I'm stuck in the highlighted part of the problem....

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- Sep 24th 2013, 01:50 AMasilvester635Solving Equations by factoring
I'm stuck in the highlighted part of the problem....

- Sep 24th 2013, 03:06 AMPlatoRe: Solving Equations by factoring
- Sep 24th 2013, 05:23 AMvotanRe: Solving Equations by factoring
- Sep 24th 2013, 05:29 AMsakonpure6Re: Solving Equations by factoring
I believe that the you can not factor 4z(z^2 - z + 6) any further... Also you are missing a negative sign before 4z

- Sep 24th 2013, 05:30 AMHallsofIvyRe: Solving Equations by factoring
There is no reason to believe there

**was**an "= 0" to be omitted! The difficulty is that asilvester635 has not told us**what**the problem is or what he is trying to do. - Sep 24th 2013, 06:08 AMvotanRe: Solving Equations by factoring
- Sep 24th 2013, 06:28 AMHallsofIvyRe: Solving Equations by factoring
I did not see that! Thank you.

asilvester635, your first line was $\displaystyle 4z^3= 4z^2+ 24z$. Your next line**should**be $\displaystyle -4z^3+ 4z^2+ 24z= 0$ (you have subtracted $\displaystyle 4z^3$ from both sides). But then you have "$\displaystyle 4z^3- 4z^2+ 24z$". My guess is that you intended to multiply by -1 but, again, you must have "= 0" and you did not multiply the "24z" by -1. You should have $\displaystyle 4z^3- 4z^2- 24z= 0$. Obviously both "4" and "z" divide each term so you can factor 4z out: $\displaystyle 4z(z^2- z- 6)= 0$. (3)(2)= 6 and their difference is 1 so we can further factor as $\displaystyle 4z(z- 3)(z+ 2)= 0$.

**Now**what? Remember the basic reason for factoring: "if ab= 0 then either a= 0 or b= 0".