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

Question

Nathan Axel · Accepted Answer

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 :)

Genesis Allen · Answer

undefined To solve the equation $x(x^2 + 2x + 3) = 4$ by factoring, follow these steps:

1. Expand the expression $x(x^2 + 2x + 3)$ to get $x^3 + 2x^2 + 3x$.
2. Rewrite the equation as $x^3 + 2x^2 + 3x - 4 = 0$.
3. Try to factor the expression $x^3 + 2x^2 + 3x - 4$ into linear factors.
4. By inspection or using techniques like synthetic division or the rational root theorem, find a root (solution) of the equation.
5. Once you find one root, use polynomial long division or synthetic division to divide the polynomial by the corresponding linear factor to find the remaining quadratic factor.
6. Factor the quadratic factor (if possible) to find the other roots.
7. Verify the solutions by substituting them back into the original equation.

This process should yield the solutions to the equation $x(x^2 + 2x + 3) = 4$.