# Thread: n=3; -5 and 4+3i are zeros; f(2) =91

1. ## n=3; -5 and 4+3i are zeros; f(2) =91

I'm really hung up on the following part of the problem:

(x+4+3i)(x-4+3i)

What I came up with:

$\displaystyle x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16$: Combine like terms to get

$\displaystyle x^2 +6xi -25$

I've also tried using -6 in place of 6xi but nope.

I then multiply by (x+5) to get a host of wrong answers.

I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.

2. ## Re: n=3; -5 and 4+3i are zeros; f(2) =91

Originally Posted by Ingersoll
I'm really hung up on the following part of the problem:

(x+4+3i)(x-4+3i)

What I came up with:

$\displaystyle x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16$: Combine like terms to get

$\displaystyle x^2 +6xi -25$

I've also tried using -6 in place of 6xi but nope.

I then multiply by (x+5) to get a host of wrong answers.

I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.
What is going on? What is to be done?
There is object stated.

3. ## Re: n=3; -5 and 4+3i are zeros; f(2) =91

Originally Posted by Ingersoll
I'm really hung up on the following part of the problem:

(x+4+3i)(x-4+3i)

What I came up with:

$\displaystyle x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16$: Combine like terms to get

$\displaystyle x^2 +6xi -25$

I've also tried using -6 in place of 6xi but nope.

I then multiply by (x+5) to get a host of wrong answers.

I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.
if $\displaystyle 4+3i$ is a zero, then its conjugate, $\displaystyle 4-3i$, is also.

$\displaystyle f(x) = k(x+5)[x-(4+3i)][x-(4-3i)]$

$\displaystyle f(x) = k(x+5)(x^2-8x+25)$

$\displaystyle f(2) = 91 = k(2+5)(4-16+25) = k(7)(13)$

$\displaystyle k = 1$

$\displaystyle f(x) = (x+5)(x^2-8x+25)$

### n=3;-5 and 4 3i are zeros;f(2)=91

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