So, since we have a cubic equation with 4 terms, the first thing we should try is factoring by grouping, so:
![12x^3+2x^2-30x+5 =0 \implies \\ 2x^2(6x+1)-5(6x+1)=0 \implies \\ (2x^2-5)(6x+1) = 0](https://tex.z-dn.net/?f=%2012x%5E3%2B2x%5E2-30x%2B5%20%3D0%20%5Cimplies%20%5C%5C%202x%5E2%286x%2B1%29-5%286x%2B1%29%3D0%20%5Cimplies%20%5C%5C%20%282x%5E2-5%29%286x%2B1%29%20%3D%200%20)
Now that we've factored our equation, we can use ZPP and break it up:
![2x^2-5 = 0 \implies \\ 2x^2=5 \implies\\ x^2=\frac{5}{2} \implies\\ x=\pm\frac{\sqrt{5}}{\sqrt{2}}=\pm\frac{\sqrt{10}}{2} \\ \\ 6x+1 =0 \implies \\ 6x=-1 \implies\\ x=\frac{-1}{6}](https://tex.z-dn.net/?f=%202x%5E2-5%20%3D%200%20%5Cimplies%20%5C%5C%202x%5E2%3D5%20%5Cimplies%5C%5C%20x%5E2%3D%5Cfrac%7B5%7D%7B2%7D%20%5Cimplies%5C%5C%20x%3D%5Cpm%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B%5Csqrt%7B2%7D%7D%3D%5Cpm%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B2%7D%20%5C%5C%20%5C%5C%206x%2B1%20%3D0%20%5Cimplies%20%5C%5C%206x%3D-1%20%5Cimplies%5C%5C%20x%3D%5Cfrac%7B-1%7D%7B6%7D%20%20)
So, our solutions are:
![x\in \{\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2},-\frac{1}{6}\}](https://tex.z-dn.net/?f=%20x%5Cin%20%5C%7B%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B2%7D%2C-%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B2%7D%2C-%5Cfrac%7B1%7D%7B6%7D%5C%7D%20)
I don't know i just want some points
Step-by-step explanation:
well thank you anyways hope someone else can help you i am sorry.
Yvette's hourly rate of pay is $12.
3+3+3+3=12
You can see the answer from the picture
Look kid, this question has been up for 16 hours think its time to let it go......... hoped i helped. ✔verified✔