Cubic polynomial missing the linear term $$x^3+4x^2+1=0$$
This is the first homework assignment for my precalculus class. The first problem I could solve:
$$x^2+4x+1=0$$
$$(x+2)^2-3=0$$
$$x=-2\pm\sqrt3$$
I've been away from maths for a bit, so I purchased a few books of self-teaching from prealgebra to precalculus (The Complete Idiot's Guide to Precalculus, Pre-Calculus for Dummies, Precalculus Mathematics in a Nutshell, Practical Algebra). They do not seem to contain a problem like mine and if they do have something close, I was able to work that problem out cleanly. 
My friend who has gone through his maths track already attempted to solve this and could not. He suggested using Wolfram Alpha, which I did, but I got back a completely wonky pair of complex roots:
$$x=-\frac43+\frac83 (1\pm i\sqrt3)\sqrt[3]{\frac2{155-3\sqrt{849}}}
+\frac16(1\mp i\sqrt3)\sqrt[3]{\frac{155-3\sqrt{849}}2}$$

Given the ugly roots of Wolfram, here a way of approximating these roots. 
First at all applying the rule of signs (Descartes) one has 
$f (x) = x^3 + 4x^2 + 1$ and $f (-x) = - x^3 + 4x^2 + 1$ have a combined $ 0 + 1 = 1 $ sign change and then the degree being $3$ there are at least $3-1$, two non-real roots. This way we know there are two non-real roots and a real root.
Besides $f(-4)=1$ and $f(-5)=-24$ then the real root $x_1$ is between $-5$ and $-4$. You can approximate
$x_1\approx -4.06$. Hence the (approximate) quotient gives
$$f(x)\approx (x+4.06)(x^2-0.06x+0.2436)$$
Solving now the quadratic equation you have
$$x_{2,3}\approx 0.03\pm0.492646i$$
Check the degree of approximation, comparing with Wolfram, is your job.
