how do you find the x-intercepts if you have 0 = x^3 -9x^2 + 15x + 30

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- Apr 2nd 2009, 06:36 AMproski117Need help finding the x-intercepts
how do you find the x-intercepts if you have 0 = x^3 -9x^2 + 15x + 30

- Apr 2nd 2009, 11:50 AMHallsofIvy
I assume you mean "if you have $\displaystyle y= x^3- 9x^2+ 15x+ 30$" so the x-intercepts are where y= 0: $\displaystyle x^3- 9x^2+ 15x+ 30= 0$. Of course, you find them by solving that equation. I recommend crossing your fingers and hoping that there is at least one

**rational**root. The rational root theorem tells us that if there is a rational root, it must be an integer that evenly divides the constant term, 30. The only possible rational roots are $\displaystyle \pm 1$, $\displaystyle \pm 2$, $\displaystyle \pm 3$, $\displaystyle \pm 5$, $\displaystyle \pm 6$, $\displaystyle \pm 10$, $\displaystyle \pm 15$, $\displaystyle \pm 30$. Try those and see if one or more satisfy the equation. If one does, you can divide by the corresponding factor to reduce to a quadratic equation. - Apr 3rd 2009, 04:20 PMstapel
As it happens, this particular polynomial has no "nice" rational roots. To find the one real root, you'll have to use

**numerical methods**.

Note that y = -44 at x = -2, but y = 5 at x = -1. Then somewhere between x = -2 and x = -1, you must have y = 0.

When x = -1.5, y = -16.125. So y = 0 must be between x = -1.5 and x = -1. Try, say, x = -1.25.

And so forth, until you've got enough decimal places of accuracy. (Wink)