mmbuckaroos
Use the Factor Theorem to determine a polynomial equation, of lowest degree, that has only the indicated roots:
0 is a root of multiplicity 5, 2 is a double root
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Mertsj
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x^5(x-2)^2=0
anonymous
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ok i cant read. start with \[x^5(x-2)^2\] and multiply out
mmbuckaroos
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so I would get \[x ^{5}+x ^{2}-4x+4\]
anonymous
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no i don't think so
anonymous
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\[x^5(x-2)^2=x^5(x-2)(x-2)=...\]
mmbuckaroos
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where does the extra (x-2)^2 come from then that you have after the x^5?
mmbuckaroos
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oh nvm.... so what am I multiplying out then? the (x-2)^2 or?
Mertsj
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\[x^5(x-)(x-2)=x^5(x^2-4x+4)=x^7-4x^6+4x^5\]
mmbuckaroos
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Oh ok so I just needed to continue multiplying to completly get rid of the parenthisis cool thank you...
so what if it is the same question but it says 1/3 is a double root and -2 is a double root? do I set it up like this:
f(x)=(x-1/3)^2(x+2)^2
Mertsj
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Yes
mmbuckaroos
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awesome thanks so much for the help!
Mertsj
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yw
mmbuckaroos
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Uh so I actually have one more silly question how do I get rid of the fraction to make that problem easier?
mmbuckaroos
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NVM I figured it out:) thanks
anonymous
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i wouldn't
anonymous
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oh looks like you got it right? start with \((3x-1)^2\)
mmbuckaroos
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Thanks:)