Create a polynomial that fits the following conditions.
Degree 4, 2 positive real zeros, 0 negative real zeros, 2 complex zeros

Question

Create a polynomial that fits the following conditions.
Degree 4, 2 positive real zeros, 0 negative real zeros, 2 complex zeros

asnaseer · Answer

are you aware of the fact that if a polynomial has a root say, $r_1$, then $(x-r_1)$ will be one of its factors?

asnaseer · Answer

e.g. if it has a root of say, 5, then (x-5) will be one of its factors.

anonymous · Answer

Yes.

asnaseer · Answer

good.

secondly, are you also aware that complex root always appear as pairs - each one being a complex conjugate of the other?

asnaseer · Answer

which means if (a+ib) is on the roots, then so is (a-ib)

asnaseer · Answer

*is one of the roots

anonymous · Answer

Yes.

asnaseer · Answer

good.

now finally you are told that the polynomial must be of degree 4 - so it has 4 roots.
two of them must be positive real roots - lets call them $r_1$ and $r_2$
and two of them must be complex - lets call them $z=a+ib$ and $\bar z =a-ib$

asnaseer · Answer

so we know this polynomial must be a product of its factors and can be written down as:$$f(x)=(x-r_1)(x-r_2)(x-z)(x-\bar z)$$now all you need to do is pick values for these variables.

asnaseer · Answer

does that make sense?

anonymous · Answer

Yes.

asnaseer · Answer

good - so you can answer the question from here now - correct?

anonymous · Answer

Not really. I knew all that stuff already, but I don't know how to answer the question.

asnaseer · Answer

ok, I showed that the polynomial in question has to be of the form:$$f(x)=(x-r_1)(x-r_2)(x-z)(x-\bar z)$$so, why don't you first pick any two positive numbers to represent $r_1$ and $r_2$.

asnaseer · Answer

just give me any two positive real numbers that come to mind

anonymous · Answer

1 and 2
$$(x-1)(x-2)$$

asnaseer · Answer

perfect - now pick a complex number of the form (a+ib)
so you need to pick values for 'a' and 'b'

asnaseer · Answer

I am /assuming/ you know that $i=\sqrt{-1}$

asnaseer · Answer

are you stuck?

anonymous · Answer

4+2i, 4-2i

asnaseer · Answer

ok, so now we know your polynomial is of the form:$$f(x)=(x-1)(x-2)(x-(4+2i))(x-(4-2i))$$

asnaseer · Answer

if we multiply the last two factors out then, because they are complex conjugates of one another, we will end up with something that does NOT contain any complex numbers.

asnaseer · Answer

$$(x-(4+2i))(x-(4-2i))=x^2-8x+20$$

asnaseer · Answer

do you understand that last step?

anonymous · Answer

$$(x^2-3x+2)(x^2-8x+20)$$

anonymous · Answer

?

asnaseer · Answer

yes, and that can be expanded out to the final polynomial as follows:$$f(x)=(x-1)(x-2)(x^2-8x+20)=x^4-11x^3+46x^2-76x+40$$although I believe it should be ok to just leave it in the form of:$$f(x)=(x-1)(x-2)(x^2-8x+20)$$

asnaseer · Answer

NOTE: you can also multiply this final polynomial by any real number without affecting the roots.
so, for example, this is also a solution:$$f(x)=10(x-1)(x-2)(x^2-8x+20)$$

asnaseer · Answer

I hope that all made sense? :)

anonymous · Answer

Yes, thanks!

asnaseer · Answer

yw :)