### All Algebra 1 Resources

## Example Questions

### Example Question #1 : Equations / Solution Sets

Which of the following displays the full real-number solution set for in the equation above?

**Possible Answers:**

**Correct answer:**

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

### Example Question #1 : Factoring Polynomials

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 #1 : How To Factor An Equation

Factor .

**Possible Answers:**

Cannot be factored any further.

**Correct answer:**

This is a difference of squares. The difference of squares formula is *a*^{2} – *b*^{2} = (*a* + *b*)(*a* – *b*).

In this problem, *a* = 6*x* and *b* = 7*y*:

36*x*^{2} – 49*y*^{2 }= (6*x* + 7*y*)(6*x* – 7*y*)

### Example Question #4 : Equations / Solution Sets

Solve the equation:

**Possible Answers:**

**Correct answer:**

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

### Example Question #1 : Equations / Solution Sets

What number is the greatest common factor of 90 and 315 divided by the least common multiple of 5 and 15?

**Possible Answers:**

**Correct answer:**

First, find the factors of 90 and 315. The greatest common factor is the largest factor shared by both of the numbers: 45.

Then, find the least common multiple of 5 and 15. This will be the smallest number that can be divided by both 5 and 15: 15.

Finally, the greatest common factor (45) divided by the least common multiple (15) = 45 / 15 = 3.

### Example Question #6 : Equations / Solution Sets

Factor the expression:

**Possible Answers:**

**Correct answer:**

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: . Thus, we can rewrite as and it follows that

### Example Question #7 : Equations / Solution Sets

Factor the equation:

**Possible Answers:**

**Correct answer:**

The product of is .

For the equation ,

must equal and must equal .

Thus and must be and , making the answer .

### Example Question #8 : Equations / Solution Sets

Solve for .

**Possible Answers:**

**Correct answer:**

This is a quadratic equation. We can solve for either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format .

To yield the first value in our original equation (), and .

To yield the final term in our original equation (), we can set and .

Now that the equation has been factored, we can evaluate . We set each factored term equal to zero and solve.

### Example Question #9 : Equations / Solution Sets

Simplify:

**Possible Answers:**

**Correct answer:**

First factor the numerator. We need two numbers with a sum of 3 and a product of 2. The numbers 1 and 2 satisfy these conditions:

Now, look to see if there are any common factors that will cancel:

The in the numerator and denominator cancel, leaving .

### Example Question #10 : Equations / Solution Sets

Factor the following expression:

**Possible Answers:**

**Correct answer:**

The general form for a factored expression of order 2 is

, which, when FOILED, gives .

Comparing this generic expression to the one given in the probem, we can see that the term should equal , and the term should equal 2.

The values of and that satisfy the two equations are and ,

so your factored expression is

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