Write a polynomial function with rational coefficients so that P(x)=0 has the given roots. #16. 2 + i and 1-√5 #15. 3i and √6
ick
i take it these are two separate problems right?
Yea
which one you want to do first? #15 is the easier of the two
Lets tackle that one first then.
ok
the conjugate of \(3i\) is \(-3i\) and the quadratic with those two roots is \[(x-3i)(x+3i)=x^2+9\] quick and easy
the conjugate of \(\sqrt6\) is \(-\sqrt6\) and the quadratic with those zeros is \[(x-\sqrt6)(x+\sqrt6)=x^2-6\]
your job is now to multiply out \[(x^2+9)(x^2-6)\] without making an algebra mistake you good for that one?
Eh..I'm stuck on multiplying the squares
\(x^2\times x^2=x^4\)
you get \[x^4-6x^2+9x^2-54\] or \[x^4+3x^2-54\]
ok for that one? the second one is harder
ya
you need a quadratic with \(2+i\) as a zero
there are several ways to find it i will show you an easy one, then a really really easy one but the really really easy one requires memorizing something
Thats fine
work backwards \[x=2+i\\ x-2=i\] then square both sides and get \[(x-2)^2=i^2\] or \[x^2-4x+4=-1\] add \(1\) and get \[x^2-4x+5\] as your quadratic
the really really easy way is to know that if \(a+bi\) is a zero, then the quadratic is \[x^2-2ax+(a^2+b^2)\] in your case you have \(2+i\) so \(a=2,b=1\) and the quadratic is \[x^2-2\times 2x+(2^2+1^2)=x^2-4x+5\]
as for \(1+\sqrt5\) without the \(i\) it is \[x^2-2x+(1^2-\sqrt5)^2=x^2-2x-4\]
your last job is to multiply out \[(x^2-4x+5)(x^2-2x-4)\] i would cheat
Okay here's another thing I'm having trouble w/ ... Finding all rational roots for P(x)=0 #26. P(x)=x^3-x^2-34x-56
cheat
does it say "find the rational roots" or "find all POSSIBLE rational roots"?
It just says "Find all rational roots"
open the link, there they are
It doesn't show me how it found the roots. Can you show me how to do one so I can do the rest?
P(x)=3x^3+x^2-8x-12
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