Intermediate Value Theorem and Synthetic Division

[opus]

Say I have a given problem that states:

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.

 

[slider142]

1) Yes.

2) No. The quotient is a polynomial, plus a fraction whose denominator is the factor $(x-k)$. In particular, if your synthetic division is:
$$\begin{array}{r|rrrrr}\frac{7}{2} & 3 & -27 & 177 & 1347 & 420\\ & & \frac{21}{2} & -\frac{231}{4} & \frac{3339}{8} & \frac{98805}{16}\\\hline & 3 & -\frac{33}{2} & \frac{477}{4} & \frac{14115}{8} & \frac{105525}{16}\end{array}$$
then the quotient only means that, when $\left(x - \frac{7}{2}\right)$ is not 0,
$$3x^4 - 27x^3 + 177x^2 + 1347x + 420 = \left(x - \frac{7}{2}\right)\left(3x^3 - \frac{33}{2}x^2 + \frac{477}{4}x + \frac{14115}{8} + \frac{105525}{16\left(x - \frac{7}{2}\right)}\right)$$
The quotient is meaningless when $x = k$, since in that particular case, you would be attempting to divide by $0$. So no information about the sign of the original polynomial can be retrieved this way, without doing a lot more work than just finding the value of the original polynomial when $x = k$ manually.

 

[opus]

Thank you for that response. I also didn't know you can write synthetic division out here, so I'll keep that in mind for asking better questions next time.

So for the second part, to try to summarize, can we say that we use synthetic division because in dividing a polynomial by $(x-k)$, the remainder is equal to $f(k)$ which will give the same solution if were were to just plug in the x=k into the function and evaluate it that way? So in both instances, we are getting the same solution, so the sign of the remainder in synthetic division, is the same sign as the x=k evaluated with the function?

I'm trying to word this in a way that it makes clear how I'm thinking about it. But ideally I want to point at things and speak about them, which is not possible through typing obviously. So I hope what I've said makes sense.

 

[slider142]

You've got it. You can use the remainder to find the sign of f(k), since the remainder is f(k), but not necessarily the sign of the quotient, which depends on the value of x.

 

[opus]

Thanks!