Category: Factoring

What is the Remainder Theorem Formula?

The remainder theorem formula applies to a polynomial P(x) that is divided by a linear binomial.

When dividing, the remainder can be calculated by performing the operations on the opposite of the constant in the binomial.

So, P(x) ÷ (x + a) = P(–a)

What is the Opposite of the Constant?

In the remainder theorem formula above, our binomial is (x + a) and our our constant is a, so the opposite of our constant is –a.

To see the free examples, please scroll to the sections below the quiz.

Remainder Theorem – Exercises

Instructions:  Use the remainder theorem formula to find the remainder in the following problems.

[WpProQuiz 26]

Remainder Theorem – Example

The remainder theorem starts off with P(x), which means a polynomial that contains x as a variable.

The polynomial is divided by a linear binomial in the form (x + a), where the constant a can be either a positive or negative number.

Example:  Divide the polynomial x³ + 3x² – 4x – 10 by x + 3.

STEP 1:

Identify the binomial and the opposite of its constant.

In our example, our binomial is x + 3

The constant in the binomial is 3.

So, the opposite of the constant in the binomial is –3.

STEP 2:

Replace x in the polynomial with the opposite of the constant (which is –3 in this question) to find the remainder.

(x³ × 1) + (x² × 1)  + (x¹ × –4) – 10

[(–3)³ × 1] + [(–3)² × 1] + [(–3)¹ × –4] + [–10] =

[(–3)³] + [(–3)² × 3] + [(–3)¹ × –4] + [–10] =

(–27) + [(9) × 3] + [(–3) × –4] – 10 =

–27 + 27 + 12 – 10 = 2

Check Your Result:

Then check this result by performing long division as shown below.

                   x²          –  4

x + 3) x³ + 3x² – 4x – 10

       –(x³ + 3x²⁾

                – ( 0 – 4x – 10)

                – (     –4x – 12)

                                      2

You should also have a look at our posts on synthetic division, factoring, and quadratics.

Then try our free algebra test.

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Like Terms Examples

There are exercises with like terms plus examples on this page. First of all, consider the following algebraic expression:

8ab + 8ay + xb + xy

The terms are the combinations of numbers and variables that make up the expression.

In the above expression, 8ab is the first term, 8ay is the second term, xb is the third term, and xy is the fourth term.

We can see that 8ab + 8ay have the number 8 and the variable a in common.

The other terms have x in common.

So, a and x are the like terms.

Like Terms - Exercises

Instructions: Group terms together in the following exercises and then factor. When you have finished, check your answers in the next section.

1)   ab²y – 3a + 9b²y – 27

2)   xy³b – 4x + 5y³b – 20

3)   a²bc + 2c – 2a²b – 4 =

4)   x²byz – 5x²b – 25 + 5yz

5)   2a²b² + 6a²b + 2a² – 3b² – 9b – 3

Like Terms - Answers

1)   ab²y – 3a + 9b²y – 27 =
(ab²y – 3a) + (9b²y – 27) =
[(a × b² × y) – (3 × a)] + [(9 × b² × y) – (9 × 3)] =
[(a × b² × y) – (a × 3)] + [(9 × b² × y) – (9 × 3)] =
a[(a × b² × y) – (a × 3)] + 9[(9 × b² × y) – (9 × 3)] =
a[(b² × y) – (3)] + 9[((b² × y) – (3)] =
a(b²y – 3) + 9(b²y – 3) =
(b²y – 3)[a(b²y – 3) + 9(b²y – 3)] =
(b²y – 3)(a + 9)

2)   xy³b – 4x + 5y³b – 20 =
(xy³b – 4x) + (5y³b – 20) =
[(x × y³ × b) – (4 × x)] + [(5 × y³ × b) – (4 × 5)] =
[(x × y³ × b) – (x × 4)] + [(5 × y³ × b) – (5 × 4)] =
x[(x × y³ × b) – (x × 4)] + 5[(5 × y³ × b) – (5 × 4)] =
x[(y³ × b) – (4)] + 5[(y³ × b) – (4)] =
x(y³b – 4) + 5(y³b – 4) =
(y³b – 4)[x(y³b – 4) + 5(y³b – 4)] =
(y³b – 4)(x + 5)

3)   a²bc + 2c – 2a²b – 4 =
a²bc – 2a²b + 2c – 4 =
[(a²b × c) – (2 × a²b)] + [(2 × c) – (2 × 2)] =
a²b[(a²b × c) – (2 × a²b)] + 2[(2 × c) – (2 × 2)] =
a²b(c – 2) + 2(c – 2) =
(c – 2)[a²b(c – 2) + 2(c – 2)] =
(c – 2)(a²b + 2)

4)   x²byz – 5x²b – 25 + 5yz =
x²byz + 5yz – 5x²b – 25 =
[(x²b × yz) + 5yz] – (5x²b + 25) =
[(x²b × yz) + (5 × yz)] – [(5 × x²b) + (5 × 5)] =
yz[(x²b × yz) + (5 × yz)] – 5[(5 × x²b) + (5 × 5)] =
yz[(x²b + 5) – 5(x²b + 5)] =
(x²b + 5)[ yz(x²b + 5) – 5(x²b + 5)] =
(x²b + 5)(yz – 5)

5)   2a²b² + 6a²b + 2a² – 3b² – 9b – 3 =
(2a²b² + 6a²b + 2a²) – (3b² + 9b + 3) =
[2a²b² + (2a² × 3 × b) + (2a² × 1)] – [3b² + (3 × 3 × b) + (3 × 1)] =
2a²[2a²b² + (2a² × 3 × b) + (2a²× 1)] – 3[3b² + (3 × 3 × b) + (3 × 1)] = 2a²(b² + 3b + 1) – 3(b² + 3b + 1) =
(b² + 3b + 1)[2a²(b² + 3b + 1) – 3(b² + 3b + 1)] =
(b² + 3b + 1)(2a² – 3)

What are Like Terms?

Like terms are the parts of an algebraic expression that have something in common. The like terms may have a number or a product in common.

Alternatively, they may have a variable or a variable raised to a certain power in common.

Grouping Like Terms - More Examples

Factoring becomes much easier when we group like terms together.

Let's look at our example expression again:

8ab + 8ay + xb + xy

You need to follow the steps shown below in order to group like terms and factor your expression.

STEP 1:

Group the terms into two parts, using two sets of parentheses.

8ab + 8ay + xb + xy =

(8ab + 8ay) + (xb + xy)

STEP 2:

Look for the greatest common factor in each group and factor each set of parentheses.

(8ab + 8ay) + (xb + xy) =

[(8a × b) + (8a × y)] + [(x × b) + (x × y)] =

8a[(8a × b) + (8a × y)] + x[(x × b) + (x × y)] =

[8a(b + y)] + [x(b + y)]

STEP 3:

Identify the new greatest common factor.

[8a(b + y)] + [x(b + y)]

From the re-written expression above, we can see that (b + y) is our new greatest common factor.

STEP 4:

Factor out the new greatest common factor.

[8a(b + y)] + [x(b + y)] =

8a(b + y) + x(b + y) =

(b + y)[8a(b + y) + x(b + y)] =

(b + y)(8a + x)

You should view the following posts:

Greatest Common Factor

Factoring Numbers

Factoring Variables

Factoring Expressions

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How to Factor the Expression Using the GCF

You will see quite a few problems on your exam that ask you to factor the expression using the GCF.

Factor the Expression Using the GCF – Quiz

Instructions:  Here is a quiz with questions that ask you to factor the expression using the GCF. If you want to see the examples on how to factor an expression using the GCF first, please look at the examples below the quiz.

[WpProQuiz 11]

What are Algebraic Expressions?

To factor the expression using the GCF, you first need to understand algebraic expressions.

Algebraic expressions are math problems that have terms that contain both numbers and variables.

For example, 5ab + 15b + 10bc is an algebraic expression.

We factor the above expression by taking out the variables and numbers that are common to each term of the expression.

The numbers that are multiplied by a variable are called coefficients.

How to Factor the Expression Using the GCF – Example

Our example expression is: 5ab + 15b + 10bc

In our example expression, 5ab is the first term, 15b is the second term, and 10bc is the third term.

Each term contains the variable b, so we can factor out this variable.

STEP 1:  Multiply out each term to isolate the common variable or variables.

5ab + 15b + 10bc =

(5 × a × b) + (15 × b) + (10 × b × c)

So, each term has the variable b.

STEP 2:  Expand the multiplication of the numbers in order to find the common coefficient.

(5 × a × b) + (15 × b) + (10 × b × c) =

(5 × a × b) + [(5 × 3) × b] + [(5 × 2) × b × c]

From the expanded expression, we can easily see that each term has a coefficient of 5.

STEP 3:  Place the common factor at the front of each expanded term.

(5 × a × b) + [(5 × 3) × b] + [(5 × 2) × b × c] =

[5b(a)] + [5b(3)] + [5b(2 × c)]

STEP 4: Place the greatest common factor at the front of the new expression.

[5b(a)] + [5b(3)] + [5b(2 × c)] =

5b[a + (3) + (2 × c)] =

5b(a + 3 + 2c) =

5b(a + 2c + 3)

Other Practice Problems

In previous posts, we have looked finding the greatest common factor and exponent laws.

The exercises on other posts in this blog are advanced-level problems that require knowledge of all of these skills.

So, you should look at the following posts before you attempt the advanced math problems.

Finding the greatest common factor

Factoring numbers

Factoring variables

You should also look at our posts on factoring quadratics and FOIL.

Free algebra practice test

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Factoring Quadratics from Standard Form to Factored Form

You will need to know how to change quadratics from the standard from to the factored form for your algebra test.

Factoring quadratics is also called “unfoiling” since it is like doing the FOIL method in reverse.

Exercises on Standard Form to Factored Form

Instructions: Factor the following quadratic equations using the steps provided in the next section. Then check your work with the solutions provided after each question. First, you may want to view the methods for factoring, which are provided below the quiz.

[WpProQuiz 14]

Factoring Quadratics – Box Method

Factoring quadratics is sometimes done using the box method.

The box method involves making a grid and then multiplying the numbers inside the boxes.

Factoring quadratics with the box method can be a bit confusing.

However, if you are confident with algebra and would like to try it, you can attempt factoring quadratics with the box method.

You should use the simplified method for factoring quadratics if you are not as confident with algebra.

We illustrate the simplified method for factoring quadratics in the following sections of this page.

Factoring Quadratics when a =1

When a = 1, you will see the quadratic equation in the following form:

x² + bx + c = 0

So, lets look at an example equation in this format and try to factor it.

 Factor the following quadratic:    x² + 3x – 28 = 0

STEP 1:

Look at the third term of the equation, which is –28 in our problem.

We need to find two integers that equal –28 when they are multiplied.

–1 × 28 = –28

–2 × 14 = –28

–4 × 7 = –28

–7 × 4 = –28

–14 × 2 = –28

–28 × 1 = –28

STEP 2:

Then look at the second term of the equation, which is 3x.

We need to be sure that our two integers also equal 3 when they are added.

–1 + 28 = 27

–2 + 14 = 12

–4 + 7 = 3

–7 + 4 = –3

–14 + 2 = –12

–28 + 1 = –27

So, we need the two integers –4 and 7 to factor our quadratic.

STEP 3:

Finally, take the two integers that meet both conditions above, and set out the factored form of the quadratic equation.

(x – 4)(x + 7) = 0

Standard Form to Factored Form – Example

You will remember from our posts on quadratic forms and the quadratic equation, that the standard quadratic equation is:

ax² + bx + c = 0

Now let’s look at a more difficult example.

 Factor the following quadratic:   2x² + 5x – 12 = 0

STEP 1:

Look at the integer in the first term of the equation, which is 2 in our problem.

We need to find two integers that equal 2 when they are multiplied.

Since 2 is a prime number, there is only one way to do the multiplication:

2 × 1 = 2

STEP 2:

Look at the third term of the equation, which is –12 in our problem.

We ignore the sign of the term since we deal with this in the last step.

So, we need to find two integers that equal 12 when they are multiplied.

Taking the order of the integers into account, we have:

1 × 12 = 12

2 × 6 = 12

3 × 4 = 12

4 × 3 = 12

6 × 2 = 12

12 × 1 = 12

STEP 3:

Take the first number (2) from the multiplication in step 1 (which was 2 × 1 = 2).

Take this number (2) and multiply by the first number from each of the multiplications in step 2.

2 × 1 = 2

2 × 2 = 4

2 × 3 = 6

2 × 4 = 8

2 × 6 = 12

2 × 12 = 24

STEP 4:

Take the second number (1) from the multiplication in step 1 (which was 2 × 1 = 2).

Take this number (1) and multiply by the second number from each of the multiplications in step 2.

1 × 12 = 12

1 × 6 = 6

1 × 4 = 4

1 × 3 = 3

1 × 2 = 2

1 × 1 = 1

STEP 5:

We need to find a result from step 3 and subtract from it a result from step 4 so that the difference will equal variable b.

Step 3 result:  2 × 4 = 8

Step 4 result:  1 × 3 = 3

Difference:  8 – 3 = 5

The difference must be equal to variable b in our original equation, which was 2x² + 5x – 12 = 0.

STEP 6:

Set out the possible factored forms of the quadratic equation.

The signs you use in factoring quadratics will depend on the signs in the original equation.

Rules for factoring quadratics:

Look at the signs on the second and third terms of the equation.

If you have + c and + b, then factor like this:   (ax  +   )(x   +   )

If you have + c and – b, then factor like this:   (ax  –   )(x   –   )

If you have –  c and + b or – b , you have to use trial and error to work out the correct combination of signs in factoring quadratics. Your result will be like this:   (nx  –  )(x   +   ) or this:  (nx  +   )(x   –   )

First

Take the two integers that we have multiplied in step 3 above (2 and 4), as well as the second number from step 4 (3).

Outside

Integers used in step 3 above:    2 × 4 = 8 (These two numbers go on the “outside” positions of the sets of parentheses.)

Inside

Second number from step 4:   1 × 3 = 3 (This number goes on the “inside” position of the first set of parentheses.)

Equation

Our original equation was: 2x² + 5x – 12 = 0, so b = + 5.

Here we have a negative sign on c and a positive sign on b.

So the possible factored forms are:  (2x – 3)(x + 4)  or  (2x + 3)(x – 4)

(2x – 3)(x + 4) = 2x² + 5x – 12 = 0

(2x + 3)(x – 4) = 2x² – 5x – 12 = 0

So our factored quadratic is: (2x – 3)(x + 4) = 0

More free algebra exercises

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Examples of Perfect Cubes

Here are a couple of examples of perfect cubes:

3³ = 27 and 2³ = 8

As you can see from the examples, perfect cubes are the third power of a number or variable. So, you get a perfect cube when you multiply a number by itself and then multiply that answer again by the first number.

Perfect Cubes – Exercises

Instructions: Factor the following algebraic expressions. To get the shortcut form for your answers, please refer to the examples below this exercise.

[WpProQuiz 13]

Perfect Cubes and Binomials

The difference of two perfect cubes is a binomial.

You will remember that a binomial is an algebraic expression that has two terms.

Here is one of our examples of an expression that shows the difference of two perfect cubes:

x³ – y³

The sum of two perfect cubes is also a binomial.

For example, we can change the expression above to show the sum of perfect cubes by using the plus sign.

x³ + y³

Difference of Two Perfect Cubes

You will need to know how to factor the difference of perfect cubes on your examination.

The second of our examples of an algebraic expression for the difference of perfect cubes is shown below.

x³ – y³

The form for factoring the difference of perfect cubes is as follows:

x³ – y³ =

(x – y)(x² + xy + y²)

You should memorize the above form for your exam.

Proof of Difference of Two Cubes

We can prove the above form by applying the distributive property of multiplication.

Multiply the first term in the first set of parentheses by all of the terms in the second set of parentheses.

x³ – y³ =

(x – y)(x² + xy + y²) =

[(x × x²) + (x × xy) + (x × y²)] – [(y) × (x² + xy + y²)] =

[(x × x²) + (x × xy) + (x × y²)] + [(–y) × (x² + xy + y²)] =

[(x³) + (x²y) + (xy²)] + [(–y) × (x² + xy + y²)] =

(x³ + x²y + xy²) × [(–y) + (x² + xy + y²)]

Then multiply the second term in the first set of parentheses by all of the terms in the second set of parentheses.

Remember to be careful with the plus or minus sign.

We are subtracting the terms in the first set of parentheses, so we need to multiply all of the terms in the second set of parentheses by (–y).

(x³ + x²y + xy²) + [(–y) × (x² + xy + y²)] =

(x³ + x²y + xy²) + [(–y × x²) + (–y × xy) + (–y × y²)] =

(x³ + x²y + xy²) + [(–yx²) + (–xy²) + (–y³)] =

(x³ + x²y + xy²) + [(–x²y) + (–xy²) + (–y³]) =

(x³ + x²y + xy²) + (–x²y + –xy² + –y³)

Then group like terms together and simplify to finish the proof.

(x³ + x²y + xy²) + (–x²y + –xy² + –y³) =

x³ + x²y + xy² – x²y – xy² – y³ =

x³ + x²y – x²y + xy² – xy² – y³ =

x³ + x²y – x²y + xy² – xy² – y³ =

x³ – y³

Sum of Two Perfect Cubes

You will need to know how to factor the sum of perfect cubes for your math test.

An algebraic expression for the sum of perfect cubes is as follows:

x³ + y³

The form for factoring the sum of perfect cubes is:

x³ + y³ =

(x + y)(x² – xy + y²)

You should also know the above above form by heart for your math test.

Proof of Sum of Two Cubes

We can also prove the above form for the sum of perfect cubes with the distributive property of multiplication.

Remember to multiply the first term in the first set of parentheses by all of the terms in the second set of parentheses.

x³ + y³ =

(x + y)(x² – xy + y²) =

[(x × x²) + (x × –xy) + (x × y²)] + [(y) × (x² – xy + y²)] =

[(x³) + (–x²y) + (xy²)] + [(y) × (x² – xy + y²)] =

(x³ + –x²y + xy²) × [(y) + (x² – xy + y²)]

Then multiply the second term in the first set of parentheses by all of the terms in the second set of parentheses.

(x³ + –x²y + xy²) + [(y) × (x² – xy + y²)] =

(x³ – x²y + xy²) + [(y × x²) + (y × –xy) + (y × y²)] =

(x³ – x²y + xy²) + [(yx²) + (–xy²) + (y³)] =

(x³ – x²y + xy²) + (x²y + –xy² + y³)

Then group like terms together and simplify to finish the proof.

(x³ – x²y + xy²) + (x²y + –xy² + y³) =

x³ – x²y + xy² + x²y – xy² + y³ =

x³ – x²y + x²y + xy² – xy² + y³ =

x³ – x²y + x²y + xy² – xy² + y³ =

x³ + y³

You may also want to look at our practice problems on factoring perfect squares.

You should also try our free practice problems on factoring and algebra.

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What is Synthetic Division?

Synthetic division is used to divide polynomials by first degree binomials.

A first degree binomial is an expression that has a variable that is not raised to any power. So, x + 5 is an example of a first degree binomial.

Free exam sample questions are provided at the end of this page.

Synthetic Division – Explanation of Terminology

To perform synthetic division, you need to identify the coefficients and constants in the polynomial.

You will remember that a polynomial is an algebraic expression that has two or more terms.

Terms can contain either numbers or variables or both.

A coefficient is the number part of a term.

A constant is a number on its own. Constants are usually placed at the end of an algebraic expression.

Synthetic Division – Example

Consider the following polynomial:

x⁴ + 5x³ – 2x² – 28x – 12

In the second term, which is 5x³, the coefficient is 5, for example.

Now look at this practice problem.

Example:  Divide the polynomial x⁴ + 5x³ – 2x² – 28x – 12 by the first degree binomial x + 3. Use synthetic division to find your answer.

STEP 1: Identify the coefficients and constants.

Remember that if you have a variable without a number in front of it, the coefficient is 1.

x⁴ + 5x³ – 2x² – 28x – 12 =

(1 × x⁴) + (5 × x³) – (2 × x²) – (28 × x) – 12 =

(1 × x⁴) + (5 × x³) + (–2 × x²) + (–28 × x) + (–12)

Our coefficients and constants are the numbers before each of the multiplication signs.

The coefficients are 1, 5, –2, and –28.

The constant is –12.

STEP 2: Identify the constant in the binomial and find its opposite.

We are going to divide by the binomial x + 3.

So, the constant in our binomial is 3.

The opposite of the constant is –3.

STEP 3: Write the problem out in synthetic division format, placing the opposite of the constant in an upside-down division sign.

Divide the numbers from step 1 by the opposite of the constant from step 2.

–3|    1    5    –2   –28   –12

STEP 4: Perform the synthetic division.

Bring down the 1 and multiply it by –3. Put that result under the 5. Then add 5 and –3 to get 2.

–3|  1     5    –2   –28   –12

     –3   –6      24      12

1     2

STEP 5: Repeat the operations from step 4 until you find the remainder.

–3|   1     5    –2   –28   –12

       –3    –6     24     12

1       2    –8     –4       0

Now try the practice problems in the next section.

Synthetic Division – Exercises

Instructions:  Perform synthetic division to find the remainder for the following problems. If you do not know how to perform synthetic division, please see the example above before completing the exercises.

1)   (x³ + 2x² + x + 5) ÷ (x + 2)

2)   (x³ – 6x² – 2x + 14) ÷ (x – 6)

3)   (x³ + x² – x – 4) ÷ (x – 2)

4)   (x⁴ + 5x³ + 10x² + 14x + 10) ÷ (x + 3)

5)   (x⁴ – 9x³ + 13x² + 4x + 21) ÷ (x – 7)

Synthetic Division – Answers

1)   The remainder is 3.

Our question was: (x³ + 2x² + x + 5) ÷ (x + 2)

The opposite of the constant in our binomial is –2.

Our coefficients and constant are:  1     2      1     5

Set up the synthetic division to solve as shown below.

–2|  1     2      1     5

     –2     0    –2 

 1     0      1      3

2)   The remainder is 2.

Our question was: (x³ – 6x² – 2x + 14) ÷ (x – 6)

The opposite of the constant in our binomial is 6.

Our coefficients and constant are: 1   –6   –2    14

Set up the synthetic division to solve as shown below.

6|  1     –6   –2    14

       6      0   –12

1     0    –2       2

3)   The remainder is 6.

Our question was: (x³ + x² – x – 4) ÷ (x – 2)

The opposite of the constant in our binomial is 2.

Our coefficients and constant are:  1     1    –1   –4

Set up the synthetic division to solve as shown below.

2|  1     1    –1    –4

      2      6     10

1     3      5       6

4)   The remainder is 4.

Our question was: (x⁴ + 5x³ + 10x² + 14x + 10) ÷ (x + 3)

The opposite of the constant in our binomial is –3.

Our coefficients and constant are:   1     5    10   14    10

Set up the synthetic division to solve as shown below.

–3|  1     5     10    14    10

     –3   –6   –12   –6

1     2      4       2      4

5)   The remainder is 0.

Our question was: (x⁴ – 9x³ + 13x² + 4x + 21) ÷ (x – 7)

The opposite of the constant in our binomial is 7.

Our coefficients and constant are:  1   –9    13    4     21

Set up the synthetic division to solve as shown below.

7|    1     –9     13     4     21

          7   –14   –7  –21

1     –2     –1    –3      0

You should also view our posts on the remainder theorem and factoring.

What are Perfect Squares?

Perfect squares are the result of multiplying a number or variable by itself.

For example, 4 and 9 are perfect squares, because they are the products of two times two and three times three, respectively.

2 × 2 = 4

3 × 3 = 9

To see the free examples, please go to the next section of this page.

Perfect Squares in Algebra

In algebra, perfect squares are the product of two equal variables.

For instance, a² and b² are examples of this, because they are the products of a times a and b times b, respectively.

a × a = a²

b × b = b²

Binomials are two term expressions.

So, the expression a² + b² is an example of a binomial.

One type of binomial is the difference between two perfect squares.

The expression a² – b² is an example of the difference between two perfect squares.

Factoring the Difference of Two Perfect Squares

If the two terms of a binomial are perfect squares and the terms are subtracted, the binomial can be factored.

The factored form is as follows:

a² – b² = (a + b)(a – b)

In other words, the variables in the first set of parentheses will be added to each other and the variables in the second set of parentheses will be subtracted from each other when factoring the difference between two perfect squares.

We can prove the form above by multiplying each of the terms and canceling out as shown below.

a² – b² =

(a + b)(a – b) =

(a × a) + (a × –b) + (b × a) + (b × –b) =

a² + –ab + ab + –b² =

a² – ab + ab – b² =

a² – ab + ab – b² =

a² – b²

In order to perform the factoring, you will need to be acquainted with the laws of square roots.

You will also need to understand how to do the FOIL method on quadratics.

You may also want to try our other problems on factoring and algebra.