13.) P(x) = x^4 - 5, Q(x)=4x+1
x^4 - 5 = 4x + 1
x^4 - 4x - 6 = 0
I can't find a way to factor this. Are you sure the equations you gave are right?
14.) P(x)=x^3 - 5x +3, Q(x) = -x^3 + 5x +3
x^3 - 5x + 3 = -x^3 + 5x + 3
2x^3 - 10x = 0
2x(x^2 - 5) = 0
x = 0
x^2 - 5 = 0
x^2 = 5
x = +- sqrt 5 = +- 2.24
2007-07-17 09:17:12
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answer #1
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answered by Mathematica 7
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the p(x) should be equal to p(x)
and the Q(x) should be equal to Q(x),
so,
2)x^4-5 = x^3-5x+3 and 1)4x+1 = -x^3+5x+3
so,
we should draw the functions of 1) and 2)
and see where they have equal points.
2007-07-17 09:08:12
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answer #2
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answered by shahab_phoenix 2
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P(x) = x^4 - 5
(x^2-sqrt(5))(x^2+sqrt(5)=0
(x^2-sqrt(5)) = 0 --> x = +/- 5^1/4 = +/- 1.50
(x^2+sqrt(5) =0 -->x = +/- 1.50i
Q(x)=4x+1
x = -1/4
P(x)=x^3 - 5x +3
x = 1.83, 0.66, - 2.49
Q(x) = -x^3 + 5x +3
x= -1.83, -.66, 2.49
2007-07-17 09:28:49
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answer #3
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answered by ironduke8159 7
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look's like calculous, good luck! haha the guy who discovered calculous did it in 2 days, figured out all the rules and everything, I heard his head was bigger than a pumpkin, well, not bigger than the pumpkin you see at the fair, could you imagine, that'd be a huge head
2007-07-17 09:01:01
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answer #4
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answered by theystolemyotherone 1
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Wish dey had dis when i was in school
2007-07-17 08:58:24
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answer #5
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answered by AntG 3
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Wow now thats a math problem . . . . .
2007-07-17 08:58:50
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answer #6
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answered by MiiSzFRESHAlOt 2
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4.20
2007-07-17 08:58:05
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answer #7
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answered by Anonymous
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