Factorising and two theorems

We are given that a stationary point occurs at x=-1. At a stationary point, the gradient is equal to zero. Hence, the derivative of f(x) is also equal to zero at this point, as the derivative works out the gradient of the function at a specific point. So, f'(x) = 0 when x = -1 =) 3x^2 + 2ax -3 = 0 when x = -1 =) 3(-1)^2 +2a(-1) -3 = 0 =) 3 -2a -3 = 0 =) -2a = 0 =) a = 0 We are also given that x-2 is a factor of f(x). Therefore, a root occurs where x = 2. We can now use polynomial division determine the value of b. I'll do my best to illustrate this process. 2 | 1 a -3 b | 2 4+2a 2+4a --------------------------- 1 2+a 1+2a b+4a+2 I hope this is clear enough. If you are not familiar with how to do polynomial division then I suggest you read up on it. It is a relatively simple process, once you learn how it works. We can see from the polynomial division that b + 4a + 2 = 0, because if x-2 is a factor (i.e. x=2 is a root) then the remainder must be zero when f(x) is divided by x-2. We already know that a = 0. So b + 2 = 0 =) b = -2 We can now factorise f(x) by plugging the values of a and b into the results we attained from the polynomial division, to get: f(x) = (x-2)[x^2 + (2+a)x + (1+2a)] = (x-2)(x^2 + 2x + 1) = (x-2)(x+1)(x+1) = (x-2)(x+1)^2
By: Wyess.

Date of comment: Fri, Dec 24th 2010