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Problem : numbers 1 and 2 are roots of polynomial P. prove P is divisible by (x-1)(x-2) given solution: since 1 is a root, P= (x-1)Q for some Q. substituting 2 for x in the equality (WHY?) we find that 2 is a root of Q (HOW?). so Q = (x-2) . R for some R. therfore P = (x-1)(x-2)R.

substituting 2 for x i get P=Q but why is 2 a root of Q? is there a reason for substituting 2.

Aditya Chintalapati
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  • No, you get $\ 0 = P(2) = (2-1)Q(2) = 0,$ so $,Q(2) = 0,$ so $,Q(x) = (x-2)R(x)\ $ therefore $\ P(x) = (x-1)Q(x) = (x-1)(x-2)R(x).\ $ Generally this works f the difference of the roots is cancellable, see the BiFactor Theorem. – Bill Dubuque Mar 22 '15 at 20:34

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Note that, since $2$ is a root of $P$, this means that $P\left(2\right)=0$. Since $P=\left(x-1\right)Q$, we then have that $0=P\left(2\right)=Q\left(2\right)$, meaning that $Q\left(2\right)=0$, which, by definition, means that $2$ is a root of $Q$.

Brian
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