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Precalc help please???:( Solving inequalities with one variable? Sign chart? *In exercises 1-6, determine the x values that cause the polynomial function to be (a)zero (b)positive (c)negative.

Mathematics
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\[f()=(x+2)(x+1)(x-5)\]
sorry it should be f(x)
First you figure out when f(x) = 0, which occurs when (x+2) =0 or (x+1) =0 or (x-5) =0. That is pretty easy, x=-2,-1,5. Your number line is -----(-2)-----(-1)------(5)----- Pick a number left of (-2) say (-5) and input it in as f(-5)= -120 therefore The part left of (-2) is negative. The next number is between (-2) and (-1) and is therefore something like (-1.5). f(-1.5) = something positive because when you put it into (x+2) you get a positive number, and into the other two parts of the equation a negative number returns. (1)(-1)(-1) = 1. Therefore that part is positive. Doing so for the other spaces (-1) to (5) and (5) to infinity results in a negative for (-1) to (5) and a positive for (5) to infinity. Therefore: ---negative--(-2)---positive---(-1)----negative----(5)-----positive----

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okay i understand that but im unsure of what to put for the answers, because in the back of the book (answers) their like this, (b) -25 (c) x<-2 or -1
That's what I get for not fully reading the question. They just want to know when that f(x) is positive, a zero, or negative. You use the sign chart to determine when it is negative and positive. So for (a) you wish to find the zeros which are at (-2), (-1), and (5). (b) however asks for when the equation is positive. It just wants the ranges. So the sign chart says that the equation is positive from (-2) to (-1) or x is greater than (-2) but less than (-1). Therefore, -2
omg thankyou so much i wouldve failed my test tomorrow lol my teacher was no help <333

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