# [SOLVED]Find a polynominal function f(x) of least degrees having only the real coeficents and zeros as given

#### karush

##### Well-known member
Find a polynominal function f(x) of least degrees having only the real coeficents and zeros as given

$$5+i \quad 4-i$$
ok I did this but don't think this is the final answer
$(x-(5+i))(x-(5-i))=x^2+26$
$(x-(4-i))(x-(4+i))=x^2-17$

#### MarkFL

Staff member
You've got the right idea, but your quadratic products aren't correct. #### karush

##### Well-known member
You've got the right idea, but your quadratic products aren't correct. $$(x-(5+i))(x-(5-i))=x^2+26$$
$$(x-(4-i))(x-(4+i))=x^2+17$$

is this as far as can go?

or is it

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

#### MarkFL

Staff member
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. #### karush

##### Well-known member
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. $$(x^2 - 10 x + 26)(x^2 - 8 x +17)=x^4-18x^3+123x^2-378x+442$$

#### Olinguito

##### Well-known member
Find a polynominal function f(x) of least degrees having only the real coeficents and zeros as given

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

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.

#### karush

##### Well-known member
What are some good online calculators for hunting down zeros

I thot EMath was nice and you
Pull latex from it.