1. 5th-degree polynomial equation

3x^5 + 4x^4 + 6x^3 - 20x^2 - 25x = 0

by the way, the answers are the following:

0, -1, 5/3 & -1±2i

*** i need a complete solution on how to solve this eqn plzzzz

2. Originally Posted by fishlord40
3x^5 + 4x^4 + 6x^3 - 20x^2 - 25x = 0

by the way, the answers are the following:

0, -1, 5/3 & -1±2i

*** i need a complete solution on how to solve this eqn plzzzz
First $x=0$ is an obvious root so that leaves us with:

$3x^4+4x^3+6x^2-20x-25=0$

The rational roots theorem tells us that if this has any rational roots they are the ratio of a factor of the constant term to a factor of the coefficient of the higest order term, so try: $x= \pm 1, \pm 5, \pm 5/3, \pm 25, \pm 25/3$. Having found all the rational roots remove the corresponding factors, which will leave you with a quadratic, and on that you use the quadratic formula.

CB

3. Originally Posted by fishlord40
3x^5 + 4x^4 + 6x^3 - 20x^2 - 25x = 0

by the way, the answers are the following:

0, -1, 5/3 & -1±2i

*** i need a complete solution on how to solve this eqn plzzzz
HI

$3x^5 + 4x^4 + 6x^3 - 20x^2 - 25x = 0$

$x(3x^4+4x^3+6x^2-20x-25)=0$

It's obvious here that x=0

Now you can use the rational root test to find the possible zeros for the polynomial inside the bracket

THe possible ones are $\pm \frac{1,5,25}{1,3}$

$\pm 1$, $\pm \frac{1}{3}$ ,

$\pm 5$ , $\pm \frac{5}{3}$ , $\pm 25$ ,

$\pm \frac{25}{3}$

whichever roots which give you 0 when you substitute them into the equation would be the zeros . In this case , they would be -1 and 5/3 .

$3x^4+4x^3+6x^2-20x-25=(x+1)(3x-5)(ax^2+bx+c)$

Use the long division to find the quadratic , and solve it where it would give you complex roots .

4. Originally Posted by fishlord40
i need a complete solution on how to solve this eqn
To learn the process (so you can do these types of exercises on your own on the test), try here.