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anonymous
 5 years ago
How do you use the Factor Theorem to show that the second polynomial is a factor of the first polynomial?
2x^3  5x^2 + 6x  2, x  1/2
anonymous
 5 years ago
How do you use the Factor Theorem to show that the second polynomial is a factor of the first polynomial? 2x^3  5x^2 + 6x  2, x  1/2

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if its got no remainder its a factor

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.5 2 5 6 2 0 1 2 2  2 4 4 0 < remainder 0, its a factor

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant really say that I know what a factor therom is tho ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factor theorem sez that if r is a zero of a polynomial \[p(x)\] then \[p(x)=(xr)q(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if you know \[p(\frac{1}{2})=0\] then you know \[p(x)=(x\frac{1}{2})q(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if f(1/2) is equal to zero then x1/2 is a factot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0more easily written as \[p(x)=( 2x1)q(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre as usual did all the work, so you even know what \[q(x)\] is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is \[2x^24x+4=2(x^2+2x+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your polynomial is \[p(x)=(2x1)(x^22x+2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops typo in other answer . it is 2x not +2x sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0115 to go and then i quit!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\(\left(\begin{array}ca&b&c\\d&e&f\\g&h&i \end{array} \right)\)
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