Answer:
4.264007e+14
Step-by-step explanation:
Hope this helps
We can one theorem to help us find rational roots of this polynomial.
2x^3 + x^2 - 4x - 2
We'll use Descartes' rule of signs.
Because there is 1 sign change, 1 of the 3 roots will be positive.
Now we can make the value of x -1 to see how many negative roots there will be.
-2 + 1 + 4 - 2
There are 2 sign changes, so we know there will be 2 negative roots.
Because of this, we should have 3 real, rational roots.
Ground: y = 0
- 16t² + 128t + 50 = 0
Apply quadratic equation.
t = 0, 8 (rounded to the nearest second)
8 seconds
![z=2x^4+8x \\ dz=(8x^3+8)\ dx=8(x^3+1)\ dx \\ \int\limits^0_{-1} {(2x^4+8x)^3*(4x^3+4)} \, dx =4* \int\limits^0_{-6} { \dfrac{z^3}{2}} \, dz \\ = 2* \left[\begin{array}{ccc} \dfrac{z^4}{4}\end{array}\right] ^0_{-6}= \left[\begin{array}{ccc} \dfrac{z^4}{2}\end{array}\right] ^0_{-6}=0-648=-648](https://tex.z-dn.net/?f=%20z%3D2x%5E4%2B8x%20%5C%5C%0Adz%3D%288x%5E3%2B8%29%5C%20dx%3D8%28x%5E3%2B1%29%5C%20dx%20%5C%5C%0A%0A%20%5Cint%5Climits%5E0_%7B-1%7D%20%7B%282x%5E4%2B8x%29%5E3%2A%284x%5E3%2B4%29%7D%20%5C%2C%20dx%20%3D4%2A%20%5Cint%5Climits%5E0_%7B-6%7D%20%7B%20%5Cdfrac%7Bz%5E3%7D%7B2%7D%7D%20%5C%2C%20dz%20%5C%5C%0A%3D%202%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cdfrac%7Bz%5E4%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%5E0_%7B-6%7D%3D0-648%3D-648)