Question

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

Answer 1

`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.