Find all the zeros of the function. (Enter your answers as a comma-separated list.)

Question

f(s) = 3s³ − 4s² + 8s + 8

s=

Write the polynomial as a product of linear factors. Use a graphing utility to verify your results graphically. (If possible, use the graphing utility to verify the imaginary zeros.)

f(s) =

Answer 1

Hi Aerin H.,

You could try to attack this problem by factoring directly, using trial factors of the form (x-c) and looking at the patterns of the remainders (if any!) after the division (in my educational era, called "synthetic division", no doubt something else now). But you'd find that there aren't any integer roots.

All right, then, you have to solve f(x) = 3x^3 - 4x^2 + 8x + 8 = 0 for its roots. The general roots of a cubic polynomial are a rather complicated expression, which you can look up on Google and start cranking on (recommended!), or use a math site such as Mathematica.

It appears to me from casual inspection that there is one real root, hence 2 complex ones. The graph of the function will tell you this, if it has some wobble (i.e. a sinusiodal component to its first or second derivative somewhere in the middle of the domain). You should know by now that a cubic equation (degree 3) has 3 roots, and that they may be all real, all imaginary, or some real and some imaginary; additionally, some roots may be multiple, i.e. if they coincide at an inflection or minimum or maximum point, as they do in y = x^2 for example -- both roots are x=0, and this is a minimum point.

What you may not yet appreciate is that the detailed appearance of the curve of the function indicates where those roots lie. For example, in f(x) = 2 + x^2, the curve never touches the x-axis, hence there are no real roots. However, if you reflect the curve vertically around its vertex (i.e. over the line y=2), the flipped curve crosses the x-axis at the points corresponding to the coefficients of the imaginary roots ( + and - (2)^0.5 ); the roots are +/-(2)^(0.5)*i .

If this seems mysterious, you may want to actually generate a few cubic functions using known complex and real roots, and look at their shape. Just a few trials, and you'll get handy at it.

-- Cheers, --Mr. d.