1. Real Zeros of Polynomial Functions

2. Solve x 3 – 2x + 1 = 0. How? Can you factor this? Can you use the quadratic formula? Now what if I tell you that one root is x =1? Then x 3 – 2x + 1 = (x – 1)( ). Use synthetic division to find the missing factor. The missing factor is x 2 + x - 1, which cannot be factored. So use the quadratic formula to get (-1 ±√5)/2.

3. and this one… Find all of the roots of f(x) = x 4 + 2x 3 – 7x 2 – 20x – 12 given that x = 3 and x = -2 are roots. Answer: x = 3, -2 (double root) and -1. Verify this using your graphing calculator.

4. But what if you don’t know any zeros and cannot factor a polynomial? Rational Zero Theorem If f(x) = a n x n + … + a 0 and has integer coefficients, then every rational zero of f(x) has the form p/q where p is a factor of a 0 and q is a factor of a n. List the possible rational zeros for f(x) = 2x 3 + 3x 2 – 4x + 6. Answer: ±1,±2,±3,±6,±1/2,±3/2

5. Now that you know the possible zeros, you can use trial and error until you get down to a quadratic. Then you can factor or use the quadratic formula. Find the x-intercepts of f(x) = x 3 + 4x 2 + x – 6 Answer: 1, -2, -3 Find the x-intercepts of f(x) = x 4 - 6x 2 + 3x + 2 Answer: 1, 2, (-3 ±√5)/2

6. Upper and Lower Bounds Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c using synthetic division. 1.If c > 0 and every number in the last row is non- negative, then c is an upper bound for the real zeros of f. 2.If c < 0 and every number in the last row alternates sign, then c is a lower bound. (Zero can be either positive or negative.)