finding zeros of p(x)=x^3-x^2-11x+15 using factor theorem

Question

Can you please tell me how to find the zeros of the function above using the factor theorem?

Answer 1

Since p(x) is a cubic polynomial, we start by finding a zero of p(x).  This is done by checking each potential zero of p(x) to see if it makes the equation p(x) = 0 true.  The potential zeros are found by taking the positive and negative factors of the constant term of p(x) and dividing by the positive and negative factors of the leading coefficient of p(x).  In this case, the positive and negative factors of the constant term are 1,-1,3,-3,5,-5,15,-15 and the positive and negative factors of the leading coefficient are 1,-1.  Therefore, the potential zeros of p(x) are 1,-1,3,-3,5,-5,15,-15.  After checking, you come to find that 3 is a zero of p(x) since p(3) = 0.  Therefore, the factor theorem implies that x - 3 is a factor of p(x).  Since x - 3 is a factor, we know that p(x) = (x - 3)(ax^2 + bx + c) for some a,b and c.  From this equation, ax^2 + bx + c = p(x)/(x - 3), and p(x)/(x - 3) can be evaluated by using long or synthetic division of polynomials.  Using synthetic division we get:

 

3 | 1  -1  -11  15     So, this tells us that a = 1, b = 2 and c = -5.  From here, we can find the remaining

          3    6  -15     zeros of p(x) if any others exist by applying the quadratic

     1   2   -5    0      formula to the equation x^2 + 2x - 5 = 0.