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
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-> 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
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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
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