Solving Special Cases

# of solutions A polynomial equation has the same number of solutions as its degree! How many solutions does each poly have? 1) 4x3 - 4x + 4 = 0 2) 3x2 + x + 3 = 0 3) 5x4 + 7x - 1 = 0

Let’s list all of the ways we have found to solve:

Solve: x3 – x2 – 2x – 12 = 0 Can we factor? Quadratic Formula? Complete the square? So…. Only way to solve is by graphing.

Graph x3 – x2 – 2x – 12 = 0 How many x-intercepts do you see Graph x3 – x2 – 2x – 12 = 0 How many x-intercepts do you see? How many should there be? What does that mean?

How do we find the other solutions? x3 – x2 – 2x – 12 = 0 1) Use the one real solution and turn it into a factor. 2) Use long division to divide. 3) Solve the remaining quadratic by one of our methods.

Try one! 2x3 + 9x2 + 25 = 0

Solve 4x3 – 8x2 + 4x = 0

Solve x3 – 8 = 0

Solve x3 – 64 = 0

Solve x3 + 27 = 0

Solve x4 – 2x2 – 8 = 0

Solve x4 + 7x2 + 6 = 0

Solve x4 – 3x2 – 10 = 0