### All SAT II Math II Resources

## Example Questions

### Example Question #8 : Factoring Polynomials

Factor the trinomial.

**Possible Answers:**

**Correct answer:**

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:

**Possible Answers:**

The polynomial is prime.

**Correct answer:**

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 #73 : Single Variable Algebra

Which of the following values of would make

a prime polynomial?

**Possible Answers:**

None of the other responses is correct.

**Correct answer:**

None of the other responses is correct.

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 #74 : Single Variable Algebra

Which of the following values of would *not* make

a prime polynomial?

**Possible Answers:**

None of the other responses is correct.

**Correct answer:**

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 #75 : Single Variable Algebra

Which of the following is a factor of the polynomial ?

**Possible Answers:**

**Correct answer:**

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 #76 : Single Variable Algebra

Give the set of all real solutions of the equation .

**Possible Answers:**

The equation has no real solution.

**Correct answer:**

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 .

### Example Question #77 : Single Variable Algebra

Define a function .

for exactly one real value of on the interval .

Which of the following statements is correct about ?

**Possible Answers:**

**Correct answer:**

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 #78 : Single Variable Algebra

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

**Possible Answers:**

The correct answer cannot be determined from the information given.

**Correct answer:**

The correct answer cannot be determined from the information given.

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 #79 : Single Variable Algebra

Define functions and .

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

**Possible Answers:**

**Correct answer:**

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 #81 : Single Variable Algebra

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?

**Possible Answers:**

Insufficient information exists to determine the polynomial.

**Correct answer:**

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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