How do you solve #4x^3 + 3x^2 + x + 2 = 0#?

1 Answer
Apr 29, 2017

Answer:

#x=-1" "# or #" "x=1/8+-sqrt(31)/8i#

Explanation:

Given:

#4x^3+3x^2+x+2 = 0#

Note that:

#-4+3-1+2 = 0#

So we can deduce that #x=-1# is a solution and #(x+1)# a factor:

#0 = 4x^3+3x^2+x+2#

#color(white)(0) = (x+1)(4x^2-x+2)#

The remaining quadratic is in the form #ax^2+bx+c# with #a=4#, #b=-1# and #c=2#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-1))^2-4(color(blue)(4))(color(blue)(2)) = 1-32 = -31#

Since #Delta < 0# this quadratic has no real zeros.

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 #i# is the imaginary unit.