How can I find an "Nth degree polynomial function"

[dove99x]

I have no idea how I would figure this out. The problem states:

Find an nth-degree polynomial function with real coefficients satisfying the given conditions.
n=3; 2 and 5i are zeros; f(1)=26

by the way, this is the answer:

My question is, how do I get this? I'm stumped.

 

[mmm4444bot]

An nth-degree polynomial has exactly n roots (considering multiplicity).

The roots of a polynomial are exactly the same as the zeros of the corresponding polynomial function. So, they say "zeros" and I'm calling them roots.

Since n = 3, you need 3 roots.

They gave you two of them: 2 and 5i.

Strict Complex roots always come in conjugate pairs, so right away we know that the third root is -5i.

Each root forms a factor (x - root) of the polynomial.

The factorized polynomial function takes this form:

f(x) = a (x - root1) (x - root2) (x - root3)

where a is the leading coefficient

Substitute the known values for f(x), x, root1, root2, root3.

Solve the resulting equation for the leading coefficient a.

If you want to expand the factorized polynomial, you can multiply it all out, at the very end. You should end up with the posted answer.

If I wrote anything that you don't understand, please ask specific questions. Otherwise, show whatever work that you can, to get more assistance.

Cheers ~ Mark

 

[dove99x]

Once I set it up this is what I get.

Am I setting it up wrong or am I multiplying it wrong? This is my setup:

This is the answer I need.

 

[galactus]

A=-1. so multiply your version by -1 and change the signs.

\(\displaystyle f(x)=a(x-2)(x+5i)(x-5i)\)

For f(1)=26:

\(\displaystyle a(1-2)\underbrace{(1-5i)(1+5i)}_{\text{this is 26}}=26\)

\(\displaystyle a(-1)(26)=26\)

\(\displaystyle a=-1\)

\(\displaystyle -(x-2)(x+5i)(x-5i)=-x^{3}+2x^{2}-25x+50\)

 

[mmm4444bot]

Am I setting it up wrong or am I multiplying it wrong?

Your multiplications are fine, but you did not substitute in all of the known values.

I posted:

f(x) = a (x - root1) (x - root2) (x - root3)

Substitute the known values for f(x), x, root1, root2, root3.

Solve the resulting equation for the leading coefficient a.

You did not substitute the two known values 26 for f(x) and 1 for x.

You can't find the value of a, without making these substitutions.

Perhaps, you didn't realize that you need to know the leading coefficient, before you're able to multiply out the polynomial which defines function f.

Otherwise, you seem to know what you're doing. I mean, your multiplications shown are all correct.

Cheers