Calcule $\frac{x^3}{3}+\frac{48}{x^2}=10\left(\frac{x}{3}-\frac{4}{x} \right)$

Question

$(\dfrac{x^5+144}{3x^2}=\dfrac{10x^3-120x}{3x^2})$

$(x^5+144=10x^3-120x)$

$x^5-10x^3+120x+144=0$

$x \approx -1.376$

Can you find the roots? No calculator please

Answer 1

You are looking for the zero's of function $$f(x)=x^5-10x^3+120x+144$$ Its first derivative $$f'(x)=5x^4-3x^2+120$$ is a quadratic in $x^2$ which does not show any real root. So, there is only one real solution. By inspection, there is one root between $-2$ since $f(-2)=-48$ and $-1$ since $f(-1)=+33$.

So, make one iteration of Newton method using $x_0=-\frac 32$ and you will get $x_1=-\frac{114}{83}\approx -1.37349$ which seems quite good.

Edit for your curiosity

One step of Newton iteration is the same as the Taylor series $$f(x)=f(a)+(x-a) f'(a)+O\left((x-a)^2\right)$$ Sooner or later, you will learn that for function approximation Padé approximants are quite better. Limiting to the simplest one, this would write $$f(x) \sim \frac { f(a)+ \left(f'(a)-\frac{f(a) f''(a)}{2 f'(a)}\right)(x-a)} {1-\frac{ f''(a)}{2 f'(a)} (x-a)}$$ making the zero as $$x\sim a+\frac{2 f(a) f'(a)}{f(a) f''(a)-2 f'(a)^2}$$ Applied to your case, still with $a=-\frac 32$, this would give $x=-\frac{9651}{7015}\approx -1.37577$ while the "exact" solution is $\approx -1.37600$.

Answer 2

First, this question can not be factorize by using Synthetic division. So I use the computer to get the answers. The answer is hare:near x=172/125, near x=-237510,999,999,999,985/10^17-15819/10^4*i*, near x=-237510,999,999,999,985/10^17+15819/10^4*i*, near x=36311/10^5-186299/10^5*i*, near x=36311/10^5+186299/10^5*i*

SO ... that's it.