SOLUTION: 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

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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