# 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: $$x=-\frac43+\frac83 (1\pm i\sqrt3)\sqrt[3]{\frac2{155-3\sqrt{849}}} +\frac16(1\mp i\sqrt3)\sqrt[3]{\frac{155-3\sqrt{849}}2}$$

• Which tools do you already have for solving cubics? – Henning Makholm Sep 8 '16 at 6:40
• MathJax in the question is completely free. And I'll give you enough reputation to post more links. – Parcly Taxel Sep 8 '16 at 6:43
• Wolfram Alpha (which is free on the web) does indeed show that the roots are pretty horrible. Unless you're supposed to have learned about Cardano's formula (which doesn't seem likely for a first exercise), are you sure you're supposed to find an exact solution rather than a numeric approximation? – Henning Makholm Sep 8 '16 at 6:49
• To be honest, I could not answer in terms of tools to solve cubics. I just don't remember and these books did not cover any complex nor edge cases that don't work out clean with sum of perfect cubes, difference of perfect cubes, or reducing the max n degree to a quadratic. Also, the method of providing the answer in the form with the arrows to and from the x variable, I do not know what that is called; surely there is a name for that method, but I can't seem to google it properly. – Collin Stevens Sep 8 '16 at 6:49
• @HenningMakholm The answer is supposed to be in a form such as the answer to the first question, which ends up with the exact solution, not an approximation. I could not get wolfram alpha to give me the exact answers and their step by step solution was quite literally pages long. We have also not learned Cardano's formula. I actually read about that before posting and it seems that's actually what wolfram is using(possibly). – Collin Stevens Sep 8 '16 at 6:50

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.