- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,716

$$5+i \quad 4-i$$

ok I did this but don't think this is the final answer

$(x-(5+i))(x-(5-i))=x^2+26$

$(x-(4-i))(x-(4+i))=x^2-17$

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,716

$$5+i \quad 4-i$$

ok I did this but don't think this is the final answer

$(x-(5+i))(x-(5-i))=x^2+26$

$(x-(4-i))(x-(4+i))=x^2-17$

- Admin
- #2

- Thread starter
- #3

- Jan 31, 2012

- 2,716

$$(x-(5+i))(x-(5-i))=x^2+26$$You've got the right idea, but your quadratic products aren't correct.

$$(x-(4-i))(x-(4+i))=x^2+17$$

is this as far as can go?

or is it

(x^2+26)(x^2+17)

- Admin
- #4

- Thread starter
- #5

- Jan 31, 2012

- 2,716

You want to first get two quadratics as you've done, and then the quartic polynomial will be their product, but you don't have the correct quadratics yet. Check your multiplications...you should have \(x\) terms.

$$(x^2 - 10 x + 26)(x^2 - 8 x +17)=x^4-18x^3+123x^2-378x+442$$

- Apr 22, 2018

- 251

A polynomial with real coefficients will have complex zeroes occurring in conjugate pairs. So if two of the zeroes are $5+i$ and $4-i$, the other ones will be $5-i$ and $4+i$. Hence $f(x)$ will be a polynomial of degree at least 4.

$$5+i \quad 4-i$$

ok I did this but don't think this is the final answer

$(x-(5+i))(x-(5-i))=x^2+26$

$(x-(4-i))(x-(4+i))=x^2-17$

A polynomial with zeroes $a\pm bi$ is $x^2-2ax+(a^2+b^2)$. Hence

$$f(x)\ =\ k(x^2-10x+26)(x^2-8x+17)$$

where $k$ is any nonzero real number.

- Thread starter
- #7

- Jan 31, 2012

- 2,716

I thot EMath was nice and you

Pull latex from it.