### All SAT II Math I Resources

## Example Questions

### Example Question #81 : 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 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.

### Example Question #72 : Solving Equations

Each of the following is a factor of the polynomial , except:

**Possible Answers:**

**Correct answer:**

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.

### Example Question #81 : 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 #84 : Single Variable Algebra

Define a function .

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 .

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 .

### Example Question #85 : Single Variable Algebra

Consider the absolute value equation

Which of the following quadratic equations has the same solution set as this one?

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #82 : Single Variable Algebra

Give the set of all real solutions of the equation .

**Possible Answers:**

**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 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 .

### Example Question #87 : Single Variable Algebra

Give the set of all real solutions of the equation .

**Possible Answers:**

**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 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 .

### Example Question #83 : Single Variable Algebra

Define a function .

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 .

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 .

### Example Question #81 : Solving Equations

Solve:

**Possible Answers:**

**Correct answer:**

Subtract 18 on both sides.

Divide by three on both sides.

The answer is:

### Example Question #82 : Solving Equations

Solve the equation:

**Possible Answers:**

**Correct answer:**

Subtract seven from both sides.

Divide both sides by negative nine.

The answer is:

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