How do you solve #4x^3 + 3x^2 + x + 2 = 0#?
1 Answer
Apr 29, 2017
Explanation:
Given:
#4x^3+3x^2+x+2 = 0#
Note that:
#-4+3-1+2 = 0#
So we can deduce that
#0 = 4x^3+3x^2+x+2#
#color(white)(0) = (x+1)(4x^2-x+2)#
The remaining quadratic is in the form
This has discriminant
#Delta = b^2-4ac = (color(blue)(-1))^2-4(color(blue)(4))(color(blue)(2)) = 1-32 = -31#
Since
We can find Complex zeros for it using the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-b+-sqrt(Delta))/(2a)#
#color(white)(x) = (1+-sqrt(-31))/8#
#color(white)(x) = 1/8+-sqrt(31)/8i#
where