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I have been doing some maths and to continue on with the question I need to somehow solve:

$$ x^3-25x^2+200x-340 = 0 $$

How do I solve this, is there a formula that I can use or a method for solving it besides grouping?

(William Jagy) The three by three matrix is $$ \left( \begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array} \right) $$

Will Jagy
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Jack Lane
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  • If this was during a homework question, it's unlikely your professor expects you to solve a cubic equation. Most likely you've made a mistake, or you're taking the wrong approach. You can certainly try Cardano's method, but depending on what you need to do with the solutions this may not be very helpful. – Jack M May 14 '18 at 05:47
  • It is a part of the process to find an Eigenspace and I have gone through the process to get to this equation 3 times and I couldn't find anything wrong with it. – Jack Lane May 14 '18 at 06:02
  • Should "besides grouping" be "besides graphing"? – hardmath May 14 '18 at 11:41
  • Jack, please edit in the entries of the three by three matrix for which you are finding eigenvalues. I put in a matrix with the entries as letters, you just need to correct each of the nine letters to the number that belongs there, finally save your edit – Will Jagy May 14 '18 at 17:32
  • Is $x$ real or complex? – Allawonder May 14 '18 at 17:40

1 Answers1

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Hint:

Set $z:=x+\dfrac{25}3$ to "depress" the polynomial to the form

$$z^3+pz+q=0.$$

Then set $z:=u-\dfrac p{3u}$ and plug in the equation:

$$0=u^3-pu+\frac{p^2}{3u}-\frac{p^3}{27u^3}+pu-\frac{p^2}{3u}+q=u^3-\frac{p^3}{27u^3}+q.$$

This is a quadratic equation in $u^3$,

$$(u^3)^2+q(u^3)-\frac{p^3}{27u^3}=0$$ which you solve in the usual way. In your case, the roots are real. Pick any of them and draw the value of $z$, then $x$.

After you have found this real root, divide the initial polynomial by $x-r$. This yields a quadratic polynomial, which will give you two conjugate complex roots, and you are done.

Note that the method is quite general, but it breaks when the equation in $u^3$ has complex roots. In this case, you need to resort to trigonometry as there is (provably) no algebraic formula.