MHB Algebra help about polynomials

[kelly]

I really need help.
f(x) is a fourth degree polynomial function
f(x) has zeros of plus or minus 2 and plus or minus 3i
f(0)=-108
Find an equation for f(x) in general form

 

[MarkFL]

Hello and welcome to MHB, kelly! :D

The general quartic having the zeroes $$x\in\{a,b,c,d\}$$ is given by:

$$f(x)=k(x-a)(x-b)(x-c)(x-d)$$ where $$k\ne0$$

Can you state the family of quartics with the given roots?

 

[evinda]

I really need help.
f(x) is a fourth degree polynomial function
f(x) has zeros of plus or minus 2 and plus or minus 3i
f(0)=-108
Find an equation for f(x) in general form

$f(x)$ is a fourth degree polynomial function, so it is of the form $f(x)=ax^4+bx^3+cx^2+dx+e$.

$f(0)=-108 \Rightarrow e=-108$What can we deduce from the fact that $f(x)$ has zeros of plus or minus $2$ and plus or minus $3i$ ?

 

[kelly]

I am unsure. That was all of the information received. I was absent and i have no idea how to even begin this problem

 

[evinda]

I am unsure. That was all of the information received. I was absent and i have no idea how to even begin this problem

$f(x)$ has zeros of plus or minus $2$ and plus or minus $3i$ means that:

$$f(2)=0 \\ f(-2)=0 \\ f(3i)=0 \\ f(-3i)=0$$

What can we get from the above relations? You just have to substitute the values.

 

[topsquark]

I really need help.
f(x) is a fourth degree polynomial function
f(x) has zeros of plus or minus 2 and plus or minus 3i
f(0)=-108
Find an equation for f(x) in general form

Hello and welcome to MHB, kelly! :D

The general quartic having the zeroes $$x\in\{a,b,c,d\}$$ is given by:

$$f(x)=k(x-a)(x-b)(x-c)(x-d)$$ where $$k\ne0$$

Can you state the family of quartics with the given roots?

Using MarkFL's method we know that, with a = 2, b = -2, c = 3i, d = -3i, we have
[math]f(x) = k(x - 2)(x + 2)(x - 3i)(x + 3i) = k(x^2 - 4)(x^2 + 9)[/math]

Then [math]f(0) = -108 = k(0^2 - 4)(0^2 + 9) = -36k[/math] and now you can solve for k.

-Dan