please help me to find the value of x?

Question

where the equation is
(x^4+3*x^3+2*x^2-3*x=0)
please show how to solve it
thank you

Answer 1

the equation can be simplified as
x * (x^3+3*x^2+2*x-3)=0
so one root of the equation is 0

taking the other part x^3+3*x^2+2*x-3 = 0, ------------ eq 1
since it won't give exact solutions follow some numerical method like newton raphson's

take intial value of x as 0,
f(x) = x^3+3*x^2+2*x-3
f dash(x) = derivative of above = 3*x^2 + 6x + 2

so x1 = x - f(x) / f dash(x)
taking x as zero first,

x1 = 1.192
x2 = 0.794.. and so on where x5 = x6 = 0.6717
hence one of the three root is 0.6717.

Now remove that root for the equation,
dividing the equation 1 by (x - 0.6717) we are left with,
x^2 + 3.6717x + 4.4463 = 0
this is a quadratic equation,
delta = b^2 - 4*a*c = -ve and hence the roots are imaginary

Answer 2

Solve,

x^4 + 3x^2 + 2x^2 - 3x = 0

Factorize,

x (x^3 + 3x + 2x - 3) = 0

One simple solution is x = 0

Note: There is one other real solution between 0 to 1 and a pair of complex solution.

Answer 3

x = 0
cos if you factorise it will be:

x(x^3 + 3x^2) = 0
anything that is multiplied to be 0 can only be 0*(something) = 0
since in this case 0 = x/3
so x * 0 = 0
3 * 0 = 0
therefore the answer is 0

Answer 4

x=0
there are more solutions but this one is the easiest to see.