# SOLUTION: P(x)=x^5+x^3+8x^2+8, Find ALL zeros of P(x).

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 Click here to see ALL problems on Rational-functions Question 631907: P(x)=x^5+x^3+8x^2+8, Find ALL zeros of P(x).Answer by lwsshak3(6760)   (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). ** 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) ============================ ...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)