Solving Equations by factoring

• September 24th 2013, 01:50 AM
asilvester635
Solving Equations by factoring
I'm stuck in the highlighted part of the problem....
• September 24th 2013, 03:06 AM
Plato
Re: Solving Equations by factoring
Quote:

Originally Posted by asilvester635
I'm stuck in the highlighted part of the problem....

"I'm stuck" what do you mean by that.
As written there is nothing else to be done with that question.

Did you fail to post the entire question?
• September 24th 2013, 05:23 AM
votan
Re: Solving Equations by factoring
Quote:

Originally Posted by asilvester635
I'm stuck in the highlighted part of the problem....

Watch the sign of 24z in the third line. Why you omitted = 0?
• September 24th 2013, 05:29 AM
sakonpure6
Re: 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
• September 24th 2013, 05:30 AM
HallsofIvy
Re: 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.
• September 24th 2013, 06:08 AM
votan
Re: Solving Equations by factoring
Quote:

Originally Posted by HallsofIvy
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.

asilvester635, your first line was $4z^3= 4z^2+ 24z$. Your next line should be $-4z^3+ 4z^2+ 24z= 0$ (you have subtracted $4z^3$ from both sides). But then you have " $4z^3- 4z^2+ 24z$". 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 $4z^3- 4z^2- 24z= 0$. Obviously both "4" and "z" divide each term so you can factor 4z out: $4z(z^2- z- 6)= 0$. (3)(2)= 6 and their difference is 1 so we can further factor as $4z(z- 3)(z+ 2)= 0$.