## anonymous 5 years ago how do you find the roots of c^3-7C^2+16C-12=0

1. anonymous

Let a=12-c^3 Solve a - 16 C + 7 C^2 = 0 for C. Replace a with 12-c^3 and simplify.$\left\{C=\frac{1}{7} \left(8-\sqrt{-20+7 c^3}\right),C=\frac{1}{7} \left(8+\sqrt{-20+7 c^3}\right)\right\}$

2. anonymous

how did you know what to set a as?

3. anonymous

or actually I see how you knew that...but your answers still have cs in them. How do you get rid of those?

4. anonymous

I have to leave in a minute. Pick a c and the corresponding values for C can be computed. There are restrictions on the values for c. Refer to the attachment from Mathematica.