SAT II Math I : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math I

varsity tutors app store varsity tutors android store

Example Questions

Example Question #81 : Single Variable Algebra

Which of the following is a factor of the polynomial  ?

Possible Answers:

Correct answer:

Explanation:

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:

Explanation:

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:

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 #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:

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 .  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:

Explanation:

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:

Explanation:

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:

Explanation:

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:

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 . 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:

Explanation:

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:

Explanation:

Subtract seven from both sides.

Divide both sides by negative nine.

The answer is:  

Learning Tools by Varsity Tutors