can someone check this for me please? :3

Question

aripotta · Answer

attachment

anonymous · Answer

3)
x^4 - x^3 + 7x^2 - 9x - 18 = 0

Rational root theorem: -18 , -9 , -6 , -3 , -2 , -1 , 1 , 2 , 3 , 6 , 9 , 18

x = -1 , 2 are solutions

(x + 1) * (x - 2) = x^2 - x - 2

(x^4 - x^3 + 7x^2 - 9x - 18) / (x^2 - x - 2) =>>

x^4 / x^2 = x^2
x^2 * (x^2 - x - 2) = x^4 - x^3 - 2x^2
x^4 - x^3 + 7x^2 - x^4 + x^3 + 2x^2 = 9x^2

9x^2 / x^2 = 9
9 * (x^2 - x - 2) = 9x^2 - 9x - 18
9x^2 - 9x - 18 - 9x^2 + 9x + 18 = 0

(x^2 - x - 2) * (x^2 + 9) = 0

x = -3i , 3i

x = -1 , 2 , -3i , 3i

aripotta · Answer

is there a such thing as -3i?? i thought it could only be positive

anonymous · Answer

yeah there is:P

aripotta · Answer

:s ok

anonymous · Answer

3i is a complex number, and yes there is such thing as ''i''

aripotta · Answer

i knew there's a such thing as i, i just thought that i couldn't be negative