P(x)=(x+2)^2(x+5)(3-x)
Look at the last one (3-x)
which can be written as (-x+3)
so P(x)=(x+2)^2(x+5)(-x+3)
so therefore you will have a negative on your leading term if you expanded this out.
-x^4-6x^3-4x^2+52x+60
2007-10-13 19:08:52
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answer #1
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answered by azianshrimp 2
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P(x) = (x + 2)^2 (x + 5)(3 - x)
Actually, I don't think the polynomial has a positive leading coefficient, because (3 - x) can be express as (-1)(x - 3). With that said, we have
P(x) = (-1) (x + 2)^2 (x + 5)(x - 3)
Which, when multiplied out, will have a negative leading coefficient.
2007-10-13 19:13:02
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answer #2
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answered by Puggy 7
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It has a negative leading coefficient, not positive. The way to tell is to look at all the x coefficients and multiply them together. The (x+2) is (1x+2), and it's squared so count it twice, and the (x+5) is (1x+5) and the tricky one is the (3-x) at the end is really (-1x+3). So multiply all the x coefficients together 1*1*1*(-1) and you get -1 as the lead coefficient.
Here's another example: P(x)=(2x-3)(-3x+1)(4x+5)
The lead coefficient is 2*(-3)*4=(-24)
2007-10-13 19:13:42
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answer #3
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answered by dogwood_lock 5
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If you multiply the 4 terms, you'll see that the leading coefficient
( of x^4) is negative.
1 times 1 times 1 times NEGATIVE 1
2007-10-13 19:09:33
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answer #4
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answered by Hk 4
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my graphing calculator is out of batteries, but i'm guessing it shows it that way because the 3-x is really -x+3, which would make it a negative.
2007-10-13 19:11:43
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answer #5
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answered by pinkchck3112 1
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your last factor is 3-x, which gives you a NEGATIVE leading coefficient. you get
-x^4 - 6x³ + 3x² + 52x + 60
and it's not cubic, it's quartic, 4th degree.
2007-10-13 19:15:04
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answer #6
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answered by Philo 7
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If you multiply that polynomial out it will have a leading (-)
2007-10-13 19:12:58
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answer #7
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answered by golffan137 3
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the term with the greatest power is -x^4
so it goes "down" at -infinity
2007-10-13 19:10:54
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answer #8
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answered by Theta40 7
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