This is a really weird question since the first root is neither fractional nor exact. Which means that you can't really 'acquire' the complex roots. Are you sure that it is 2x^3 + 2x^2 and not 2x^3 - 2x^2?
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Saad Mughal
Aly Emran
Umm i forgot the question 
maybe this is 2x^3-2x^2 but how to solve this cubic equation :S
maybe this is 2x^3-2x^2 but how to solve this cubic equation :S
Saad Mughal
Find first root using trial and error, i.e. put arbitrary values for x
Put x = -1,
-2 - 2 - 1 + 5 = 0
Therefore, first root => x = -1
(x+1) (ax^2+bx+c) = 2x^3 - 2x^2 + x + 5
Comparing x^3,
a = 2,
Comparing x^2,
a + b = -2
b = -4
Comparing constants,
c = 5
Therefore 2x^3-2x^2+x+5 = (x+1)(2x^2 - 4x + 5)
Considering the equation 2x^2 - 4x + 5 = 0
b^2 - 4ac = 16 - 40 = -24 = 4.89i (since i = root -1)
Put x = -1,
-2 - 2 - 1 + 5 = 0
Therefore, first root => x = -1
(x+1) (ax^2+bx+c) = 2x^3 - 2x^2 + x + 5
Comparing x^3,
a = 2,
Comparing x^2,
a + b = -2
b = -4
Comparing constants,
c = 5
Therefore 2x^3-2x^2+x+5 = (x+1)(2x^2 - 4x + 5)
Considering the equation 2x^2 - 4x + 5 = 0
b^2 - 4ac = 16 - 40 = -24 = 4.89i (since i = root -1)
Saad Mughal
Now, using quadratic formula,
x = [-(-4) +/- (4.89i)]/4
x = 1 + 1.22i
or
x = 1 - 1.22i
x = [-(-4) +/- (4.89i)]/4
x = 1 + 1.22i
or
x = 1 - 1.22i
Aly Emran
ahan thanks soo much btw i got that ans thanks anyway i appreciate ur help
thanks anyway i appreciate ur help