Given that b, c and d are non-zero integers together with q and p we can deduce that the expression of the right hand side is an integer for all allowed values when . Hence, the left hand side of the equation, , is also a whole number. It is also known that the ratio p and q have no common factors. Therefore the ratio as well as are in lowest terms as is multiple of p. Similarly to part (b), we argue that if q weren’t a factor of a then for values in which the division of does not give an integer, the left hand side expression would not be a whole number, contradicting the equation.
Hence, it follows that q must be dividing a to get an integer which multiplied by gives us the whole number value of the expression in the right. Thus it is proved that q is a factor of a. e) Generalize results (b) and (d) to prove the following result: Rational-zero theorem Suppose all the coefficients of the polynomial function are integers with . Let be a rational number in its lowest terms. If is a zero of F, the p is a factor of and q is a factor of . For the polynomial at we have:
Which is equal to: Multiplying through by qn: From here we can proceed with our generalization for both cases, firstly for that of the p being a factor of : Factorizing by p in LHS: Dividing by p: Following the same line of reasoning as that encountered in part (b) into our generalization we can deduce that, because the right hand side term is an integer and, as qn and p have no common factors the only way to obtain this result is if p divides a0, therefore p must be a factor of a0 when is a zero of F.
From the equation we can see that if a0 takes a zero value, p would also be 0, and, as the root has the form of , it would be calculated to be at commonly known result which explains under this perspective why the y-intercept of the curve is found at the origin when the last term of the polynomial has a null value. For our proof of q is a factor of we have: Subtracting : Factorizing the LHS by q: Dividing by q: We arrive to a generalized form of the expression treated in part (d).
As we saw before the right hand side results in an integer, therefore, the left hand side is also a whole number. Because p and q have no common factors, the same is true for p and qn, therefore their division would in most cases, as outlined before, result in a non-integer quantity. Therefore the only way to obtain the result is if p divides an, hence p must be a factor of an when is a zero of F and . The hypothetical case in which an = 0 can be considered for the sake of clarity.
Firstly, in terms of the roots of the polynomial we deduce that if an = 0 then q = 0. However, when making F a zero we realize that we are encountered with division by zero for all roots, hence arriving to a non-computable value for the x-intercepts of the given polynomial, implying that it has no roots. Nonetheless, if we consider this case less narrowly we find that the null value of an simply reduces the degree of our polynomial by 1, and thus to find the roots we would need to consider as the new an , which is perfectly reasonable given the case.
Thus we prove that it is senseless to speak of an being 0 as this denomination would simply be transposed to the next term coefficient of the polynomial in order to use Rational-zero theorem. 4. List all possible candidates for rational zeros of Determine if P has any rational zeros, if so, find them and all other remaining zeros. We know that and that p must be a factor of 6 and q a factor of 2. Therefore the possible values for p are: And those of q are: So the possible candidates for rational zeros are (in ascending order): Thus we need to test for each individual value in the function and determine if it gives us a zero.