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Factorisation of cubic equation

can we factories \(x^3+x+1=0\)

Note by Chinmay Sangawadekar
2 years, 10 months ago

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Compare it with the identity \((u+v)^3-3uv(u+v)-(u^3+v^3)=0\).We get the following system of equations: \[u+v=x\\-3uv=1\rightarrow uv=\frac{-1}{3}\rightarrow u^3v^3=\frac{-1}{27}\\u^3+v^3=-1\] Construct an equation in variable \(z\) having roots \(u^3\) and \(v^3\). \[(z-u^3)(z-v^3)=z^2-(u^3+v^3)z+u^3v^3=z^2+1z-\frac{1}{27}=0\rightarrow 27z^2+27z-1=0\] Plugging the values in the quadratic formula,we get: \[\frac{-27\pm\sqrt{27^2-4(27)(-1)}}{2(27)}\\=\frac{-27\pm\sqrt{729+108}}{54}\\=\frac{-27\pm\sqrt{837}}{54}\\=\frac{-27\pm3\sqrt{93}}{54}=\frac{-9\pm\sqrt{93}}{18}\] So \(u^3=\frac{-9+\sqrt{93}}{18}\rightarrow u=\sqrt[3]{\frac{-9+\sqrt{93}}{18}}\approx 0.329452338\)and \(v^3=\frac{-9-\sqrt{93}}{18}\rightarrow v=\sqrt[3]{\frac{-9-\sqrt{93}}{18}}\approx -1.011780141\)and \(x=u+v=0.329452338-1.011780141=-0.682327803\).Dividing by \((x+0.682327803)\) and discarding the remainder (since we are working with approximations there will be some very small remainders which would be 0 had we divided by the exact value) we get \(x^2-0.682328x+1.46557\) which has roots \(0.341164\pm1.16154i\).So the cubic can be factored as \((x+0.682327803)(x-(0.341164+1.16154i))(x-(0.341164-1.16451i))\). Abdur Rehman Zahid · 2 years, 4 months ago

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niceone Rowegie Lambojon · 2 years, 10 months ago

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