Factoring and Finding Roots

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 .

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.

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:

.

Let be a sixth root of 729. The question is to find a solution of the equation

.

Subtracting 64 from both sides, this equation becomes

729 is a perfect square (of 27) The binomial at left can be factored first as the difference of two squares:

27 is a perfect cube (of 3), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

The equation therefore becomes

.

By the Zero Product Principle, one of these factors must be equal to 0.

If , then ; if , then . Therefore, and 3 are sixth roots of 729. However, these are not choices, so we examine the other polynomials for their zeroes.

If , then, setting in the following quadratic formula:

If , then, setting in the quadratic formula:

Therefore, the set of sixth roots of 729 is

Of the choices given, is the one that appears in this set.

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:

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 the endpoints - . By the IVT, , and , for some .

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 .

From the discriminant, , we know that this equation will have two solutions:

Next, factor the equation .

Finally, solve for .

Factor the trinomial.

The multiples of the first term is .

The multiples of the third term is .

We can then factor using these terms.

Set the equation to zero.

This means that each product will equal zero.

The roots are either

The answer is:

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.