### All Algebra II Resources

## Example Questions

### Example Question #101 : Intermediate Single Variable Algebra

Factor:

**Possible Answers:**

Cannot be factored.

**Correct answer:**

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 #55 : Factoring Polynomials

Factor this polynomial:

**Possible Answers:**

**Correct answer:**

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?

### Example Question #56 : Factoring Polynomials

Factor completely:

**Possible Answers:**

**Correct answer:**

Before we do anything, we notice that both terms in the expression have a common factor of 4. Thus, we can factor it out, leaving us with: . We recognize that the expression inside the parentheses is a difference of squares, and factors as such: . Finally, we are done.

### Example Question #57 : Factoring Polynomials

Factor the following polynomial into its simplest form:

**Possible Answers:**

**Correct answer:**

We notice that the polynomial has a **greatest** common factor (i.e. the biggest multiplicatve "part") of .

First , we put the terms in ascending order of degree:

Since the number is the biggest number that can fit into all of the terms, we can pull it out front of the parentheses.

This results in times a reducible quadratic factor.

is of a special class of polynomial in which the middle term squared is equal to the last term. in this case, it can be factored into the form

consider:

so, the final, simplest factorized form of the polynomial is:

Note:In the original form of this polynomial there are other common factors. One could factor out or but the greatest common factor is because it reduces

### Example Question #102 : Intermediate Single Variable Algebra

factor the following polynomial:

**Possible Answers:**

**Correct answer:**

The first thing to notice is that each term in the polynomial has a common factor of . We can pull that out and see what we end up with.

Now, in the parentheses, we have a reducible quadratic factor. We see that it fits a special class of quadratic since *half the middle number* squared is equal to the last number. Hence, it can be factored as such:

Which is our answer because there are no more common terms and the contents of the parentheses can't be simplified anymore.

To check if this is true, we can expand, using FOIL, the answer to see if we get the original polynomial.

Which is our original polynomial.

### Example Question #51 : Factoring Polynomials

Factor the following polynomial to its simplest form:

**Possible Answers:**

**Correct answer:**

The first thing we notice is that there's a common factor of in both terms. We can factor it out in front.

Next, we notice that the polynomial in the parentheses can be factored, since it is the difference of two squared numbers: and . Hence it can be factored in the following way:

which is our answer.

To see that this is the right answer, we can re-expand what we came up with to see if the result is the original polynomial.

expanding the two linear binomial factors using FOIL:

the s cancel out and we're left with

Destributing the 3 out, we end up with the original polynomial.

### Example Question #60 : Factoring Polynomials

Factor the following polynomial:

**Possible Answers:**

**Correct answer:**

We notice that this is the difference of two squared numbers: and .

Hence, we can follow the rule that the difference of two perfect squares is equal to

To see this a little better, we can FOIL out the answer:

the s cancel out and we're left with the original expression:

=

### Example Question #61 : Factoring Polynomials

Factor the following polynomial into the simplest form possible:

**Possible Answers:**

**Correct answer:**

We look for the comon factors of each term in the polynomial (that means, the things that each term has in common).

It turns out that each term in the polinomial has a factor (multiplicative "part") of . In this case could be a number or a variable but it doesn't matter in this case. To factor, we need to synthisize what's common with all the factors and put it in front of parentheses. Factoring like this helps to simplify equations and expressions down the road.

Noticing that is the common factor, we can take it outside of the parentheses, writing within what's left over after we take out the .

### Example Question #62 : Factoring Polynomials

Factor the following polynomial:

**Possible Answers:**

**Correct answer:**

The polynomial fits a special class of polynomials because the last number, , is the square of half the middle number, . i.e. the middle number is the last number* squared.*

to see if it's right, we can expand it all over again:

usinig FOIL to expand the expression,

Which was our original polynomial.

### Example Question #63 : Factoring Polynomials

Factor the following polynomial:

**Possible Answers:**

**Correct answer:**

We start with the polynomial

and, putting the summands in ascending order of degree, we have

We notice that the comon factor is because it "fits" in each of the terms.

Notice

so, we factor out and we have

.

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