Edward C. answered • 04/02/20
Tutor
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Caltech Grad for math tutoring: Algebra through Calculus
There are several different ways to do this problem. I don't know what methods you have learned.
One way is to factor m(x) to see what the zeros are.
Another way is to test the different solutions (-7,-4,4, and 7) to see which ones make m(x) = 0.
Yet another way is to multiply out each of the factored forms in the choices to see which one is equal to m(x).
Edward C.
tutor
Well if you multiply out the last one you get -(x^2 + 3x -28) = -x^2 - 3x + 28, which isn't equal to m(x), so that can't be right.
Another way to do the problem is to draw a graph of m(x) and see where it crosses the x-axis. I'll get you started - when x = 0 then m(x) = -28, so the point (0,-28) is on the graph. When x = 1 then m(x) = -18 so the point (1,-18) is on the graph. See if you can plot a few more points on the graph and see where it crosses the x-axis.
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04/02/20
Jalixa A.
I don't understand I really don't know much about graphing
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04/02/20
Edward C.
tutor
Ok well if you don't know how to graph it, you can't factor it, and you can't multiply out the factored forms, then you're going to have to test each value to see which ones make m(x)=0. I'll do -7 for you. When x = -7, then m(-7) = -(-7)^2 + 11*(-7) -28 = -49 - 77 - 28 = -154. This is not equal to 0, so -7 is not a zero of m(x). See it you can evaluate m(-4), m(4) and m(7) to see which ones make m(x) = 0.
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04/02/20
Jalixa A.
i think its the last one04/02/20