x - 3 a = 2, b = -5 , c = -3 Multiplying a and c = 2 * - 3 = -6 Now the sign on c (-3) is negative so you must subtract the possible combinations of ac to get b. Possible combinations of ac = 1*6, 2*3 1-6 = -5 which is b so we use that 2x^2 + x - 6x - 3 x(2x + 1) - 3 (2x + 1) (2x+1)(x-3) But dont forget the -4 we took common at the begining so the final answer is -4(2x+1)(x-3) NOTE: if this is still very difficult i suggest you choose the alternate way of doing this using the formula: x = (-b +- root(b^2 - 4ac) / 2a) This should also give you the same answer. Hope this helps lol. I tried as hard as I can. Sorry if it got too complicated.