In mathematics, we learn algebraic expressions using the alphabet. We use symbols to perform calculations such as addition, subtraction, multiplication, and division.

Parentheses are often used in math calculations. When multiplying parentheses by parentheses, we use factoring formulas.

Once you learn factoring formulas, you will also be able to factorize them. Factorization is the way to convert polynomials into multiplication equations using parentheses. In calculating polynomials, factorization is more difficult than expanding an expression using factoring formula, so you need to understand how to calculate it.

Learning factoring formula and factorization together will make the math easier to understand. We will explain how to use the factoring formula and solve factorization problems.

Table of Contents

## Proof of Multiplication of Polynomials by the Distributive Property

In polynomial calculations, there are so many problems that require us to expand equations. Removing the parentheses so that the equation takes the form of addition and subtraction of monomials is called expanding (or developing). In expanding equations, multiplying two parentheses can be complicated.

Specifically, we do the following calculation.

- $(a+b)(c+d)$

In other words, multiply $(a+b)$ by $(c+d)$. How do we do this kind of polynomial calculation?

When multiplying two polynomials, expand the expression as follows.

Why is it possible to develop an equation in this way? In mathematics, we need to understand why. So, let’s try to prove that this formula is valid.

As we know from learning algebraic expressions, numbers can be replaced with letters. So let’s replace $(c+d)$ with $N$. Then $(a+b)(c+d)$ becomes $(a+b)N$. Therefore, by using the distributive property (or distributive law), we can calculate the following.

- $(a+b)N=aN+bN$

$N$ is $(c+d)$. So, we substitute $N=(c+d)$. The result is the following calculation.

$aN+bN$

$=a(c+d)+b(c+d)$

$=ac+ad+bc+bd$

From this result, we can prove that $(a+b)(c+d)=ac+ad+bc+bd$ holds. In the multiplication of two polynomials, we can use this formula to expand the equation.

However, we don’t have to prove the above formula every time when we do math calculations. So understand that this formula holds and remember how to calculate it. Learn how to expand the equation.

### Expanding the Equation Using the Four Factoring Formulas

By using the formula we just discussed, we can understand that the factoring formulas are established. Factoring formulas are formulas that are used to expand or factor.

There are four important factoring formulas. Each of them is as follows.

- $(x+a)(x+b)=x^2+(a+b)x+ab$
- $(x+a)^2=x^2+2ax+a^2$
- $(x-a)^2=x^2-2ax+a^2$
- $(x+a)(x-a)=x^2-a^2$

You must remember all four of these factoring formulas. If you don’t, you won’t be able to calculate polynomials efficiently.

It is easy to prove why these formulas are valid. Using the formulae discussed previously, we can create factoring formulas. We will check each of them.

**-Factoring Formula: Part 1**

$(x+a)(x+b)$ can be calculated as follows.

By expanding the equation, we can see that $(x+a)(x+b)=x^2+(a+b)x+ab$. This is the most frequently used factoring formula in mathematics and must be memorized.

For example, when solving the problem $(x+1)(x+2)$, we can use the factoring formula to expand the equation as follows.

$(x+1)(x+3)$

$=x^2+(1+3)x+1×3$

$=x^2+4x+3$

**-Factoring Formula: Part 2**

In the same way, we can get the factoring formula by expanding the polynomial. $(x+a)^2$ is as follows.

In the previous factoring formula, the numbers of $a$ and $b$ are different. This factoring formula, on the other hand, squares $(x+a)$. So the answer is $2ax$ instead of $(a+b)x$. Also, it is $a^2$, not $ab$.

**-Factoring Formula: Part 3**

In contrast, $(x-a)^2$ is as follows.

In contrast to $(x+a)^2$, the sign is negative in $(x-a)^2$. Although it is just a plus sign becoming a minus sign, it is important in factoring formulas. Note that $(x+a)^2$ has $a^2$ in the same way.

**-Factoring Formula: Part 4**

The factoring formula for $(x+a)(x-a)$ can be calculated as follows.

$-ax+ax=0$. Therefore, calculating $(x+a)(x-a)$ gives us $x^2-a^2$.

Compared to other factoring formulas, $(x+a)(x-a)=x^2-a^2$ is used less often. However, you must remember it as a factoring formula.

### If You Forget the Factoring Formulas, You Can Still Calculate

In some cases, you may forget the four factoring formulas. Even in that case, there is no problem. You just need to calculate $(a+b)(c+d)=ac+ad+bc+bd$.

For example, let’s say you want to calculate $(x-3)^2$. Using the factoring formulas mentioned earlier, you can quickly get the answer to $x^2-6x+9$. But even if you forget the factoring formulas, you can still expand the equation by multiplying each of them. It is as follows.

$(x-3)^2$

$=(x-3)(x-3)$

$=x^2-3x-3x+9$

$=x^2-6x+9$

If you remember the factoring formulas, you can skip these calculations and get the answer. However, even if you forget, you can still do the math to get the answer.

## Factorization: Making Parentheses Equations

After learning about equation development in math, the next step to study is factorization. It is the expansion of an equation that removes the parentheses. On the other hand, what is factorization? Factorization is the opposite of expansion of an equation. In other words, you make parentheses equations from non-parentheses equations.

The multiplication equation is called a factor. We decompose an equation that is made up of addition and subtraction so that it is a multiplication of factors by each other.

In factorization, we need to find common factors. For example, we get the following.

Factorization is more difficult than equation development. In equation expansion, we know clearly which numbers to multiply. In factorization, on the other hand, we need to find common factors. Since the common factors are not written, we have to think about what would be the common factors.

### Factorization Has Multiple Options

Also, in the case of factorization, many of the questions are presented using the factoring formulas. This means that if you don’t remember the factoring formulas, you cannot factor it. In equation expansion, even if you forget the factoring formulas, you can still get the answer. On the other hand, factorization requires you to remember the factoring formulas.

In addition, there are multiple options in the factorization using the factoring formulas. You have to find the best answer from among them.

For example, how can we factor the following formula?

- $=x^2+4x+3$

As we explained earlier, there are four factoring formulas. In other words, you must choose appropriate factoring formula from among the four options. Among the factoring formulas, the following formula is the one most often used in factorization.

- $(x+a)(x+b)=x^2+(a+b)x+ab$

Comparing $x^2+4x+3$ with $x^2+(a+b)x+ab$, it must be as follows.

- $a+b=4$
- $ab=3$

We have to find factors that satisfy these two. Note that there is the best way of factorization. That is, we need to focus on multiplication.

**-Addition and Subtraction Have Infinite Answers**

If we apply $a$ and $b$ so that $a+b=4$, there are infinitely many answers. For example, we have the following.

- $2+2=4$
- $10-6=4$
- $-100+104=4$

So when we factorize, we don’t focus on $a+b$ first. Rather, we focus on $ab$, which is multiplication. There are only two integers with $ab=3$.

- $1×3=3$
- $-1×(-3)=3$

That is, there are two possible answers when we factorize. Which of these is the correct answer? So, we need to find numbers that fit $a+b=4$. The result is as follows

- If $a=1,b=3$, then $1+3=4$
- If $a=-1,b=-3$, then $-1+(-3)=-4$

If $a=1,b=3$, then both $a+b=4$ and $ab=3$ are satisfied. Therefore, we can factorize it as follows.

Thus, in factorization, we have to figure out the correct answer from multiple candidates. In this example we used the factorization of a simple equation as an example. However, when the equation becomes more complex, the factorization becomes more difficult.

### Factorization Using the Four-Pattern Formula

As mentioned above, there are four factoring formulas. Therefore, the first time you learn factorization, you have to choose which of the four factoring formulas to use. Although there are other formulas for factorization, those should be learned when learning more challenging math, such as high school.

One way to distinguish between the factoring formulas used in factorization is to make sure that they contain powers: Of the four factoring formulas, the following three contain powers.

- $(x+a)^2=x^2+2ax+\textcolor{red}{a^2}$
- $(x-a)^2=x^2-2ax+\textcolor{red}{a^2}$
- $(x+a)(x-a)=x^2-\textcolor{red}{a^2}$

If powers are included, these factoring formulas may be used to factorize them. For example, how can we factor the following polynomial?

- $x^2+6x+9$
- $x^2-6x+9$

$3^2=9$. So we can see that these polynomials contain powers. Also, if we double 3, we get 6. Therefore, we can factorize them as follows.

If it can be powered, and further doubling the number yields $2ax$ (or $-2ax$), then it is possible to factor it using the factoring formula.

On the other hand, how can we factor the following case?

- $x^2-25$

It is $5^2=25$ and this equation contains a power. It also has a negative sign and the $2ax$ or $(a+b)x$ does not exist. If these conditions are met, the following factorization is possible.

In this way, we can factorize using the factoring formulas.

**-Factorization in Cases Where Power Cannot Be Applied**

In contrast, there are often cases where the three factoring formulas explained earlier cannot be used, such as not including powers in the equation. In such cases, the following factoring formula can be used to factorize the equation.

- $(x+a)(x+b)=x^2+(a+b)x+ab$

We have already discussed how to factorize using this multiplication formula. We will use this factoring formula most often in factorization. In any case, you should be able to answer from more than one option in the factorization.

## Exercises: Factoring Formulas and Factorization Calculations

**Q1:** Expand the following equation.

- $(x-3)(x+5)$
- $(x+4)^2-(2x+3)(x-6)$

**A1:** Answers.

The factoring formulas can be used to develop the equations.

**(a)**

$(x-3)(x+5)=x^2+2x-15$

**(b)**

$(x+4)^2-(2x+3)(x-6)$

$=x^2+8x+16-(2x^2-12x+3x-18)$

$=x^2+8x+16-2x^2+12x-3x+18$

$=-x^2+17x+34$

**Q2:** Factorize the following equation.

- $x^2-6x-16$
- $x^2y-49y$
- $(2x+3)(2x-3)-5x(x-2)$

**A2:** Answers.

You can factor the equation by using the factoring formulas. You may also factor the equation after expanding it.

**(a)**

In factorization, we first focus on multiplication. So let’s check the -16 of $x^2-6x\textcolor{red}{-16}$. Here are some candidates for numbers that can be multiplied to -16

- $1×(-16)=-16$
- $-1×16=-16$
- $2×(-8)=-16$
- $-2×8=-16$
- $4×(-4)=-16$

Which of these candidate numbers satisfies $a+b=-6$? If $a=2,b=-8$, then $a+b=-6$. Therefore, we can factorize it as follows.

- $x^2-6x-16=(x+2)(x-8)$

**(b)**

In factorization, we don’t always use the factoring formulas first. In some cases, the factoring formula is used after the common factors are put together.

In $x^2\textcolor{red}{y}-49\textcolor{red}{y}$, $y$ is the common factor. So the first factorization is as follows.

- $x^2y-49y=y(x^2-49)$

It is important to note that $(x^2-49)$ can be further factored. Therefore, we can factor it as follows.

- $y(x^2-49)=y(x+7)(x-7)$

**(c)**

Sometimes, after expanding the expression, we factor it out. So, do the following calculations first

$\textcolor{red}{(2x+3)(2x-3)}-5x(x-2)$

$=\textcolor{red}{4x^2-9}-5x^2+10x$

$=-x^2+10x-9$

After expanding the equation in this way, we factor it out. The factorization of this equation is as follows.

$-x^2+10x-9$

$=-(x^2-10x+9)$

$=-(x-1)(x-9)$

**Q3:** Factorize the following equation.

- $(x+2)(x-2)+y(2x+y)$

**A3:** Answers.

We must first expand the equation. Therefore, we have the following.

$(x+2)(x-2)+y(2x+y)$

$=x^2-4+2xy+y^2$

$=x^2+2xy+y^2-4$

How should we factor after this? In this equation, we separate into two parts as follows.

- $\textcolor{red}{x^2+2xy+y^2}-4$

If we focus on $x^2+2xy+y^2$, we can factorize it as follows

$\textcolor{red}{x^2+2xy+y^2}-4$

$=(x+y)^2-4$

However, the equation is not yet just a multiplication equation. That means we have to factor it further. So if we check the equation, we can notice that $4=2^2$.

Also, $(x+y)$ is hard to understand, so let’s replace it with $(x+y)=N$. Then we get the following.

- $\textcolor{red}{(x+y)}^2-4=\textcolor{red}{N}^2-2^2$

By substituting the equation in this way, notice that we can factorize it using the factoring formula. We can factor it as follows.

- $N^2–2^2=(N+2)(N-2)$

However, $N=x+y$. Then, by substituting $N=x+y$, we can complete the factorization as follows.

- $(N+2)(N-2)=(x+y+2)(x+y-2)$

## Use Factoring Formulas and Factorization to Solve Polynomial Problems

In mathematics, factoring formulas and factorization should be understood as a set of contents. So you need to be able to factor as well as develop equations using the factoring formulas.

The factoring formulas must be memorized; you have to be able to use the four factoring formulas. In equation development, you don’t need to memorize the factoring formulas to get an answer. On the other hand, in factorization, you have to remember the factoring formulas to get the answer.

Compared to equation expansion using the factoring formulas, factorization has many more candidate numbers and is more difficult to answer. However, in the meaning of using the factoring formulas, both equation expansion and factorization are the same. Also, in order to factorize, there are many cases where you need to expand the equation first, as shown in the exercise.

One of the most important contents in mathematics is equation expansion and factorization using factoring formulas. Make sure you understand how to use the factoring formulas and how to expand and factor the equations.