 Quandaries and Queries Question: Could you please solve the following for x x2+6=5x x2+16=8x 2x+21=3x2 Hi, I will do a similar problem, x2 + 2x = 8 First move all the terms to one side of the equation so that the x2 term is positive. x2 + 2x - 8 = 0 Now I am going to factor the left side. Since the constant term (-8) is negative I know that, if I can factor the expression the factorization will have the form (x + a)(x - b). Expanding this I get (x + a)(x - b) = x2 + (a - b)x - ab Comparing this to x2 + 2x - 8 I can see that ab = 8 and a - b = 2. The problem thus is reduced from factoring an algebraic expression to factoring a number, 8. I want two factors of 8 (a and b) which have a difference of 2. 8 can be factored as 8 = 8 1 but 8 - 1 = 7. 8 can also be factored as 8 = 4 2. In this case 4 - 2 = 2 so these factors will work, that is a = 4 and b = 2. Thus my factorization is x2 + 2x - 8 = (x + 4)(x - 2) At this point you should check by expanding the right side of the equation above to verify that it is equal to the right side. Thus the equation x2 + 2x - 8 = 0 can be rewritten (x + 4)(x - 2) = 0 This implies that either x + 4 = 0 or x - 2 = 0. The first gives x = -4 and the second, x = 2. Penny Go to Math Central