Question

what is a quartic polynomial function in standard form with zeros 1, 4, 1, and 1

Answer 1

A polynomial function with zeros \(\large\color{black}{\color{blue}{a},~\color{red}{b},~\color{green}{c},~\color{darkgoldenrod}{d}}\)
would be: \(\large\color{black}{\color{black}{f(x)=\color{blue}{(x-a}})\color{red}{(x-b)}\color{green}{(x-c)}\color{darkgoldenrod}{(x-d)}}\)

Answer 2

because when x is either, a, b, c, or d, then the f(x) is equal to zero.

Answer 3

can you give me the answer and steps to get there?

Answer 4

So for example if you had zeros of:
\(\large\color{black}{\color{black}{\color{blue}{-3}},~\color{red}{4},~\color{green}{-4},~\color{darkgoldenrod}{2}}\)

\(\large\color{black}{\color{black}{f(x)=\color{blue}{(x+3}})\color{red}{(x-4)}\color{green}{(x+4)}\color{darkgoldenrod}{(x-2)}}\)

Answer 5

I am not allowed to give you steps and answers on this website, because I am supposed to be teaching you not doing your work. But I hope that the example makes sense.

Answer 6

okay, now how would i do the next step? because thats not one of my answer choices..

Answer 7

I did an example.I didn't use your numbers as zeros, but just made up and problem and showed how to solve it.

Answer 8

made up *a problem.

Answer 9

I'll be back as soon as I can, after you reply...

Answer 10

thats not in any of my options though. so what is the next step

Answer 11

I am saying again, I just gave you an example, that was NOT your problem.

Answer 12

okay, well how do i get the quartic polynomial function?

Answer 13

thats my question

Answer 14

what i meant is that format isnt in any of my options

Answer 15

Well, you can expand the (x-a)(x+b).... etc.,

Answer 16

use the problem you had to show me how to get it in quartic polynomial form

Answer 17

lets say my zeros are: \(\large\color{black}{\color{black}{\color{blue}{1}},~\color{red}{-2},~\color{green}{3},~\color{darkgoldenrod}{2}}\)

\(\large\color{black}{\color{black}{f(x)=\color{blue}{(x-1}})\color{red}{(x+2)}\color{green}{(x-3)}\color{darkgoldenrod}{(x-2)}}\)

expand and you get:
\(\large\color{black}{ f(x)=(x-1)(x+2)(x-3)(x-2) }\)
\(\large\color{black}{ f(x)=(x^2+2x-x-2)(x-3)(x-2) }\)
\(\large\color{black}{ f(x)=(x^2+x-2)(x-3)(x-2) }\)
\(\large\color{black}{ f(x)=(x^2+x-2)(x^2-2x-3x+6) }\)
\(\large\color{black}{ f(x)=(x^2+x-2)(x^2-5x+6) }\)
\(\large\color{black}{ f(x)=x^4-5x^3+6x^2+x^3-5x^2+6x-2x^2+10x-12 }\)
\(\large\color{black}{ f(x)=x^4-5x^3+x^3+6x^2-5x^2-2x^2+6x+10x-12 }\)
\(\large\color{black}{ f(x)=x^4-4x^3-x^2+16x-12 }\)

Answer 18

And using the problem you have as an example, would be basically doing your work for you. On this website. I am not allowed to do this, apologize.

Answer 19

can you show it to me with all the zeros being positive so it can make more sense?

Answer 20

I lost connection again-:(

Answer 21

Okay, sure:
when the zeros are: \(\large\color{black}{ \color{blue}{1},~\color{green}{2},~\color{red}{3},~\color{darkgoldenrod}{4} }\)
then, the function is:
\(\large\color{black}{ f(x)=\color{blue}{(x-1)} \color{green}{(x-2)}\color{red}{(x-3)}\color{darkgoldenrod}{(x-4)} }\)
then expand the function.
\(\large\color{black}{ f(x)=(x-1)(x-2)(x-3)(x-4) }\)
\(\large\color{black}{ f(x)=(x^2-2x-x+2)(x-3)(x-4) }\)
\(\large\color{black}{ f(x)=(x^2-3x+2)(x-3)(x-4) }\)
\(\large\color{black}{ f(x)=(x^2-3x+2)(x^2-4x-3x+12) }\)
\(\large\color{black}{ f(x)=(x^2-3x+2)(x^2-7x+12) }\)
\(\large\color{black}{ f(x)=x^4-7x^3+24x^2-3x^2+21x-36x+2x^2-14x+24 }\)
\(\large\color{black}{ f(x)=x^4-7x^3-3x^2+24x^2+2x^2+21x-36x-14x+24 }\)
\(\large\color{black}{ f(x)=x^4-10x^3+26x^2-29x+24 }\)