I Solved A Nice Cubic Equation | Algebra

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[@TheOldeCrowe]

(1) Factored 308 to look for rational roots
(2) Found rational root -4
(3) Used long division to get
(x + 4)(x² - 4 - 77) = 0
(4) Factored quadratic to get
(x + 4)(x - 11)(x + 7) = 0
So the roots are -4, 11, and -7

[@knutthompson7879]

Super easy using RRT. 11,-7,-4. Falls right out

[@StaR-uw3dc]

Nice solutions.
Minor remark to the second method: ab=31 not 93

[@lawrencejelsma8118]

Good error find. (31) = (93/3) So his missed (31)^3 step gives his correct 29791 and a quadratic equation c of solutions he doesn't evaluate. He worked it out correctly and recorrects in where did 93^3 become 29791 actual recorrection step.

[@lhdill2009]

After taking care of the missing factor of 3, I find a = (11 + sqrt(3) i ) /2 and b = (11 - sqrt(3) i ) /2 (or vv). That a and b are complex makes this solution method brutal compared to the rational root theorem or the approach used by Syber!

[@phill3986]

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[@nasrullahhusnan2289]

x³-98x-308=0
x³-98x=308
x(x²-93)=4×7×11
=11×28
=11(121-98)
=11(11²-98) --> x=11

[@user-uz3hv2fx7f]

the solutions are:-7,-4,11

[@DonRedmond-jk6hj]

Technically its Fior's method. Also Cardano was able to extend Tartaglia's method somewhat.