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 $\displaystyle x=0$ 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
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 .