Question

How do I create a function with these characteristics?
(pre-calc)

Has rational coefficients and roots of 3, 3, and 3-i

Answer 1

The 3-i is mainly what throws me off...

Answer 2

well, the issue being... that say when you use the quadratic formula, and say you have a negative discriminant, then you'd end up with 2 COMPLEX solutions

recall \(\bf \text{quadratic formula}\\
x= \cfrac{ - b \color{red}{\pm} \sqrt { b^2 -4ac}}{2a}\)

what does that mean? well, it means that the complex root there, is not all by her lonesome, she's got a "conjugate" companion too, that is 3+i, so those are your roots

Answer 3

\(\bf x=3\implies x-3=0\implies (x-3)=0\\ \quad \\
x=3\implies x-3=0\implies (x-3)=0\\ \quad \\
x=3-i\implies x-3+i=0\implies (x-3+i)=0\\ \quad \\
x=3+i\implies x-3-i=0\implies (x-3-i)=0\\ \quad \\ \quad \\
\color{green}{(x-3)(x-3)(x-3+i)(x-3-i)=0}\\ \quad \\ \quad \\
(x-3+i)(x-3-i)\implies [(x-3)-(i)][(x-3)+(i)]\\ \quad \\
\textit{recall that}\quad \color{blue}{(a-b)(a+b) = a^2-b^2}\qquad thus\\ \quad \\\
[(x-3)-(i)][(x-3)+(i)]\implies [(x-3)^2-i^2]
\)

Answer 4

and as you recall what \(\bf i^2\) is, then expand the binomial, cancel out like terms, and that'd be the binomial the complex came from

Answer 5

the (x-3)(x-3) is just a multiple root, or a root with multiplicity of 2

Answer 6

that all just confused me a lot /: do i just multiple (x - 3)(x - 3)(x - 3 - i)?

Answer 7

*multiply

Answer 8

hhehe

Answer 9

ok... what part you didn't get?

Answer 10

almost all of it lol

Answer 11

have you even done quadratics at all? do you know how to factor them?

Answer 12

I have, but still pretty new at them

Answer 13

you know the difference of 2 squares, right? \(\bf a^2-b^2 = (a-b)(a+b)\)

Answer 14

I recall something like that, yes

Answer 15

... have you covered the quadratic formula?

Answer 16

I have that written down in my notes

Answer 17

so yes

Answer 18

say for example, in a quadratic.... you end up with say using the quadratic formula and looks like \(\bf x= \cfrac{ - 5 \pm \sqrt { 3^2 -4(2)(3)}}{2(2)}\implies x= \cfrac{ - 5 \pm \sqrt { 9-24}}{4}\implies x= \cfrac{ - 5 \pm \sqrt {-15}}{4}\\ \quad \\
x=\cfrac{ - 5 + \sqrt {-15}}{4}\qquad and\qquad x=\cfrac{ - 5 - \sqrt {-15}}{4}\)

Answer 19

as you can see, from there, you'd end up with 2, not 1, 2 imaginary or complex roots

so the idea being that, above there you have 3 roots, but 1 is complex, but complex roots never come alone, they come with their conjugate companion
so you really have 4 roots to work with, in order to get the function or polynomial

Answer 20

hmm actually .... I .... 5 should have been 3... but anyhow, just to exemplify, that's the reason, why a complex will not come alone

Answer 21

So then do I multiply (x-3)(x-3)(x-3-i)(x-3+i)?

Answer 22

so the roots you were given PLUS the conjugate of the given complex will yield \(\bf \bf x=3\implies x-3=0\implies (x-3)=0\\ \quad \\
x=3\implies x-3=0\implies (x-3)=0\\ \quad \\
x=3-i\implies x-3+i=0\implies (x-3+i)=0\\ \quad \\
x=3+i\implies x-3-i=0\implies (x-3-i)=0\)

Answer 23

yeap, which is what I showed above \(\bf \color{green}{(x-3)(x-3)(x-3+i)(x-3-i)=0}\\ \quad \\ \quad \\
(x-3+i)(x-3-i)\implies [(x-3)-(i)][(x-3)+(i)]\\ \quad \\
\textit{recall that}\quad \color{blue}{(a-b)(a+b) = a^2-b^2}\qquad thus\\ \quad \\\
[(x-3)-(i)][(x-3)+(i)]\implies [(x-3)^2-i^2]\)

Answer 24

Ugh so many numbers lol it's overwhelming. So wait does that mean the equation would be [(x-3)^2 - i^2]?

Answer 25

hehe

Answer 26

Uggggh I'm sorrry lol

Answer 27

I'm not usually this bad at math

Answer 28

I'm giving you a medal cuz I know you sincerely are trying lol

Answer 29

yes, lemme show it a bit more.... \(\bf \color{green}{(x-3)(x-3)(x-3+i)(x-3-i)=0}\\ \quad \\ \quad \\
(\color{red}{x-3}+\color{blue}{i})(\color{red}{x-3}-\color{blue}{i})\implies [(\color{red}{x-3})-(\color{blue}{i})][(\color{red}{x-3})+(\color{blue}{i})]\\ \quad \\
\textit{recall that}\quad \color{blue}{(a-b)(a+b) = a^2-b^2}\qquad thus\\ \quad \\\
[(x-3)-(i)][(x-3)+(i)]\implies [(\color{red}{x-3})^2-\color{blue}{i}^2]\)

Answer 30

I just.... I don't understand what any of that means, like I can't comprehend it

Answer 31

well.... do you know how I got \(\bf [(x-3)^2-i^2]\quad ?\)

Answer 32

is really just plain terms grouping all I've done

Answer 33

Yes, I think

Answer 34

You used that rule

Answer 35

But that's not the equation?

Answer 36

the equation is obtained by multiplying its roots

Answer 37

so far we've multiplied the 2 complex ones, so now we know that \(\bf (x-3)(x-3)(x-3+i)(x-3-i)=0\\ \quad \\
\implies (x-3)(x-3)[(x-3)^2-i^2]\)

Answer 38

so... can you recall what \(\bf i^2\) is?

Answer 39

Ohhh, I see! And i^2... uggggh I know it's 1 or -1 or something like that, right?

Answer 40

\(\bf i^2\implies \sqrt{-1}\cdot \sqrt{-1}\implies (\sqrt{-1})^2\implies \sqrt{(-1)^2}\implies -1\)

Answer 41

Okay, so it's -1

Answer 42

\(\bf (x-3)(x-3)[(x-3)^2-i^2]\implies (x-3)(x-3)[(x-3)^2-(-1)]\\ \quad \\
(x-3)(x-3)[(x-3)^2+1]\implies (x-3)^2[(x-3)^2+1]\)

Answer 43

so now all you have to do is, expand both binomials, and multiply their terms, then cancel any like-terms

Answer 44

as you can see, it'd be a 4th degree polynomial, thus 4 degree, 4 roots

Answer 45

well (x - 3)(x - 3) is x^2 - 6x + 9, and I could multiply that by itself (because there's another (x-3)^2) and then 1?

Answer 46

well, you're right... but .... \(\bf (x-3)^2[(x-3)^2+1]\implies (x^2-6x+9)[x^2-6x+9+1]\\ \quad \\
(x^2-6x+9)[x^2-6x+10]\)

Answer 47

I got x^4 - 12x^3 + 55x^2 - 114x + 90

Answer 48

yeap, that's what I got :)

Answer 49

yay! lol thank you so much for your time

Answer 50

yw