AS Maths ->Expresss x^2 + 3x + 2 in the form (x+p)^2 + q... where p and q are rational number

So we are meant to rearrange the given equation in a way that it looks like: (x+p)^2 + q This should look familiar. It is obtained by completing the square. Which will be the method I will describe below: First group the X terms with brackets. (x^2 + 3x) + 2 Then we can write the expression inside the brackets in the following way: (x +(3/2))^2 and take away (3/2)^2 Resulting on the following: (x+ (3/2))^2 - (3/2)^2 + 2 (Dont forget this number - we had it in the first expression) SO WHAT HAPPENED? Essentially the coefficient in front of the X term must be always halved and an X term is taken away from both (x^2) and (x) because we will square the expression inside. Then we will need to TAKE AWAY the SQUARE OF the value that we halved -> (3/2)^2 You can manually check why. If you expand (x + (3/2))^2 you will get: (x^2) +(3x) +(3/2)^2 but we dont want that value at the end because it does not appear in the initial expression...so take it away. We ended up having: (x+ (3/2))^2 - (3/2)^2 + 2 SIMPLIFY THE TWO VALUES AT THE END TO OBTAIN THE ANSWER: (x + (3/2))^2 - (1/4) p = 3/2 q = -1/4