Question

Use the p over q method and synthetic division to factor the polynomial P(x). Then solve P(x)=0 P(x)=x3-6x2+3x-10

Answer 1

Is the "p over q method" perhaps the rational root theorem? Basically, the rational roots of that equation are the factors of the constant term (as the leading term conveniently has a coefficient of 1). Test each candidate r to see if P(r) = 0 — if so, (x - r) is a root and you can divide it off via synthetic division to get a simpler polynomial with the same roots as the remaining roots of the original polynomial.

Answer 2

Are you sure that \[P(x) = x^3-6x^2+3x-10\]? I think \[P(x)=x^3\color{red}{+}6x^2+3x-10\] is more likely...

Answer 3

im sure its correct. what is the intermediate value theorem? i think you might have to use it for this problem

Answer 4

Well, \[x^3-6x^2+3x-10\]has 1 positive real root and 2 complex roots. Are you expecting one of your factors to be anything like \[x-(2+\frac{1}{3} \left(270-27 \sqrt{73}\right)^{1/3}+\left(10+\sqrt{73}\right)^{1/3})\]

If not, you might want to recheck the polynomial.

Answer 5

Somebody told me the answer was appr. 5.7803. At this x value P(x) does equal zero but they used the intermediate value theorem to get this answer and I dont understand it.

Answer 6

Well, yes, one of the roots is approximately 5.78029. But the other two are complex numbers. You can't get any of these with the p over q method and synthetic division. You can, however, do so if you change the sign of the 6x^2 term (I didn't investigate beyond that, possibly changing one of the other signs also turns it into a polynomial that can be factored).

I think you have the wrong polynomial there.

Answer 7

The polynomial is correct but what do you mean by complex numbers?

Answer 8

involving \(i = \sqrt{-1}\)

Answer 9

One of the complex roots will be\[2-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(270-27 \sqrt{73}\right)^{1/3}-\frac{1}{2} \left(1+i \sqrt{3}\right) \left(10+\sqrt{73}\right)^{1/3}\]

Still convinced you have the correct polynomial?

Answer 10

Changing the sign on the constant term will also make it into a polynomial you can factor.

Answer 11

I understand what you are trying to say. I typed to graph onto a graphing calculator and the only zero i get is the 5.78. Do you know what the intermediate value theorem is? That is all I am having trouble with right now?

Answer 12

The intermediate value theorem just says that if you have one value of x such that P(x) > 0 and another value of x such that P(x) < 0, there's a value of x in between them where P(x) = 0 because the curve can't just hop over the x-axis...

Answer 13

The problem statement tells you to do something that is impossible with this polynomial. It also is notable for its absence of any instruction to use the IVT.

Answer 14

Humor me. Change the polynomial in one of the ways that I suggested, and do the problem as requested.

Answer 15

just did it. i got 1,-2, and -5

Answer 16

Very good. Doesn't that seem more likely to be what was desired?

Answer 17

It does. Hold on a second. i would just like to show you the work that someone did for the original polynomial

Answer 18

sure...

Answer 19

By the way, here's a graph of the 3 polynomials (the one originally posted, and the two I suggested). You can clearly see that one of them (the blue line) only crosses the x-axis 1 time; that's the one as posted.

Answer 20

http://tullyschools.org/hsteachers/dneuman/APCalc/APNotes/01September/004PrecalcReview/002Topic7Sol.pdf

Please look at 3

Answer 21

Sure. If you look at my graph, for the blue curve, you can see that at x = 6, P(x) >0, and at x=5, P(x) <0, and that's what the synthetic division positive/negative remainder is telling you, too. The IVT says that if you have a continuous function running between those two points, there's some point between them at which P(x) = 0. With a calculating device of some sort you could narrow down the region in which that axis crossing occurs by dividing that region in half, looking to see if P(x) still had different signs between the two trial endpoints, and moving in closer and closer.

Answer 22

Given that there's only one such problem, I'm not entirely convinced it wasn't just a mistake, but I have no way of proving that. It is useful to understand what is going on here, after all, but why torture students by asking them to do something that cannot be done? Better to teach them how to recognize the situation and how to make the best of it!

Answer 23

Thank you and yes i also agree that it was a mistake but there is no way of trying to convince my teacher of this during summer so i gotta answer the question somehow

Answer 24

Well, if you want to really impress your teacher, read this page and find the roots :-)
http://mathworld.wolfram.com/CubicFormula.html

Answer 25

Thanks!