Factorise the quadratic into two brackets to make (x-3)(x-5)=0 Because the product of these brackets is zero then the value of one of the brackets must equal zero. In order for one of the brackets to have a value of 0, we need to substitute positive 3 or positive 5 into the brackets so that the equation is correct. This means that the graph x2 - 8x + 15 crosses the x axis at the points 3 and 5.