Please help medal and fan will be given. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 2, -4, and 1 + 3i
f(x) = x4 - 2x2 + 36x - 80 f(x) = x4 - 3x3 + 6x2 - 18x + 80 f(x) = x4 - 9x2 + 36x - 80 f(x) = x4 - 3x3 - 6x2 + 18x - 80 these are the answer choices
the first part is easiest you know if the zeros are \(2\) and \(-4\) then two factors are \[(x-2)(x+4)\] then you have to find the quadratic whose zeros are \(1\pm3i\)
there are a couple ways to do this
yeah I did the first part and got x^2+2x-8 ...... not sure what to do with the second half
one way is to work backwards \[x=1+3i\\ x-1=3i\\ (x-1)^2=(3i)^2=-9\\ x^2-2x+1=-9\\ x^2-2x+10=0\]
so your quadratic is \(x^2-2x+10\)
okay so now I multiply the one I got by the one you got?
the snap way is to memorize that if \(a+bi\) is a zero, the quadratic is \[x^2-2ax+(a^2+b^2)\] in your case \(a=1,b=3\) you get \[x^2-2x+1^2+3^2=x^2-2x+10\]
your last job is to multiply \[(x-2)(x+4)(x^2-2x+10)\] without making an algebra mistake i would cheat
how would you cheat?
Thank you! Can you help me with one more problem?
Using the given zero, find all other zeros of f(x). -2i is a zero of f(x) = x4 - 45x2 - 196 2i, 14i, -14i 2i, 7, -7 2i, 14, -14 2i, 7i, -7i
Hey I just need help with this problem down here. I have no idea.
you know that \(-2i\) is a zero and therefore so is \(2i\) one factor is \((x-2i)(x+2i)=x^2+4\)
so \[x^4 - 45x^2 - 196 =(x^2+4)(\text{some quadratic)}\] and your job is to find the quadratic
is it (x^2+49)?so the answer would be D?
not too hard for this one the first term has to be \(x^2\) and the last term has to be \(49\)
yeah that is what i get
okay awesome you are the best!
course you can always solve \[ x^4 - 45x^2 - 196 \]by treating it as \[z^2-45z-196=0\] where \(z=x^2\) and once you have \(z\) you can solve for \(x\)
thanks!
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