How would you find zeros for the polynomial $f(x)=-0.006x^4+0.140x^3-0.053x^2+1.79x$ with a degree of $4$?
I know a zero is $x=0$, but how do I find the other ones, and are there any imaginary zeros? I am able to see an actual $x$-int on the graph, but how would I find that with this function? Thank you.
As MathYobani commented, you are left with $$6 x^3-140 x^2+53 x-1790=0$$ Use the method for cubic equations. Not very pleasant because working with huge numbers but the key result is $$\Delta=-21275507448$$ which means that there is one real root and two complex roots.
For the real root, you can use the hyperbolic solution and get the nasty $$x=\frac{70}{9}+\frac{\sqrt{18646}}{9} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{853400 \sqrt2}{(9323)^{\frac 32}}\right)\right)\approx 23.4977$$