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Find the exact solutions of $x^3 + 5x^2 -2x -15 =0$.

While making notes for my students (in high school), I came across this problem. Using the Rational Root Theorem there don't seem to be any rational roots that work. Is it possible to have 3 real solutions when not even one is rational?

user128534
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  • here an example of calculationg the root with Cardanos formula. But this is rather labourious. Maybe there is a typo in the equation? You can find the exact solution using a CAS like Wolfram Alpha. – miracle173 Feb 13 '14 at 15:43

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Just like quadratic polynomials, cubic polynomials have a discriminant: $$ D=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2 $$ where $$ p(x)=ax^3+bx^2+cx+d. $$ If $D>0$, then $p$ has three real roots. If $D<0$ then $p$ has one real root and two complex roots that are conjugate. If $D=0$ then $p$ has three real roots and one of them is a multiple root. In your case $D=4257$. So, the roots are all real. You eliminated the possibility that they are rational.

J126
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Well, you just found one, so it would seem possible indeed. (You can graph it and see where the roots are, roughly.)

If we're talking about a textbook-for-HS problem, I would bet that the authors wanted $x^3+5x^2-3x-15=0$ (which is factorable), unless this is in one of those "enrichment" sections where they make the point that things don't always work.

tabstop
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