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

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- Oct 18th 2009, 11:53 PMfishlord405th-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 - Oct 19th 2009, 12:08 AMCaptainBlack
First $\displaystyle x=0$ is an obvious root so that leaves us with:

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

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: $\displaystyle x= \pm 1, \pm 5, \pm 5/3, \pm 25, \pm 25/3$. 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 - Oct 19th 2009, 12:12 AMmathaddict
HI

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

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

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 $\displaystyle \pm \frac{1,5,25}{1,3}$

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

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

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

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 .

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

Use the long division to find the quadratic , and solve it where it would give you complex roots . - Oct 19th 2009, 04:49 AMstapel
To learn the process (so you can do these types of exercises on your own on the test), try

**here**. (Wink)