A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity  is counted  times. Since its lead term is , we know by the Factor Theorem that

where the  terms are the four zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is a zero, so is its complex conjugate . Also,  has only three distinct zeroes, so one of the three known zeroes must have multiplicity 2; there is only one unknown zero, and imaginary zeroes must occur in conjugate pairs, so it follows that the zero with multiplicity 2 must be . Therefore, setting

, , and ,

the polynomial, in factored form, is

This can be rewritten as

 can be calculated using the binomial square pattern:

 can be calculated by way of the difference of squares and binomial square patterns:

Thus, 

Multiply:

Multiply both sides of the equation by the least common denominator of the expressions, which is :

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:

or 

 The solution set is the following:

A polynomial is divisible by  if and only if the alternating sum of its coefficients is 0- that is, if every other coefficient is reversed in sign and the sum of the resulting numbers is 0. For each polynomial, add the coefficients, reversing the signs of the  and  coefficients:

:

:

:

:

The alternating sum of the coefficients of this last polynomial add up to 0, so this is the correct choice.

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity  is counted  times. Since its lead term is , we know by the Factor Theorem that

where the  terms are the four zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is a zero of multiplicity 2, so is its complex conjugate . We can set  and , and 

We can rewrite this as

or

Multiply these factors using the difference of squares pattern, then the square of a binomial pattern:

Therefore, 

Multiplying:

Four of the choices can be eliminated quickly using the Rational Zeroes Theorem. By this 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 84 has as its factors 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84, 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 . 5 is not a possible zero of , since it does not appear in the above set; therefore,  can be eliminated as a possible linear factor of . , , and  can be eliminated for the same reason. 

This leaves , which can be confirmed to be a factor of  by showing that 7 is a zero of  - that is,. Setting :

 can be seen to fit the perfect square trinomial pattern:

The equation can therefore be rewritten as 

Multiply both sides of the equation by the least common denominator of the expressions, which is :

Distributing and collecting like terms, we get

Multiplying both sides by :

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 2 and 12, 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: 

Or:

Both solutions can be confirmed by substitution; the solution set is 

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity  is counted  times. Since its lead term is , we know that

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is a zero, so is its complex conjugate ; similarly, since  is a zero, so is its complex conjugate . Substituting these four values for the four  values: 

This can rewritten as

or

Multiply the first two factors using the difference of squares pattern, then the square of a binomial pattern:

Multiply the last two factors similarly:

Thus,

Multiply:

________________

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 

However, this does not hold for any real value of . No solution is yielded here.

Or:

Substituting back:

, or 

,

the only solution. This is not among the choices.

Two numbers of like sign have a positive quotient.

Therefore,  has as its solution set the set of points at which  and  are both positive or both negative.

To find this set of points, we identify the zeroes of both expressions. 

Since  is nonzero we have to exclude ;  is excluded anyway since it would bring about a denominator of zero. We choose one test point on each of the three intervals  and determine where the inequality is correct.

Choose :

 - True.

Choose :

 - False.

Choose :

 - True.

The solution set is 

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, add 7 to each side.

Now, divide both sides by 2.