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

Sep 28, 2015

First multiply out and write in standard polynomial form to obtain

${x}^{3} + 2 {x}^{2} + 3 x - 4 = 0$

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 $f \left(x\right) = {x}^{3} + 2 {x}^{2} + 3 x - 4$, then we may for example use Newton's method of root finding in numerical analysis to find the first root by inspection.
Note that $f ' \left(x\right) = 3 {x}^{2} + 4 x + 3$
Starting with initial value ${x}_{0} = 0$ for example, we obtain that
${x}_{1} = {x}_{o} - f \frac{{x}_{o}}{f ' \left({x}_{0}\right)} = 0 - - \frac{4}{3} = \frac{4}{3}$
${x}_{2} = {x}_{1} - f \frac{{x}_{1}}{f ' \left({x}_{1}\right)} = \frac{4}{3} - f \frac{\frac{4}{3}}{f ' \left(\frac{4}{3}\right)} = \frac{4}{3} - \frac{\frac{160}{27}}{\frac{41}{3}} = \frac{332}{369} \cong 0 , 9$

Now since this is not an integer value, we cannot long divide $f \left(x\right)$ by $\left(x - 0 , 9\right)$ 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 :)