Can someone help me with this? Precalc solving inequalities with one variable? The 3x is throwing me off

- anonymous

- jamiebookeater

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- anonymous

\[f(x)=(x-7)(3x+1)(x+4)\]

- anonymous

do you need to simplify?

- anonymous

You actually have 2 unknowns here. Is there more to this question?

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## More answers

- anonymous

No, you use a sign chart. @Brittni0605 And yes sorry it says *In the exercises 1-6 determine the x values that cause the polynomial functions to be (a)zero (b) positive and (c)negative @ChmE

- anonymous

I don't see a chart, ill let @ChmE help you.

- anonymous

Oh no its a sign chart that you make to solve it with lol but okay(:

- anonymous

so we set it equal to zero, then less than for negative, then greater than for positive. Ok.
Let's start with setting it equal to zero. Do not distribute, leave the equation as is. Go to each ( ) and think what will x need to be to make this zero. Only one needs to be zero for the whole thing to be zero. There should be 3 answers. I'll give you a sec.
Let me know if you didn't understand what I just said

- anonymous

I got 7, -1/3 and -4 ....but the -1/3 doesnt look right to me:/

- anonymous

Thats correct! So that solves our zero.

- anonymous

0

- anonymous

But here is where it is tricky. Say we use the number 0 which satisfies the first but not the last two. Lets see if we get a positive number.
0<(0-7)(3(0)+1)(0+4)
0<-7*1*4
0<-28
no.
So the answer for 0<.... is x<-4
because that is the lowest.
***I missed typed above this is for the positive****

- anonymous

Now what do you think the answer will be for the negative?

- anonymous

ummm im a little confused now :(

- anonymous

I don't know how to make a sign chart. I don't ever remember having to do that.

- anonymous

Umm well im not sure i think i kinda get it a little better? Im sorry this is taking long btw, just my teacher never helped me lol

- anonymous

Oh okay I got it now(:

- anonymous

WAITTTTTTT!!!!!!

- anonymous

Plug in zero

- anonymous

o-o? ok lol

- anonymous

0>-7*1*4 negative. OMG. Now I need to rethink some things. Im so sorry. Lets try to make a sign chart.

- anonymous

lol okay|dw:1352178275962:dw| would i put the 1/3 on it though? i dont know lol Its bothering me since its a fraction lol

- anonymous

|dw:1352178252794:dw|

- anonymous

so when it is negative is x<-4, -1/37

- anonymous

okay thats easier for me to understand(:

- anonymous

How to solve problems like this.
Set equal to zero. Solve each term for the zeros.
Put those points on a number line. Pick easy numbers between those points. Test to see if it is positive or negative.
Follow the number line like a map to get the solutions for neg and pos.

- anonymous

Sorry for leading you estray but sometimes working through a problem and finding mistakes helps us understand a little more about the correct way in terms of what is not right.
I think a sign graph is very helpful, I'd continue to use that.
Do you think you could do a problem like this by yourself now?
Thanks for refreshing me as well. It took a while but we got it !!

- anonymous

Umm yes i think so(: ill practice a little more before i goto sleep lol but thanks a bunch <3

- anonymous

I'm in differential equations right now which is Calc 4, and I almost got this question wrong, humbling. Good luck and check your work if you have time, it's very beneficial.

- anonymous

Oh wow but alright(: thanks again <3

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