We'll learn in the Curve Drawing section that a cubic curve can have either:

The best thing to do is to sketch the curve, but let's assume that's too hard for now. We can determine where at least one of the roots occurs by substituting values of `x` until we see a change in the sign for `y` (which would indicate the curve must have passed through the `x`-axis in that interval).

Starting at `x=0`, we find that `y=2`, which is positive.

Next, try `x=1`, and we get `y=1`, which is still positive. (Of course, the curve could have dipped below the `x`-axis between `0` and `1`, but without a sketch, we don't know.)

Try `x=2` and we get `y=0`. This is a lucky guess, and gives us one of the roots.

Trying more positive values (`x=3,4,5...`) gives us larger and larger values for `y` (which are `17,70,177,...`), so we are clearly moving away from the root.

We continue to stumble around in the dark, and assume maybe there is a root near `x=1`. Let's try the algorithm with `x_0=1` and see what happens.

We'll need the derivative:

`dy/dx = 9x^2-18x+5`

So

`x_1=x_0-(f(x_0))/(f’(x_0))`

` = 1-(f(1))/(f’(1))`

` = 1-1/(-4)`

` = 1.25`

Another step:

`x_2=x_1-(f(x_1))/(f’(x_1))`

` = 1.25-(f(1.25))/(f’(1.25))`

` = 1.26363636363`

Continuing on we get:

x3 = 1.26376260461638

x4 = 1.26376261582597

x5 = 1.26376261582597

x6 = 1.26376261582597

Since there is no more change to our value, we can assume we've found the second root, at `x=1.26376261582597`.

You can use the interactive graph applet to find the third root.

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