The vertex is the minimum or maximum of a parabola.

Write the vertex formula for the polynomial .

Substitute the coefficients.

The answer is:  

In order to factorize this quadratic, we will need to identify the roots of the first and last term and order it into the two binomials.

We know that it will be in the form of:

The value of  can be divided into , and is the only possibility to be replaced with  and .

Substitute this into the binomials.

Now we need to determine  such that it will equal to 12, and satisfy the central term of .

The roots of 12 that can be interchangeable are:

The only terms that are possible are  since

.

Remember that we must have a positive ending term!

This means that .  

Substitute the terms.

The answer is:  

By the Rational Zeroes Theorem (RZT), if a polynomial has only integer coefficients, then any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the coefficient of greatest degree. In other words, the RZT only gives a set of rational numbers of which any rational zeroes must be an element; it does not dictate what rational zeroes, if any, must actually be zeroes of the polynomial.

For example, examine the polynomial 

,

which satisfies the given characteristics in that its first and last terms are  and . If  is a zero, then 

.

Substitute  for :

 is not a zero of .

Therefore, it does not hold in general that a polynomial with the given highest- and lowest-degree terms has  as a zero.

By the Rational Zeroes Theorem (RZT), if a polynomial has only integer coefficients, then any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the coefficient of greatest degree. These integers are, respectively, 24, which as as its factors 1, 2, 3, 4, 6, 8, 12, and 24, and 4, which has as its factors 1, 2, and 4.

The complete set of quotients of factors of the former and factors of the latter is derived by dividing each element of  by each element of . The resulting set is 

,

so any rational zero must be an element of this set.  is not an element of this set, so by the RZT, it cannot be a zero of the polynomial.

The easiest way to answer this question is arguably as follows:

Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the alternating sum of its coefficients (accounting for minus symbols) is 0.

To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In , there are two such terms - the fifth-degree and first-degree (linear) term, so alternating coefficient sum is  

.

It follows that  is divisible by .

The easiest way to answer this question is arguably as follows:

Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the sum of its coefficients (accounting for minus symbols) is 0.  has 

as its coefficient sum, so  is not divisible by .

One way to answer this question is as follows:

Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the sum of its coefficients (accounting for minus symbols) is 0.  has 

as its coefficient sum, so  is indeed divisible by .

If a polynomial has only integer coefficients, it is possible for its zeroes to be real or imaginary; however, by the Complex Conjugate Roots Theorem, any imaginary zeroes occur in conjugate pairs. For example, if  is a zero, so is . Therefore, such a polynomial must have an even number of imaginary zeroes, distinct or otherwise. But by the Fundamental Theorem of Algebra, a polynomial of degree 7 has 7 zeroes, distinct or otherwise. Therefore, the polynomial has six distinct imaginary zeroes at most.

By the Fundamental Theorem of Algebra, a polynomial of degree  has  zeroes that may or may not be distinct. By the Conjugate Zeroes Theorem, if the polynomial has only rational coefficients, then its imaginary zeroes occur in conjugate pairs, so there must be an even number of them. If the polynomial has degree six, then there are three pairs, and they may be distinct, for a total of six distinct imaginary zeroes. 

In fact, we can construct an example of a polynomial with six distinct zeroes: , as follows:

If a polynomial has  as a zero, then its factorization, if taken down to linear binomials, includes the factor .

Examine the polynomial 

,

which has as its set of zeroes  - six distinct imaginary numbers. Applying the sum/difference pattern of binomial multiplication three times:

This polynomial has, by construction, integer coefficients and six distinct imaginary zeroes.

Let . By a corollary of the Factor Theorem,  is divisible by  if and only if the alternating sum of its coefficients (accounting for minus symbols) is 0.

To find this alternating sum, it is necessary to reverse the symbol before all terms of odd degree. In , there are no such terms, (the degree of the constant is 0), so the alternating coefficient sum is  

,

and  is divisible by .