Use the -method to split the middle term into the sum of two terms whose coefficients have sum  and product . These two numbers can be found, using trial and error, to be  and .

and

Now we know that is equal to .

Factor by grouping.

Since both terms are perfect cubes , the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

We substitute  for  and 7 for :

The latter factor cannot be factored further, since we would need to find two integers whose product is 49 and whose sum is ; they do not exist. This is as far as we can go with the factoring.

 is the cube of . Therefore, if  is a perfect cube, the expression  is factorable as the sum of two cubes. All four of the choices are perfect cubes - 8, 27, 64, and 125 are the cubes of 2, 3, 4 and 5, respectively. The correct response is that none of the choices are correct.

 is a perfect square term - it is equal to . All of the values of  given in the choices are perfect squares - 25, 36, 49, and 64 are the squares of 5, 6, 7, and 8, respectively. 

Therefore, for each given value of , the polynomial is the sum of squares, which is normally a prime polynomial. However, if  - and only in this case - the polynomial can be factored as follows:

.

Call .

By the Rational Zeroes Theorem, since  has only integer coefficients, any rational solution of  must be a factor of 18 divided by a factor of 1 - positive or negative. 18 has as its factors 1, 2, 3, 6, 9, and 18; 1 has only itself as a factor. Therefore, the rational solutions of  must be chosen from this set:

.

By the Factor Theorem, a polynomial  is divisible by  if and only if  - that is, if  is a zero. By the preceding result, we can immediately eliminate  and  as factors, since 4 and 5 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that  is the factor by evaluating :

By the Factor Theorem, it follows that  is a factor.

As for the other two, we can confirm that neither is a factor by evaluating  and :

Set . Then . 

 can be rewritten as

Substituting  for  and  for , the equation becomes 

,

a quadratic equation in .

This can be solved using the  method. We are looking for two integers whose sum is  and whose product is . Through some trial and error, the integers are found to be  and , so the above equation can be rewritten, and solved using grouping, as

By the Zero Product Principle, one of these factors is equal to zero:

Either:

Substituting  back for :

Taking the positive and negative square roots of both sides:

.

Or:

Substituting back:

Taking the positive and negative square roots of both sides, and applying the Quotient of Radicals property, then simplifying by rationalizing the denominator:

The solution set is .

Define . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if  is a continuous function, and  and  are of unlike sign, then  for some .   is a continuous function, so the IVT applies here.

Evaluate  for each of the following values: 

Only in the case of  does it hold that  assumes a different sign at each endpoint - . By the IVT, , and , for some .

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Two imaginary zeroes are given that are each other's complex conjugate -   and . Since the polynomial is cubic - of degree 3 - it has one other zero, which must be real. However, no information is given about that zero. Therefore, the polynomial cannot be determined.

Define 

Then if ,

it follows that

,

or, equivalently,

.

By the Intermediate Value Theorem (IVT), if  is a continuous function, and  and  are of unlike sign, then  for some . 

Since polynomial  and exponential function  are continuous everywhere, so is , so the IVT applies here.

Evaluate  for each of the following values: :

Only in the case of  does it hold that  assumes a different sign at both endpoints - . By the IVT, , and , for some .

A cubic polynomial has three zeroes, if a zero of multiplicity  is counted  times. Since its lead term is , we know that, in factored form,

, 

where , , and  are its zeroes.

Since 2 is a zero of multiplicity 1, its only other zero, 3, must be a zero of multiplicity 2.

Therefore, we can set , ,  in the factored form of , and 

,

or

To rewrite this, firs square  by way of the square of a binomial pattern:

Thus,

Multiplying:

________

,

the correct polynomial.