0
$\begingroup$

What is the best method to find the solutions to this equation?

$x^3(x^2+2)=-297$

I have obtained the solution from the back of the book but have no idea as to the method to get the solution. Is it merely a matter of plugging in numbers?

$\endgroup$
4
  • 1
    $\begingroup$ If you were to assume that $x$ is an integer, then you can look at the factorization of $297=3^3\cdot 11$ and use this to reach your conclusion that $x=-3$ is one solution. $\endgroup$
    – JMoravitz
    Apr 20, 2019 at 21:09
  • $\begingroup$ If you do not make the assumption that $x$ must be an integer, then you'll find there are potentially five different complex number solutions to the equation. In some special cases you could find them explicitly, however there is no guarantee you can for fifth degree or higher polynomials. $\endgroup$
    – JMoravitz
    Apr 20, 2019 at 21:11
  • $\begingroup$ Note also that because $x^5$ and $2x^3$ are strictly increasing functions of $x$ (for $x\in \mathbb R$) so is their sum, and it is easy to deduce from this that there will be precisely one real solution. $-4^5=-1024$ and the left-hand side is zero for $x=0$ so the solution will be between $-4$ and $0$. Because the equation has integer coefficients, and the leading coefficient is $1$, any rational solution will be an integer. Any integer solution will be odd. So it is easy to constrain the search space. $\endgroup$ Apr 20, 2019 at 21:56
  • $\begingroup$ You might want to look at the rational root theorem if you don't know it already. $\endgroup$
    – NickD
    Apr 21, 2019 at 0:36

1 Answer 1

1
$\begingroup$

Hint: $$x^3(x^2+2)=-297$$ or $$x^3(x^2+2)=(-3)^3(11)$$ $$x^3(x^2+2)=(-3)^3((-3)^2+2)$$

$\endgroup$
3
  • $\begingroup$ you're right. That was the answer on the back of the book. But as for the method to find that, you just recognise that -297 can be factored by 11 and -27 or 11 and $-3^3$? Is this just a matter of experience with such problems that cradle to see the factors of large numbers and plug-in the solution accordingly? $\endgroup$
    – esc1234
    Apr 20, 2019 at 21:13
  • $\begingroup$ I think that the idea of ​​this question is to focus on factoring $\endgroup$
    – E.H.E
    Apr 20, 2019 at 21:21
  • $\begingroup$ yeah, i guess I'll get better at seeing such factoring patterns with time and experience. Thanks for the feedback! $\endgroup$
    – esc1234
    Apr 20, 2019 at 21:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .