Birthday Polynomial

[BadCo55]

Hi, I have a project about making a polynomial based on my birthday which is January 6, 1997.


I came up with: (1x+6)(1x+9)2(9x-7); it has to be in factored form.
From there I got: (x^2+9x+6x+54)(18x-14)
When I added all those I get: 18x^3+256x^2+762x-756
I then simplified and god: 9x^3+128x^2+381x-378


The instructions tell me that it must contain at least 3 factors. One zero must have a multiplicity of 2 or 3 and at least two zeroes must be imaginary. I'm not sure how to find the last part.


Can you please tell me 1. Is my polynomial correct and 2. How do I find the last part? Thank you!!

 

[Ishuda]

Hi, I have a project about making a polynomial based on my birthday which is January 6, 1997.


I came up with: (1x+6)(1x+9)2(9x-7); it has to be in factored form.
From there I got: (x^2+9x+6x+54)(18x-14)
When I added all those I get: 18x^3+256x^2+762x-756
I then simplified and god: 9x^3+128x^2+381x-378


The instructions tell me that it must contain at least 3 factors. One zero must have a multiplicity of 2 or 3 and at least two zeroes must be imaginary. I'm not sure how to find the last part.


Can you please tell me 1. Is my polynomial correct and 2. How do I find the last part? Thank you!!

(1x+6)(1x+9)2(9x-7) is a polynomial based on your birthday but it doesn't satisfy the conditions so let's look at those:
(2) One zero must have a multiplicity of 2 or 3.
This means that , written in factored form, one of the factors must repeat itself 2 or 3 times. If you are attending a classical class (going to a classroom with students pretty much your own age), everyone in your class was probably born in the 1990's, so every one could use the 9's for the repeating zeros and get
...(x-9) (x-9)...
for the multiplicity of 2. Since your birthday was in January, you could use the 1's
(3) At least two zeroes must be imaginary.
This mean that two of the zeros must be of the form x = a i where a is a real number and i² = -1.

As an example, suppose I had been born Feb. 2, 1998, I could choose to use, among the many available,
(x - 2 i) (x + 2 i) (1 x) (x-9) (x-9) 8
= (x² + 4) x (x² - 18 x + 81) 8
= 8 x (x⁴ - 18 x³ + 85 x² - 72 x + 324)
= 8 x⁵ - 144 x⁴ + 680 x³ - 576 x² + 2592 x
unless, of course, I've made some dumb misteak in simple arithmetic.

Oh yes, and that first requirement (must contain at least 3 factors), well both of our examples do.

 

[BadCo55]

Why the "2"; isn't only 1 6 1 9 9 7 allowed?

Both of the examples that my teacher gave me had the two in that spot, that's why I put it there.

 

[BadCo55]

(1x+6)(1x+9)2(9x-7) is a polynomial based on your birthday but it doesn't satisfy the conditions so let's look at those:
(2) One zero must have a multiplicity of 2 or 3.
This means that , written in factored form, one of the factors must repeat itself 2 or 3 times. If you are attending a classical class (going to a classroom with students pretty much your own age), everyone in your class was probably born in the 1990's, so every one could use the 9's for the repeating zeros and get
...(x-9) (x-9)...
for the multiplicity of 2. Since your birthday was in January, you could use the 1's
(3) At least two zeroes must be imaginary.
This mean that two of the zeros must be of the form x = a i where a is a real number and i² = -1.

As an example, suppose I had been born Feb. 2, 1998, I could choose to use, among the many available,
(x - 2 i) (x + 2 i) (1 x) (x-9) (x-9) 8
= (x² + 4) x (x² - 18 x + 81) 8
= 8 x (x⁴ - 18 x³ + 85 x² - 72 x + 324)
= 8 x⁵ - 144 x⁴ + 680 x³ - 576 x² + 2592 x
unless, of course, I've made some dumb misteak in simple arithmetic.

Oh yes, and that first requirement (must contain at least 3 factors), well both of our examples do.

So would (x-1i)(x+1i)(6x+1)(x-9)(x-9)7 be sufficient in my case?