# SOLUTION: I am suppose to find the number of real zeros and the number of imaginary zeros for the polynomial. x^4+4x^3+5x^2+4x+4=f(X) The answers I got were 1 real zero and 3 imaginary

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: I am suppose to find the number of real zeros and the number of imaginary zeros for the polynomial. x^4+4x^3+5x^2+4x+4=f(X) The answers I got were 1 real zero and 3 imaginary      Log On

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 Click here to see ALL problems on Polynomials-and-rational-expressions Question 363587: I am suppose to find the number of real zeros and the number of imaginary zeros for the polynomial. x^4+4x^3+5x^2+4x+4=f(X) The answers I got were 1 real zero and 3 imaginary zeros?Answer by vasumathi(46)   (Show Source): You can put this solution on YOUR website!Solution: Let x = -2 .Plug in this in the given polynomial x^4+4x^3+5x^2+4x+4 (-2)^4 + 4 (-2)^3+5(-2)^2 +4(-2)+ 4=16-32+20-8+4=0 so x = -2 is a root and (x+2) is a factor Let us divide the given polynomial with (x+2) we get x^3+2x^2+x+2 again plug in x = -2 so (x+2) is a factor again and divide x^3+2x^2+x+2 by (x+2) we get x^2 + 1 after division and the roots of x^2 +1 = 0 are x = + i and x = -i so the roots are x = -2 repeated twice so there are two real roots -2 and -2 and there are two complex roots .They are i and -i