(ii)
Testing for the smallest positive root we obtain .
(iii)
We obtain the smallest positive root as .
These set of results corroborate the assertion on the previous section, while the second and fourth term are kept constant, the numerator of the smallest positive root remains the same, while the denominator assumes the value of the coefficient of the first term.
b) (i)
From the graph we obtain that the smallest positive root is .
This curve has the same smallest root as the previous one, we observe once again that what actually defines this value are not the individual values of the coefficients but their proportion, more specifically, that of the first and fourth coefficients. In the curve of section 2(b)(i) we see that this proportion is of 1 : 6, the same occurs with those in this curve: 30 : 180 = 1 : 6.
(ii)
For this curve we get the value of the smallest positive root as
(iii)
Finally the smallest positive root for this curve is
These two last curves support our result in a different manner: the ratio of the two roots is equal to that of the two coefficients in the fourth term. We see that the first term is constant, as well as the denominator in the roots, therefore we can assert at this stage that the numerator is independent of the first term coefficient and dependant on the fourth, while the denominator is independent of the fourth term coefficient and while dependant on the first.
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Suppose a, b, c and d are integers with, and thatis the polynomial defined by . Suppose, in addition, that p and q are positive integers with no common factors.
-
Show that the equation can be rewritten in the form
and
-
Factorizing the left hand side by p:
-
Hence show that p must be a factor of d.
-
Dividing the expression by p we obtain:
We know from definition that a, b and c are non-zero integers, as well q and p. It therefore follows that the left hand side expression of the equation is also an integer for all allowed values when . As a result of this, we deduce that the right hand side term is also an integer.
However, we know from definition that the ratio is in lowest terms, this is, p and q have no common factors. Therefore the ratio as well as are in lowest terms as is multiple of q.
As d is also an integer, we come to the conclusion that p cannot be diving : whereas for some values of p and q the division would result in the needed integer (e.g. equation 1(a)(ii) would give us the whole number 8), for some others the value would be a decimal (e.g. for equation 1(c) the result is ).
If p weren’t a factor of d this would contradict the result of the left hand side expression. Hence, it follows that p must be dividing d to get an integer which multiplied by gives us the whole number value of the expression in the left.
Thus it is proved that p is a factor of d.
-
Show that the equation can also be written in the form
- We begin, as in the previous example, with,:
and
-
Factorizing the right hand side by q:
-
Hence show that q is a factor of a.
- We use a similar method as part (b) but beginning with:
-
Dividing the expression by q we obtain:
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.
- 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.
- 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.
For
NOT A ROOT
For
NOT A ROOT
For
NOT A ROOT
For
NOT A ROOT
For
NOT A ROOT
For
NOT A ROOT
For
ROOT
For
ROOT
For
NOT A ROOT
For
NOT A ROOT
Therefore the only real roots are found at
Summing up, the Rational-zero theorem which we have been able to prove lets us formulate possibilities for rational roots of polynomials of any given degree, whose number would only depend on the amount of factors that the first and last term of the polynomial have. We have found out that the denominator of the root expressed as a fraction must be a factor of the first term coefficient while the numerator a factor of the last term coefficient.
However, this method is limited only to integer coefficients because, as we have seen in section 1(b), the values specifically related to a given term will vary from numerator to denominator or vice-versa, depending on the individual context and therefore cannot be generalized in this manner. This method additionally cannot be used to find irrational roots of polynomials.
