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

1 Answer
Bill K.
May 22, 2015

Distribute: #x^3+2x^2+3x=4#

Rearrange: #x^3+2x^2+3x-4=0#

It would be great if we could now factor the left-hand side. Unfortunately, the roots of this are an irrational real number as well as two complex numbers with irrational real and imaginary parts.

Using a computer or calculator gives approximate roots:

#x\approx 0.776045#, #x\approx -1.38802\pm 1.79659i#

There is a method/formula for finding exact expressions for these roots (involving roots that are irrational), but it's not pleasant:
general method/formula for cubic roots on Wikipedia

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