Hi Cici,

Are you sure you have this problem stated correctly?

The term real zero, as it relates to functions, is asking for the x-intercepts of the graph. At every x-intercept, the graph of the polynomial function can cross through the point, cross through the point while be tangent to it or simply touch the x axis and not cross it. In you case, you only want the positive real zeros or x-intercepts to the right of the y-axis.

To find your zeros, simply set f(x)=0, solve for x by factoring then disregard any negative answers. Unfortunately the function f(x) = x⁴ + x³ - 7x - 1 can't be factored.

A second technique is to use the remainder theorem http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/sean5.html and synthetic division http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/edward1.html

Again this technique fails for f(x) = x⁴ + x³ - 7x - 1.

You can see that f(0) = -1 and f(2) = 16 so there is a zero between 0 and 2 and hence there is at least one positive real zero. If you know Descarte's Rule of Signs you can use it to show that there is only one positive real zero. Likewise you can use caculus to show this.

Janice and Harley