Let `P(x)` and `Q(x)` be polynomials. Show that `P(x)` can be divided by `Q(x)` with a remainder `R(x) -= 0` , and `Q(x)` can be divided by `P(x)` with a remainder `r(x)-=0` if and only if `P(x)=c*Q(x)` , where c is a constant.

Prove using the notion of the polynomial degree.

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If `P ( x ) ` is divisible by `Q ( x ) ` without a remainder (or with zero remainder, which is the same), then `P ( x ) = A ( x ) Q ( x ) , ` where `A ( x ) ` is a polynomial.

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If `P ( x ) ` is divisible by `Q ( x ) ` without a remainder (or with zero remainder, which is the same), then `P ( x ) = A ( x ) Q ( x ) , ` where `A ( x ) ` is a polynomial.

Similarly, if `Q ( x ) , ` in turn, is divisible by `Q ( x ) ` with zero remainder, then `Q ( x ) = B ( x ) P ( x ) , ` where `B ( x ) ` is a polynomial.

Substitute the expression `Q ( x ) = B ( x ) P ( x ) ` into the expression `P ( x ) = A ( x ) Q ( x ) :`

`P ( x ) = A ( x ) Q ( x ) = A ( x ) ( B ( x ) P ( x ) ) = ( A ( x ) B ( x ) ) P ( x ) .`

Here we used the associativity of polynomial multiplication (which can be derived from the associativity of number multiplication).

Now we have the equality `P( x ) = C ( x ) P ( x ) , ` where `C ( x ) = A ( x ) B ( x )` is again a polynomial. Use the notion of polynomial degree: if the degree of `P` is `n ` and the degree of `C ` is `m , ` then the degree of `C P ` is `m + n . ` Which immediately gives `n = m + n ; ` in other words, `m = 0 , ` or `C ( x ) ` is a (nonzero) constant.

The properties of the degree also tell us that the degree of `A ` is zero and the degree of `B ` is zero, which is what we want.

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