- #1

- 717

- 131

Does the Intermediate Value Theorem guarantee that the following equation has a real solution between ##(\frac{7}{2})## and ##(\frac{9}{2})##?

$$3x^4-27x^3+177x^2+1347x+420=0$$

Now what I want to do is determine the sign of x=##(\frac{7}{2})## and x=##(\frac{9}{2})## at the given equation.

I can do this two ways: one way is to just evaluate ##f(7/2)## and ##f(9/2)##. This gives two positive results, so the Intermediate Value Theorem cannot guarantee that there's a zero between these points. This leads me into my fist question:

**1)**In plugging the two given x values into the given equation, the left side

*does not*equal zero. So this is telling me that the two given x values are not solutions to the equation (meaning the graph doesn't cross the x-axis at these points). However there is a

*point*where the graph does have the x-value of x=##(\frac{7}{2})## and x=##(\frac{9}{2})## and at these points, the graph is positive.

**Is this a true statement?**

Another way to do this is to use synthetic division, dividing the polynomial by k=##(7/2)## and k=##(9/2)## This works because of the Remainder Theorem- where if we divide a polynomial ##p(x)## by ##(x-k)##, the remainder is ##p(k)##. Since x=k, we find the value of the function ##3x^4-27x^3+177x^2+1347x+420=0## evaluated at x=##(\frac{7}{2})## and x=##(\frac{9}{2})##. In this case, the

*remainders*are both positive, and is the second part of my question.

When dividing polynomials, if the remainder is positive, does that mean that the entire quotient is positive? In other words, if the final digit in the synthetic division is positive, is the entire quotient positive? This question doesn't seem very intuitive to me as it's not with concrete number values, but with expressions with variables and graphs that can fluctuate between positive and negative values.