Figure A-10.2(a), a graph of the cubic polynomial $p\left(x\right)$, suggests that the equation $p\left(x\right)\=0$ has three real roots. The graph is drawn with code hidden in the cell in which the figure appears. It could also be drawn with the Plot Builder, launched from the Context Panel for the polynomial.

The Context Panel for the polynomial will contain the Solve option, and Maple will assume that the polynomial is set equal to zero.

Choosing the Solve≻Solve option results in the three exact solutions shown (compressed) in Table A-10.2(a).

$3\+A\/6\+58\/A$, where $A\={\left(4860\+12 i \sqrt{128643}\right)}^{1\/3}$

$3-29\left(1 \pm i \sqrt{3}\right)\/A-A \left(1\mp i \sqrt{3}\right)\/12$

Table A-10.2(a)   Compressed form of the exact roots of $x^3-9x^2-2 x\+15\=0$  

$3\+A\/6\+58\/A$

$3\+\left(u\+ v\right) \cos \left(B\right)\+i \left(u-v\right) \sin \left(B\right)$, where

$B\=\arctan \left(\sqrt{128643}\/405\right)\/3$
$u\={696}^{1\/3}{87}^{1\/6}\/6$ and $v\={696}^{2\/3}{87}^{5\/6}\/1044$

$3\+2\sqrt{29\/3} \cos \left(\arctan \left(\sqrt{385929}\/405\right)\/3\right)$

$9.037651011$

Table A-10.2(b)   Schematic for transforming the first root to the form $a\+b i$, where b = 0.

$x^3-9x^2-2 x\+15$

$x^3-9x^2-2x\+15$

$\stackrel{\text{solve}}{\to }$

$-1.307265533\,1.269614522\,9.037651011$

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$\left[9.037651012\,-1.307265533\,1.269614521\right]$

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