Show that the funtion (x-3)(x^2+3x+1) has two stationary points and give the co-ordinates of these points

Stationary points are points where the gradient of a function is equal to 0. In this question the product rule can be used to find the gradient of the given function. The product rule is given by u'v+uv'= (uv)' and so the differential in this question is (1)(x^2+3x+1)+(x-3)(2x+3). Setting this equal to 0 we can rearrange to get (x^2+3x+1)=-(x-3)(2x+3). We then expand the brackets on the right hand side of the equation. (x^2+3x+1)=-(2x^2-6x+3x-9) = (-2x^2+3x+9). Cancelling terms and reaaranging for x we can find the number of stationary points. 3x^2 = 8 -> x = (8/3)^1/2, as we know that a positive number always has two roots (one positive and one negative) we can see that there are two stationary points at +/- (8/3)^1/2. These x coordinates are then substituted into the original equation to give the y coordinate of the stationary point.