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How to find the roots of this cubic equation ?


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Well, one of them is $\;7\;$ . How did I find it?! – DonAntonio Dec 12 '13 at 21:10
@DonAntonio First WA and then rational root theorem? – Git Gud Dec 12 '13 at 21:11
You knew $7^3+16(7)=455$ ? – K. Rmth Dec 12 '13 at 21:12
@GitGud What is WA? – Harald Hanche-Olsen Dec 12 '13 at 21:12
Nop, @GitGud : only the rational root theorem. Pretty easy to apply here, as $\;455=5\cdot7\cdot 13\;$ and the only one that seems to fit is $\;7\;$ . All the rest too small or too big – DonAntonio Dec 12 '13 at 21:12
up vote 12 down vote accepted

Another solution, using algebraic manipulation :


$x^4+16x^2=65\cdot 7x$, add $49x^2$ both sides, then we get

$x^4+65x^2+(\frac{65}{2})^2=49x^2+65\cdot 7x+(\frac{65}{2})^2$



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Amazing. Thank you guy ! – Joel Dec 12 '13 at 21:19
@Joel You are very welcome ! – Ewin Dec 12 '13 at 21:22

Use rational root theorem to find if the cubic equation has a rational root. If it has you can easily bring it to a quadratic equation, by dividing the polynomial with $(x-x_0)$, where $x_0$ is its root.

Anyway if the cubic equation doesn't have a rational root, you can use Cardano's method to find the roots or you can use resolvents.

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By the Rational root theorem we find that $7$ is a root of this cubic equation. Can you take it from here?

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Alternately: ...the Factor Theorem states that if $f(a) = 0$, then $(x-a)$ must be a factor of the polynomial. Plugging in values into $f(x)$, you'll discover

$f(7) = (7)^3 +16(7) -455 = 0 \implies (x-7)$ is a factor of $f(x)$.

A little time consuming with all the calculations, but a pretty simple method nonetheless.

Dividing $f(x)$ by $(x-7)$ will get you the other root, $x^2+7x+65$.

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Equation $$ax^3 + bx^2 + cx +d =0$$ with $b = 0$ can be in general quite easily solved as described here (Look at section Cardano's method).

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One of them is 7. I found it from this calculator.

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