# SAT II Math II : Factoring and Finding Roots

## Example Questions

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### Example Question #52 : Polynomials

Factor the trinomial.

Explanation:

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.

### Example Question #1 : Factoring And Finding Roots

Factor completely:

The polynomial is prime.

Explanation:

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.

### Example Question #2 : Factoring And Finding Roots

Which of the following values of  would make

a prime polynomial?

None of the other responses is correct.

None of the other responses is correct.

Explanation:

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.

### Example Question #3 : Factoring And Finding Roots

Which of the following values of  would not make

a prime polynomial?

None of the other responses is correct.

Explanation:

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:

.

### Example Question #4 : Factoring And Finding Roots

Which of the following is a factor of the polynomial ?

Explanation:

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 :

### Example Question #5 : Factoring And Finding Roots

Give the set of all real solutions of the equation .

The equation has no real solution.

Explanation:

Set . Then

can be rewritten as

Substituting  for  and  for , the equation becomes

,

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 .

### Example Question #6 : Factoring And Finding Roots

Define a function .

for exactly one real value of  on the interval .

Which of the following statements is correct about ?

Explanation:

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 .

### Example Question #7 : Factoring And Finding Roots

A cubic polynomial  with rational coefficients whose lead term is  has  and  as two of its zeroes. Which of the following is this polynomial?

The correct answer cannot be determined from the information given.

The correct answer cannot be determined from the information given.

Explanation:

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.

### Example Question #8 : Factoring And Finding Roots

Define functions  and .

for exactly one value of  on the interval . Which of the following is true of ?

Explanation:

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 .

### Example Question #9 : Factoring And Finding Roots

A cubic polynomial  with rational coefficients and with  as its leading term has 2 and 3 as its only zeroes. 2 is a zero of multiplicity 1.

Which of the following is this polynomial?

Insufficient information exists to determine the polynomial.

Explanation:

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.

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