The first step will be to plug our given variable into the equation to get 

.  

Then you do the multiplication first so it is now, 

.  

Finally, subtract  from  to get .

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

, 

where , , and  are its zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is one of its zeroes, so is its complex conjugate, . It has one other known zero, 2.

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

To rewrite this, first multiply the first two factors with the help of the difference of squares pattern and the square of a binomial pattern:

Thus,

Distributing:

Call 

By the Rational Zeroes Theorem, since  has only integer coefficients, any rational solution of  must be a factor of 54 divided by a factor of 1 - positive or negative. 54 has as its factors 1, 2, 3, 6, 9, 18, 27 , 54; 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  - that is,  and  - as factors, since  and  have been eliminated as possible zeroes.

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

, so , or, equivalently, , is a factor of . 

Of the remaining two choices,  and , both can be proved to not be factors by showing that  and  are both nonzero:

, so  is not a factor.

, so  is not a factor. 

Set 

By the Rational Zeroes Theorem, a rational number can be a zero of a polynomial  if and only if it belongs to the set of numbers that are the quotient of a factor of the constant and a factor of the leading coefficient (positive or negative). Constant 105 has as its factors 1, 3, 5, 7, 15 21, 35, and 105, and lead coefficient 1 has only factor 1, so any rational zeroes must come from the set

By the Factor Theorem, a linear binomial  is a factor of  if and only if  is a zero of . From the set , only 2 can be eliminated as a zero by the above result, so, of the five choices,  is the one that is eliminated as a factor of . 

The other four can be confirmed to be factors by demonstrating that 

,

but this is not necessary.

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 .

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 .  and  are both continuous everywhere, so  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 both endpoints - . By the IVT, , and , for some .

The absolute value equation  can be rewritten as the compound equation

Each simple equation can be solved by subtracting 7 from both sides:

The absolute value equation has solution set .

Suppose  is a polynomial. Then, by the Factor Theorem,  is a factor of  if and only if   - that is, if  is a solution of . We are seeking a quadratic equation with solution set the same as that of the absolute value equation, , we want polynomial  to have as its linear binomial factors , or , and . 

The correct polynomial will be the product of these two:

Using the FOIL technique to rewrite this:

The correct choice is the equation .

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 4, 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:

, or 

Cubing both sides:

Or:

Substituting back:

, or 

Cubing both sides:

.

The solution set is .

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 4, 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:

However, there are no real numbers whose squares are negative. Since we are looking only for real solutions, none are yielded here.

Or:

Substituting back:

Taking both square roots of both sides and simplifying using the Quotient of Radicals Rule:

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 . As a polynomial,  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 different signs at the endpoints - . By the IVT, , and , for some .