### All College Algebra Resources

## Example Questions

### Example Question #1 : Factoring Polynomials

Factor the polynomial:

**Possible Answers:**

**Correct answer:**

First, begin by factoring out a common term, in this case :

Then, factor the terms in parentheses by finding two integers that sum to and multiply to :

### Example Question #3 : Factoring Polynomials

Factor the following expression:

**Possible Answers:**

**Correct answer:**

Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.

Only the last two terms have so it will not be factored out. Each term has at least and so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial:

### Example Question #2 : Factoring Polynomials

Factor the expression:

**Possible Answers:**

**Correct answer:**

To find the greatest common factor, we must break each term into its prime factors:

The terms have , , and in common; thus, the GCF is .

Pull this out of the expression to find the answer: .

### Example Question #1 : How To Factor An Equation

Factor .

**Possible Answers:**

**Correct answer:**

First pull out 3u from both terms.

3*u*^{4} – 24*uv*^{3 }= 3*u*(*u*^{3} – 8*v*^{3}) = 3*u*[*u*^{3} – (2*v*)^{3}]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is *a*^{3} – *b*^{3} = (*a* – *b*)(*a*^{2} + *ab* + *b*^{2}). In our problem, *a* = *u* and *b* = 2*v:*

3*u*^{4} – 24*uv*^{3 }= 3*u*(*u*^{3} – 8*v*^{3}) = 3*u*[*u*^{3} – (2*v*)^{3}]

= 3*u*(*u* – 2*v*)(*u*^{2} + 2*uv* + 4*v*^{2})

### Example Question #3 : Factoring Polynomials

Factor:

**Possible Answers:**

**Correct answer:**

### Example Question #61 : Intermediate Single Variable Algebra

Factor the following expression:

**Possible Answers:**

**Correct answer:**

To factor, we are looking for two terms that multiply to give and add together to get .

Possible factors of :

Based on these options, it is clear our factors are and .

Our final answer will be:

### Example Question #64 : Intermediate Single Variable Algebra

Factor the following expression:

**Possible Answers:**

**Correct answer:**

To factor, we are looking for two terms that multiply to give and add together to get . There are numerous factors of , so we will only list a few.

Possible factors of :

Based on these options, it is clear our factors are and .

Our final answer will be:

### Example Question #4 : Factoring Polynomials

Factor the trinomial .

**Possible Answers:**

**Correct answer:**

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with and the other of which begins with . This is the only way the binomials will multiply to give us .

The next part, however, is slightly more difficult. The last part of the trinomial is , which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be , the 1 must be multiplied with the and be negative, and the 2 must be multiplied with the and be positive. This would give us , or the that we are looking for.

In other words, our answer must be

to properly multiply out to the trinomial given in this question.

### Example Question #5 : Factoring Polynomials

Factor this polynomial:

**Possible Answers:**

**Correct answer:**

Factor out the largest quantity common to all terms:

Factor the simplified quadratic:

### Example Question #6 : Factoring Polynomials

Factor the polynomial

**Possible Answers:**

**Correct answer:**

can be looked as . When A=1, as it does in this case, we can ignore it. So now we need to look at factors of "C" that add up to "B."

Factors of 20 are:

1 20

2 10

4 5

Of these three options, the 4 & 5 will add to 9, so we write