Question

The following graph shows a seventh-degree polynomial:

Part 1: List the polynomial’s zeroes with possible multiplicities.
Part 2: Write a possible factored form of the seventh degree function.

Answer 1

I was sure it was (x+5) (x+5) (x+1) (x-4) (x-4) (x-4) (x-7), but when I graphed it, I got a different graph

Answer 2

I agree that the roots have values \(-5,-1, 4, 7\)

A root where the curve crosses the \(x-\)axis will have an odd multiplicity. A root where it touches and reverses course will have an even multiplicity. Which roots will be duplicated?

Answer 3

I already know about what multiplicities mean, and I'm sure it must be one of the following:
(x+5) (x+5) (x+1) (x-4) (x-4) (x-4) (x-7)
(x+5) (x+5) (x+1) (x+1) (x+1) (x-4) (x-7)
(x+5) (x+5) (x+1) (x-4) (x-7) (x-7) (x-7)
(x+5) (x+5) (x+5) (x+5) (x+1) (x-4) (x-7)

However, none of these resembles the graph I was provided.

Answer 4

Don't forget that there's a constant factor in front, often assumed to be \(1\). I think if you multiply the correct choice by \(\dfrac{1}{5488}\) you will get something very similar to the graph shown, as I did:

Answer 5

If you really know about the multiplicities, you should strike some of your possible answers from consideration. For example, the root at \(x=4\) crosses the \(x-\)axis, so that must be an odd multiplicity, yet some of your possible polynomials have an even power for the \((x-4)\) term...

Answer 6

Oh, my lesson never mentioned anything about a constant factor.

Which one had an even power for the (x-4) term?

Answer 7

the first one...

Answer 8

sorry, I misread it, there are 3 there on first one, nvm

Answer 9

There are 3 of them. I don't know why I wrote it out like that but it should be:
(x+5)^2(x+1)(x-4)^3(x-7)

Answer 10

You got me worried for a moment

Answer 11

so I just wrote out the roots from left to right, then made the one that touched but did not cross be an even power. I also had to decide where to add the "extra" multiplicities, and the spot where it goes up to the \(x\) axis at \(x=4\) and then crosses it was the obvious spot. Looking at the graph, the root at \(x=-3\) looked like an easy one to read off the resulting \(y\) value, so I plugged \(x=-3\) into the equation and choose a value for the constant multiplier to make the whole thing equal the value I eyeballed at that point (5). That constant turns out to be the fraction I gave you.

Answer 12

In general, any polynomial \(P(x)\) with rational roots \(x = r_1, r_2, ... r_n\) can be written in factored form \[P(x) = a(x-r_1)(x-r_2)...(x-r_n)\] and if you think about it for a moment or two, you'll realize that there are an infinite number of polynomials of the same form that pass through a given set of roots, all controlled by the value of \(a\).

Answer 13

I graphed -1/5600(x+5)(x+5)(x+1)(x-4)(x-4)(x-4)(x-7) and got the original graph too. However, when I plugged in 1/5488, I got something different

Answer 14

Oh, nvm that last statement

Answer 15

Sorry, I dropped the negative sign...

Answer 16

Thanks for all the help!

Answer 17

Those are all graphs of \[a(x-1)(x-3)(x+2)\]for different values of \(a\): \(1,1/2, 3, -2\)