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Algebra/Algebra 2 Cumulative Review

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Question
The question says "Solve using your knowledge of inequalities, the zero product rule, chart and sign analysis, and domain intervals. Graph the solution set and write the solution in interval notation. Show work." I don't know where to begin. The inequality is:
x(x+3)(x-2)(3x+2)<0

Answer
The first thing to note is where each term is zero, the function is 0.
The function is f(x) = x(x+3)(x-2)(3x+2).

It is 0 at x=0, x+3=0, x-2=0, and 3x+2=0.
That is, at x=0, x=-3, x=2, and x=-2/3.

In order, the zeros of the function are at x=-3, x=-2/3, x=0, and x=2.
When the function is negative, those are the points we are trying to find.

The function will be negative where an odd number of terms are negative.
This is true when x is less than all the points { four negatives multiply to a positive }, x is between -2/3 and 0 { for two are positive and two are negative, so multiplying gives a positive }, and when 2<x { all four terms are positive, so the result is positive }.

Thus, we know that the function is positive when x<-3, -2/3<x<0, and 2<x.

At this point, I would put the function is Excel and see what it looks like.
Put the minimum x in at -4 and add 1/6 until 3 is gotten to.  I have done that for you below, so all you have to do is highlight, cut, and paste into Excel.

A set of points is here:
-4.000 240
-3.833 177.025463
-3.667 124.6666667
-3.500  81.8125
-3.333   47.40740741
-3.167   20.45138889
-3.000      9.32587E-14
-2.833  -14.83564815
-2.667  -24.88888889
-2.500  -30.9375
-2.333  -33.7037037
-2.167  -33.85416667
-2.000  -32
-1.833  -28.69675926
-1.667  -24.44444444
-1.500  -19.6875
-1.333  -14.81481481
-1.167  -10.15972222
-1.000  -6
-0.833  -2.55787037
-0.667  -1.84215E-14
-0.500     1.5625
-0.333   2.074074074
-0.167   1.534722222
0.000   1.79856E-14
0.167  -2.418981481
0.333  -5.555555556
0.500  -9.1875
0.667  -13.03703704
0.833  -16.77083333
1.000  -20
1.167  -22.28009259
1.333  -23.11111111
1.500  -21.9375
1.667  -18.14814815
1.833  -11.07638889
2.000  -1.06581E-13
2.167   15.8587963
2.333   37.33333333
2.500   65.3125
2.667   100.7407407
2.833   144.6180556
3.000   198

Highlight the points, then cut them, open up an Excel file, and paste them in.
Then Insert a Scatter Chart.  It can be seen that the function is only slightly positive between -2/3 and 0.

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Scott A Wilson

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