# Thread: Solving Equations by factoring

1. ## Solving Equations by factoring

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

2. ## Re: Solving Equations by factoring

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?

3. ## Re: Solving Equations by factoring

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?

4. ## 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

5. ## 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.

6. ## Re: Solving Equations by factoring

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 $\displaystyle 4z^3= 4z^2+ 24z$. 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 $\displaystyle 4z(z- 3)(z+ 2)= 0$.