SOLUTION: Find a polynomial f(x)of degree 3 that has the indicated zeros and satisfies the given condition. -2i,2i,3; f(1)=20 (I get scared when I see imaginary numbers, because i have

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Find a polynomial f(x)of degree 3 that has the indicated zeros and satisfies the given condition. -2i,2i,3; f(1)=20 (I get scared when I see imaginary numbers, because i have      Log On


   

Question 589115: Find a polynomial f(x)of degree 3 that has the indicated zeros and satisfies the given condition.
-2i,2i,3; f(1)=20
(I get scared when I see imaginary numbers, because i have no clue what to do with them.)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x+=+-2i+
+x+=+2i+
+x+=+3+
+f%281%29+=+20+
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Note that +i%5E2+=+-1+
+%28+x%5E2+%2B+4+%29%2A%28+x+-+3+%29+=+0+
+x%5E3+%2B+4x+-+3x%5E2+-+12+=+0+
+k%2A%28+x%5E3+-+3x%5E2+%2B+4x+-+12+%29+=+k%2A0+
+y+=++k%2A%28+x%5E3+-+3x%5E2+%2B+4x+-+12+%29+
+y%281%29+=+k%2A%28+1+-+3+%2B+4+-+12+%29+
+y%281%29+=-10k+
+-10k+=+20+
+k+=+-2+
+y+=++-2%2A%28+x%5E3+-+3x%5E2+%2B+4x+-+12+%29+
+y+=+-2x%5E3+%2B+6x%5E2+-8x+%2B+24+
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I think of imaginaries as a way to get off
the real number line by rotating 90 degrees
each time you multiply a real by +i+
So, if you multiply twice, that puts you
back on the real number line going
180 degrees the other way, or
multiplying by +-1+