There are formulas for the real and/or complex roots of a cubic, but they are so complicated that they are rarely used. Instead, various other strategies are employed. My favorite is the simplest--let a graphing calculator show you the zeros.

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Descartes observed that the sign changes in the coefficients can tell you the number of real roots. This expression has two sign changes (+-+), so has 0 or 2 positive real roots. If the odd-degree terms have their signs changed, there is only one sign change (-++), so one negative real root.

It can also be informative to add the coefficients in both cases--as is, and with the odd-degree term signs changed. Here, the sum is zero in the first case, so we know immediately that x=1 is a zero of the expression. That is sufficient to help us reduce the problem to finding the zeros of the remaining quadratic factor.

Using synthetic division (or polynomial long division) to factor out x-1 (after removing the common factor of 4), we find the remaining quadratic factor to be x²+x-4.

The zeros of this quadratic factor can be found using the quadratic formula:

a=1, b=1, c=-4

x = (-b±√(b²-4ac))/(2a) = (-1±√1+16)/2

x = (-1 ±√17)2

The zeros are 1 and (-1±√17)/2.

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The graph shows the zeros of the expression. It also shows the quadratic after dividing out the factor (x-1). The vertex of that quadratic can be used to find the remaining solutions exactly: -0.5 ± √4.25.

The given expression factors as ...

4(x -1)(x² +x -4)

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