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Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (mertsj):

x^5(x-2)^2=0

OpenStudy (anonymous):

ok i cant read. start with \[x^5(x-2)^2\] and multiply out

OpenStudy (anonymous):

so I would get \[x ^{5}+x ^{2}-4x+4\]

OpenStudy (anonymous):

no i don't think so

OpenStudy (anonymous):

\[x^5(x-2)^2=x^5(x-2)(x-2)=...\]

OpenStudy (anonymous):

where does the extra (x-2)^2 come from then that you have after the x^5?

OpenStudy (anonymous):

oh nvm.... so what am I multiplying out then? the (x-2)^2 or?

OpenStudy (mertsj):

\[x^5(x-)(x-2)=x^5(x^2-4x+4)=x^7-4x^6+4x^5\]

OpenStudy (anonymous):

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

OpenStudy (mertsj):

Yes

OpenStudy (anonymous):

awesome thanks so much for the help!

OpenStudy (mertsj):

yw

OpenStudy (anonymous):

Uh so I actually have one more silly question how do I get rid of the fraction to make that problem easier?

OpenStudy (anonymous):

NVM I figured it out:) thanks

OpenStudy (anonymous):

i wouldn't

OpenStudy (anonymous):

oh looks like you got it right? start with \((3x-1)^2\)

OpenStudy (anonymous):

Thanks:)

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