# If P(x) is a cubic polynomial such that P(0)=0 P(2)=-4 and P(x)is positive only when x> 4 find P(x)

If P(x) is a cubic polynomial such that P(0)=0 P(2)=-4 and P(x)is positive only when x> 4 find P(x)
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phravincegrln2
eventually given that numbers greater than 4 would turn out to be positive, P cubed would have to be positive... and you would have to subtract a multiple of P squared... so far the equation would look like ${P}^{3}-{P}^{2}$ Now just look for a number that would work there (where _ is) with 2 as a given for P and -4 as the answer... ${2}^{3}=8$ $8-\left(\ast {2}^{2}\right)=-4$ $8-\left(\ast 4\right)=-4$
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Intomathymnma
since P(0)=0 and P(x) is positive only when x > 4 then (0,0) must be a double root with the function staying below the x axis.
since P(x) > 0 when x > 4 then a root must exist at x=4
now you have $y=a{x}^{2}\left(x-4\right)$
use the given point (2,-4) to find a
$-4=a{2}^{2}\left(2-4\right)$
-4=a (-8)
a=1/2
therefore cubic polynomial is $y=\left(1/2\right){x}^{2}\left(x-4\right)$