Algebra II : Polynomials

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #24 : Factoring Polynomials

Solve   algebraically.

Possible Answers:

 or 

 or 

 or 

Correct answer:

 or 

Explanation:

Subtract 12 from both sides:

Factor:

Then, by the zero-product property,

 or .

Solve:

 or 

Example Question #25 : Factoring Polynomials

Factor the following equation:

Possible Answers:

Correct answer:

Explanation:

Factor:

The last term is positive, which indicates that both factors should be positive:

The sum of the factors must be , and their product must be .

This means the factors are and

Therefore:

Example Question #26 : Factoring Polynomials

Factor.

Possible Answers:

Correct answer:

Explanation:

When factoring trinomials with larger coefficients, rather than guess and check, there is a 4 step process that will always yield a correct answer.

Step 1 - Multiply the first and third terms

 

Step 2 - Find two factors that multiply to equal the result of step 1 () and add up to equal the second/middle term.

The product  and 

Their sum is      This works!!

So,  and  are our new factors

 

Step 3 - Replace our middle term () with our two new factors.

 

Step 4 - Factor by grouping to find your answer

   Answer!!!

 

To check your work, FOIL, and see if you get our original trinomial 

 

Example Question #27 : Factoring Polynomials

Factor the following polynomial:

Possible Answers:

Correct answer:

Explanation:

This can be solved by looking at all of the answers and multiplying them and comparing to the answer. However this is time consuming. You can start by noting that the term  can be a result of , where one of the terms is negative, so one answer can be eliminated. It is also clear that  must be the result of  multiplied by , so two additional answers can be eliminated. Looking at the last two answers and multiplying through, the correct answer can be determined.

Example Question #71 : Polynomials

Which of the following expressions is a factor of this polynomial: 3x² + 7x – 6?

Possible Answers:

(x – 3) 

(3x – 6)

(x + 3)

(x – 3x)

(3x + 2) 

Correct answer:

(x + 3)

Explanation:

The polynomial factors into (x + 3) (3x - 2).

3x² + 7x – 6 = (a + b)(c + d)

There must be a 3x term to get a 3x² term.

3x² + 7x – 6 = (3x + b)(x + d)

The other two numbers must multiply to –6 and add to +7 when one is multiplied by 3.

b * d = –6 and 3d + b = 7

b = –2 and d = 3

3x² + 7x – 6 = (3x – 2)(x + 3)

(x + 3) is the correct answer.

Example Question #72 : Polynomials

Factor the polynomial .

Possible Answers:

y = (x – 3)(x + 2)

y = (x – 2)(x + 3)

y = (x + 3)(x + 2)

y = (x + 5)(x + 1)

y = (x + 6)(x + 1)

Correct answer:

y = (x + 3)(x + 2)

Explanation:

The product of the last two numbers should be 6, while the sum of the products of the inner and outer numbers should be 5x. Factors of six include 1 and 6, and 2 and 3. In this case, our sum is five so the correct choices are 2 and 3. Then, our factored expression is (x + 2)(x + 3). You can check your answer by using FOIL.

y = x2 + 5x + 6

2 * 3 = 6 and 2 + 3 = 5

(x + 2)(x + 3) = x2 + 5x + 6

Example Question #30 : Factoring Polynomials

Two consecutive odd numbers have a product of 195. What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

You can set the two numbers to equal variables, so that you can set up the algebra in this problem. The first odd number can be defined as  and the second odd number, since the two numbers are consecutive, will be .

This allows you to set up the following equation to include the given product of 195:

Next you can subtract 195 to the left and set the equation equal to 0, which allows you to solve for :

You can factor this quadratic equation by determining which factors of 195 add up to 2. Keep in mind they will need to have opposite signs to result in a product of negative 195:

Set each binomial equal to 0 and solve for . For the purpose of this problem, you'll only make use of the positive value for :

Now that you have solved for , you know the two consecutive odd numbers are 13 and 15. You solve for the answer by finding the sum of these two numbers:

Example Question #4641 : Algebra 1

Factor:

Possible Answers:

Correct answer:

Explanation:

The common factor here is . Pull this out of both terms to simplify:

Example Question #32 : Factoring Polynomials

Find the LCM of the following polynomials:

 

, ,

Possible Answers:

Correct answer:

Explanation:

LCM of

LCM of 

and since

The LCM 

 

Example Question #33 : Factoring Polynomials

Factor completely: 

Possible Answers:

The polynomial cannot be factored further.

Correct answer:

The polynomial cannot be factored further.

Explanation:

We are looking to factor this quadratic trinomial into two factors, where the question marks are to be replaced by two integers whose product is  and whose sum is 

We need to look at the factor pairs of  in which the negative number has the greater absolute value and the sum is :

None of these pairs have the desired sum, so the polynomial is prime.

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