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| Question 887448:  Hi,
 I have been having severe trouble with these polynomial/rational functions and would like help with this problem:
 P(x) = 4x^4+4x^3+5x^2+4x+1 --> Directions: Find all the zeros of the polynomial.
 --thank you to whomever decides to help me :)
 Answer by nerdybill(7384)
      (Show Source): 
You can put this solution on YOUR website! Because: P(x) = 4x^4+4x^3+5x^2+4x+1
 is a polynomial of degree 4, you must use synthetic division to reduce to a quadratic.
 .
 Possible zeros:
 1/4 = (+-) (1)/(1,2,4)
 possible zeros: +- 1, 1/2, 1/4
 try x=-1/2:
 -.5 | 4 4  5  4  1
 -2 -1 -2 -1
 ---------------
 4 2  4  2  0
 .
 so, now you have:
 P(x) = (x+1/2)(4x^3+2x^2+4x+2)
 .
 possible zeros are (+-) factors of 2/4
 try x = -1/2
 -.5 | 4  2 4  2
 -2 0 -2
 -------------
 4  0 4  0
 .
 so, now we have:
 P(x) = (x+1/2)(x+1/2)(4x^2+4)
 P(x) = 4(x+1/2)(x+1/2)(x^2+1)
 setting
 x^2+1 =0
 x^2 = -1
 x = {-i, i}
 .
 x = {-i, i, -1/2}  (two complex zeros and one real zero)
 
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