Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #202 : Equations / Inequalities

Solve for x.

Possible Answers:

x = –5, 5

x = –5

x = –2/5, –5

x = –5/2, –5

x = –2/3, –3

Correct answer:

x = –5/2, –5

Explanation:

1) The first step would be to simplify, but since 2, 15, and 25 have no common factors greater than 1, simplification is impossible.

Now we factor. Multiply the first coefficient by the final term and list off the factors.

2 * 25 = 50

Factors of 50 include:

1 + 50 = 51

2 + 25 = 27

5 + 10 = 15

2) Split up the middle term to make factoring by grouping possible.

Note that the "2" and the "10," and the "5" and the "25," have to go together for factoring to come out with integers. Always make sure the groups actually have a common factor to pull.

3) Pull out the common factors from both groups, "2x" from the first and "5" from the second.

4) Factor out the "(x+5)" from both terms.

5) Set each parenthetical expression equal to zero and solve.

2x + 5 = 0,  x = –5/2

x + 5 = 0, x = –5

Example Question #91 : Polynomials

Possible Answers:

Correct answer:

Explanation:

Example Question #5 : How To Factor The Quadratic Equation

Factor the following quadratic equation. 

Possible Answers:

 

Correct answer:

Explanation:

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as  the factored numbers are  and . Since , we know the factors for 1 are 1 and 1. So we know the terms will be

Looking at our constant, , we see a positive 6. So 6 factors into either 2 and 3  or 1 and 6 (since  and ). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the  part of the quadratic. We are looking for 2 numbers which add up to our . So, 1 and 6 do not work, since . But, 2 and 3 are perfect since .

But, since our  is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

 

or

Example Question #92 : Polynomials

Factor the following:  

Possible Answers:

Correct answer:

Explanation:

Using the FOIL rule, only  yields the same polynomial as given in the question.

Example Question #93 : Polynomials

Factor the following polynomial:

Possible Answers:

Can't be factored

Correct answer:

Explanation:

When asked to factor a difference of squares, the solution will always be the square roots of the coefficients with opposite signs in each pair of parentheses.

Example Question #13 : How To Factor A Variable

Solve for  when 

Possible Answers:

Correct answer:

Explanation:

First, factor the numerator: .  

Now your expression looks like 

Second, cancel the "like" terms -  - which leaves us with .  

Third, solve for , which leaves you with 

Example Question #4652 : Algebra 1

Factor the following polynomial: .

Possible Answers:

Correct answer:

Explanation:

Because the  term has a coefficient, you begin by multiplying the  and the  terms () together: 

Find the factors of  that when added together equal the second coefficient (the  term) of the polynomial: 

There are four factors of , and only two of those factors, , can be manipulated to equal  when added together and manipulated to equal  when multiplied together:  

Example Question #51 : Factoring Polynomials

Solve for :

Possible Answers:

None of the other answers.

Correct answer:

Explanation:

Step 1: Subtract 30 from both sides of the equation in order to make the equation equal to 0.

Step 2: Factor out a 3.

Step 3: Factor the trinomial.

At this step you set both factors equal to 0 and solve for .

 

Example Question #1234 : Algebra Ii

Factor:  

Possible Answers:

Cannot be factored.

Correct answer:

Explanation:

The method to factor is to find the roots of  of the polynomial in standard form.

The factors are:  

Select the set that will most likely add or subtract to achieve the coefficient of the middle term.

The set  will satisfy.  Since  is a positive number, the signs inside the binomials will be negative.

Example Question #101 : Intermediate Single Variable Algebra

Factor this polynomial: 

Possible Answers:

Correct answer:

Explanation:

To factor this problem correctly you must recognize the similarity in all three terms. What is common to all three? They all have x yes. But is there a common factor amongst the constants? To factor correctly you factor out as much as you can from each term.

First factor out as much from each constant as you can. It is similar to dividing (at least that is a way to think about what you are doing). In essence you are just removing quantities.

Now factor out as many x's as you can from each term:

Now factor what is left over in the paranthesis. A way to think about this is what two x terms can I multiply to get the first x term and what two constants multiplied by each other would equal the third term and what sign must they have to add or subtract to get the middle term. You are just doing the reverse of the FOIL method.

See how the the x terms in the parentheses multiply to equal X^2? And how 2*1 gives the desired 2?

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