Question 631907: P(x)=x^5+x^3+8x^2+8, Find ALL zeros of P(x).
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! P(x)=x^5+x^3+8x^2+8, Find ALL zeros of P(x).
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Use rational roots theorem to solve:
...0...|.....1.......0.......1........8......0........8
...1..|......1.......1.......2.......10....10.....18 (1 is upper limit, all numbers>0)
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...0...|.....1.......0.......1.......8......0.........8
.-1...|.....1.....-1.......2.......6.....-6........2
.-2...|.....1.....-2.......3.......2.....-4........0 (-2 is a root)
.-3...|.....1.....-3.......8....-16.....48.....-144 (-3 is lower limit, numbers alternate in sign)
P(x)=(x+2)(x^4-2x^3+3x^2+2x-4)
Function has only one real zero=-2.
Other 4 zeros are 2 pairs of non-real or imaginary roots. (complex conjugates)
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