What are the solutions of (x^3)+6 = 2(x^2)+5x given x = 3 is a solution?
Firstly we will put this into a form equal to zero, by rearranging to get x3-2x2-5x+6 = 0. This is because in order to solve a polynomial we first need to set it equal to zero. We now know that the values of x such that x3-2x2-5x+6 = 0 are also the values of x that solve the equation in its original form. We are given that x=3 is a solution, so by putting 3 into the equation, we see that 33-2(32)-5(3)+6 = 27-2(9)-5(3)+6 = 27-18-15+6 = 0. Since x = 3 is a solution, we know that (x - 3) is a factor of the equation. So by factorising, we get (x-3)(x2+x-2) = 0 . Then by factorising the quadratic further, we see that (x-3)(x-1)(x+2) = 0 . This occurs in three ways, when x-3 = 0, which is the solution we were given, or when x-1 = 0, giving x = 1, and when x+2 = 0, giving x = -2. So the equation is solved by x = 3 , x = 1 , x = -2 .
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