Question

How do you solve `x (x^2 + 2x + 3)=4` by factoring?

Answer 1

First multiply out and write in standard polynomial form to obtain

`x^3+2x^2+3x-4=0`

This is now a 3rd degree cubic equation and has 3 roots.

We may use the remainder theorem, which involves first obtaining a single root by inspection, and then long dividing the corresponding factor into the cubic and then factorizing the resultant quadratic by factors.

Now if we let `f(x)=x^3+2x^2+3x-4`, then we may for example use Newton's method of root finding in numerical analysis to find the first root by inspection.
Note that `f'(x)=3x^2+4x+3`
Starting with initial value `x_0=0` for example, we obtain that
`x_1=x_o-f(x_o)/(f'(x_0))=0--4/3=4/3`
`x_2=x_1-f(x_1)/(f'(x_1))=4/3-f(4/3)/(f'(4/3))=4/3-(160/27)/(41/3)=332/369~=0,9`

Now since this is not an integer value, we cannot long divide `f(x)` by `(x - 0,9)` as we will battle to obtain factors.

So the best way will be to continue using Newton's method for the other 2 roots as well, and then rewrite the original polynomial in factor form like that.
I leave the details as an exercise :)