# Solve cubic with no rational zeros

x^3-8x+10=0

## The Attempt at a Solution

By dividing by x-1 (1 is a factor of the solution), I got (x-1)(x2+x-7)+3+10=0

which equals x^3-8x+20=0

I think the solution would be imaginary but I used a graphing calculator and there is one solution but I don't know how to calculate it.

Last edited by a moderator:

Mark44
Mentor

## Homework Statement

x^3-8x+10=0
What is the complete problem statement?

## The Attempt at a Solution

By dividing by x-1 (1 is a factor of the solution), I got (x-1)(x2+x-7)+3+10=0
x - 1 is NOT a factor of x^3 - 8x + 10.
which equals x^3-8x+20=0

I think the solution would be imaginary but I used a graphing calculator and there is one solution but I don't know how to calculate it.

We can't help you if we don't know what it is that you're supposed to do. That's where the complete problem statement would be useful information.

LCKurtz
Homework Helper
Gold Member

x^3-8x+10=0

## The Attempt at a Solution

By dividing by x-1 (1 is a factor of the solution), I got (x-1)(x2+x-7)+3+10=0

which equals x^3-8x+20=0

I think the solution would be imaginary but I used a graphing calculator and there is one solution but I don't know how to calculate it.

Unless you know how to solve a cubic by radicals, you won't be finding the exact answer for the real root. Maple gives it as:

-(1/3)*(135+3*sqrt(489))^(1/3) - 8/(135+3*sqrt(489))^(1/3)

and a decimal approximation to that is -3.318628218.