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In mathematics, a factor is an entity(a number, a polynomial, a function, a matrix, etc) that divides another entity, leaving no remainder. Factorization or factoring is defined as the breaking or decomposition of an entity. Factorization provides a standard method for solving higher-order equations and simplifying complicated expressions. It is also useful in graphing functions.

Let’s understand what the factorization of polynomials is and what are the different methods of factoring the polynomials.

## What is Factoring of Polynomials?

Factoring polynomials is actually the reverse process of the multiplication of polynomials.

For example $(x + 1)(x – 2)(x + 3) = x^3 + 2x^2 – 5x – 6$, then writing $x^3 + 2x^2 – 5x – 6$ in the form $(x + 1)(x – 2)(x + 3)$ is factoring of the polynomial $x^3 + 2x^2 – 5x – 6$.

Factoring polynomials means decomposing the given polynomial into a product of two or more polynomials. Factoring polynomials help in simplifying the polynomials easily. It also helps in finding the values of the variables of the given expression or finding the zeros of the polynomial expression.

For example, if we want to find the zeroes or the roots of the polynomial $x^3 + 2x^2 – 5x – 6$, then we factor the polynomial as $(x + 1)(x – 2)(x + 3)$ and then can find the zeroes or the roots easily by equating each of the terms to zero.

$(x + 1) = 0 => x = -1$, $(x – 2) = 0 => x = 2$, and $(x + 3) = 0 => x = -3$. So, we get zeroes of the polynomial $x^3 + 2x^2 – 5x – 6$ as $x = -1$, $x = 2$, and $x = -3$.

The process of decomposing polynomials into their prime factors and writing the polynomials as a product of their prime factors is called factorization of polynomials.

## How to Find Factors of a Polynomial?

The steps followed to find factors of a polynomial are

- Factor out if there is a factor common to all the terms of the polynomial.
- Identify the appropriate method for factoring polynomials. You can use regrouping or algebraic identities to find the factors of the polynomial.
- Write the polynomial as the product of its factors.

## Methods of Factoring Polynomials

There are various methods of factoring polynomials, based on the expression. The method of factorization depends on the degree of the polynomial and the number of variables included in the expression. The four most commonly used methods of factoring polynomials are

- Method of Common Factors
- Grouping Method
- Factoring by splitting terms
- Factoring Using Algebraic Identities

### 1. Method of Common Factors

This is the simplest method of factoring a polynomial. In this method, factorization is done by taking common factors of each of the terms of the given expression.

The steps involved in this method are

**Step 1:** Find the GCF(Greatest Common Factor) of all the terms in the polynomial.

**Step 2:** Express each term as a product of the GCF and another factor.

**Step 3:** Use the distributive property to factor out the GCF.

### Examples

**Ex 1:** Factorize $3x^5 – 12x^3$

First of all find GCF of each term of $3x^5 – 12x^3$

So, the GCF of $3x^5$ and $12x^3$ is $3 \times x \times x \times x = 3x^{3}$.

Now, expressing each term as a product of $3x^{3}$.

$3x^5 = 3x^3 \times x^2$ and $12x^3 = 3x^3 \times 4$.

So, the polynomial can be written as $3x^3 \times x^2 – 3x^3 \times 4$

Now, in the last step factor out the GCF $3x^3 \left(x^2 – 4 \right)$.

**Ex 2:** Factorize $10x^2 + 25x + 15$

$10x^2 + 25x + 15 = 5 \times x \times x + 5 \times 5 \times x + 5 \times 3$

$=5 \left(2x^2 + 5x + 3 \right)$

**Ex 3:** Factorize $3xy^2 + 12x^2y$

$3xy^2 + 12x^2y = 3 \times x \times y \times y + 3 \times 4 \times x \times x \times y$

$= 3xy \times y + 3xy \times 4x = 3xy \left(y + 4x \right)$

**Ex 4:** $4pq^2 – 20pq – 8p^2q$

$4pq^2 – 20pq – 8p^2q = 4 \times 2 \times p \times q \times q – 4 \times 5 \times p \times q – 4 \times 2 \times p \times p \times q$

$= 4pq \times q + 4pq \times (-5) + 4pq \times (-2p)$

$=4pq(q – 5 – 2p)$

### 2. Grouping Method

Factorization by grouping means that before factoring, we must first group the terms with common factors and then take the common out of them.

The steps involved in this method are

**Step 1:** Check whether there is any common factor between two pairs of adjacent terms (like $1^{st}$ and $2^{nd}$, $3^{rd}$ and $4^{th}$, $5^{th}$ and $6^{th}$, and so on)

**Step 2**: If there are no common factors between two adjacent terms, then rearrange the terms

**Step 3:** Group the first two terms together and then the last two terms together

**Step 4:** Factor out a GCF from each separate binomial

**Step 5:** Factor out the common binomial.

### Examples

**Ex 1:** $3xy + 21x – 2y – 14$

For first two terms, GCF is $3x$ and for next two terms GCF is $-2$

Now, group first, second and third, fourth terms

$3xy + 21x – 2y – 14 = (3xy + 21x) + (- 2y – 14)$

Now take out the common factor of each pair

$(3xy + 21x) + (- 2y – 14) = 3x(y + 7) – 2(y + 7)$

Each of the two terms in the resultant expression has a common term $(y + 7)$.

Take out $(y + 7)$ from each of the two terms.

$3x(y + 7) – 2(y + 7) = (y + 7)(3x – 2)$

Therefore, factorization of $3xy + 21x – 2y – 14$ is $(y + 7)(3x – 2)$ and the prime factors of $3xy + 21x – 2y – 14$ are $(y + 7)$ and $(3x – 2)$.

**Ex 2: **Factorize $a^3 – 2a^2 + 5a – 10$

$a^3 – 2a^2 + 5a – 10 = (a^3 – 2a^2) + (5a – 10)$

$(a^3 – 2a^2) + (5a – 10) = a^2(a – 2) + 5(a – 2) = (a – 2)(a^2 – 5)$

Therefore, factorization of $a^3 – 2a^2 + 5a – 10$ is $(a – 2)(a^2 – 5)$ and the prime factors are $(a – 2)$ and $(a^2 – 5)$.

**Ex 3:** Factorize $-b^2 + 12ac + 6ab – 2bc$

Notice that there is no common factor between the first two terms, i.e., $-b^2$ and $12ac$.

So rearrange the terms.

$-b^2 + 12ac + 6ab – 2bc = -b^2 + 6ab + 12ac – 2bc$

$-b^2 + 6ab + 12ac – 2bc = (-b^2 + 6ab) + (12ac – 2bc)$

$= -b(b – 6a) – 2c(-6a + b) = -b(b – 6a) – 2c(b – 6a)$

$=(b – 6a)(-b – 2c) = (b – 6a)(-(b + 2c)) = -(b – 6a)(b + 2c) = (6a – b)(b + 2c)$

Therefore, factorization of $-b^2 + 12ac + 6ab – 2bc$ is $(6a – b)(b + 2c)$ and the prime factors are $(6a – b)$ and $(b + 2c)$.

**Ex 4:** Factorize $amx^2 + bmxy – anxy – bny^2$

$amx^2 + bmxy – anxy – bny^2 = (amx^2 + bmxy) + (– anxy – bny^2)$

$= mx(ax + by) – ny(ax + by) = (ax + by)(mx – ny)$

### 3. Factoring by Splitting Terms

We can make pairs of terms only when a polynomial has an even number of terms. But when a polynomial has an odd number of terms, then we split any of the terms of the polynomial to get an even number of terms. This method of splitting a term for factoring is called factoring by splitting terms.

Once we get an even number of terms, we use the grouping method to group the terms and take out the common factor.

The method of factoring by splitting the terms is commonly used to factorize the trinomials, especially the quadratic polynomials, where it is referred to as **splitting the middle-term** method.

These are the different cases that can exist if a polynomial is a trinomial.

- Quadratic polynomials of the form $ax^2 + bx + c$
- Polynomials reducible to the form $ax^2 + bx + c$
- Polynomials that are not quadratic polynomials

#### 3. 1 Factoring of Quadratic Polynomials of the Form $ax^2 + bx + c$

The steps involved in factoring of quadratic polynomials of the form $ax^2 + bx + c$ are as follows

**Step 1:** Find two numbers $p$ and $q$ such that $b = p + q$ and $ac = pq$

**Step 2:** Replace $bx$ by $px + qx$, i.e, split $b$ into two numbers $p$ and $q$

**Step 3:** Make pairs of the adjacent terms

**Step 4:** Take a common factor from each of the pairs of terms

**Step 5**: Express polynomial as a product of prime factors

#### Examples

**Ex 1:** Factorize $x^{2} + 5x + 6$

Comparing $x^{2} + 5x + 6$ with $ax^{2} + bx + c$, we get $a = 1$, $b = 5$, and $c = 6$.

$b = 5 = 3 + 2$ and $ac = 1 \times 6 = 6 = 3 \times 2$

Therefore, $x^{2} + 5x + 6 = x^{2} + 3x + 2x + 6 = (x^{2} + 3x) + (2x + 6)$

$= x(x + 3) + 2(x + 3) = (x + 3)(x + 2)$

Therefore, factorization of $x^{2} + 5x + 6 = (x + 3)(x + 2)$ and its prime factors are $(x + 3)$ and $(x + 2)$.

**Ex 2:** Factorize $x^{2} + x – 12$

Comparing $x^{2} + x – 12$ with $ax^{2} + bx + c$, we get $a = 1$, $b = 1$, and $c = -12$.

$b = 1 = 4 – 3$ and $ac = 1 \times (-12) = -12 = 4 \times (-3)$

Therefore, $x^{2} + x – 12 = x^{2} + 4x – 3x – 12 = x(x + 4) – 3(x + 4)$

$=(x + 4)(x – 3)$

Therefore, factorization of $x^{2} + x – 12 = (x + 4)(x – 3)$ and its prime factors are $(x + 4)$ and $(x – 3)$.

**Ex 3:** Factorize $6x^2 + 17x + 12$

Comparing $6x^{2} + 17x + 12$ with $ax^{2} + bx + c$, we get $a = 6$, $b = 17$, and $c = 12$.

$b = 17 = 8 + 9$ and $ac = 6 \times 12 = 72 = 8 \times 9$

Therefore, $6x^{2} + 17x + 12 = 6x^{2} + 8x + 9x + 12$

$= 2x(3x + 4) + 3(3x + 4) = (3x + 4)(2x + 3)$

Therefore, $6x^2 + 17x + 12 = (3x + 4)(2x + 3)$

**Ex 4:** Factorize $x^2 + 3 \sqrt{3}x + 6$

Comparing $x^2 + 3 \sqrt{3}x + 6$ with $ax^{2} + bx + c$, we get $a = 1$, $b = 3 \sqrt{3}$, and $c = 6$.

$b = 3 \sqrt{3} = 2 \sqrt{3} + \sqrt{3}$ and $ac = 1 \times 3 \sqrt{3} = 3 \sqrt{3} = 2 \sqrt{3} + \sqrt{3}$.

Therefore, $x^2 + 3 \sqrt{3}x + 6 = x^2 + \sqrt{3}x + 2 \sqrt{3}x + 6$

$= x \left(x + \sqrt{3} \right) + 2 \sqrt{3} \left(x + \sqrt{3} \right)$

$= \left(x + \sqrt{3} \right) \left(x + 2 \sqrt{3} \right)$

Therefore, $x^2 + 3 \sqrt{3}x + 6 = \left(x + \sqrt{3} \right) \left(x + 2 \sqrt{3} \right)$.

#### 3. 2 Factoring of Polynomials Reducible to the Form $ax^2 + bx + c$

We often see polynomials that do not look in the form $ax^2 + bx + c$, but can be converted or reduced to the form $ax^2 + bx + c$.

Let’s take some examples of such polynomials.

#### Examples

**Ex 1:** Factorize $ \left(a^2 – 3a \right)^2 – 5 \left(a^2 – 3a \right) + 4$

The given polynomial is $ \left(a^2 – 3a \right)^2 – 5 \left(a^2 – 3a \right) + 4$

Observe that the expression $\left(a^2 – 3a \right)$ is repeated two times – one with power $2$ and the other with power $1$.

Let’s replace $\left(a^2 – 3a \right)$ with $u$.

So, we have $ \left(a^2 – 3a \right)^2 – 5 \left(a^2 – 3a \right) + 4 = u^2 – 5u + 4$ which is in the form $ax^2 + bx + c$, where $a = 1$, $b = -5$ and $c = 4$.

$u^2 – 5u + 4 = u^2 – u – 4u + 4 = u(u – 1) – 4(u -1) = (u – 1)(u – 4)$

Now, replace $u$ with the original expression $\left(a^2 – 3a \right)$

$\left(a^2 – 3a – 1 \right) \left(a^2 – 3a – 4 \right)$

Therefore, factorization of $ \left(a^2 – 3a \right)^2 – 5 \left(a^2 – 3a \right) + 4$ is $\left(a^2 – 3a – 1 \right) \left(a^2 – 3a – 4 \right)$.

**Ex 2:** Factorize $2x^{4} + 9x^2 + 14$

$x^{4} + 9x^2 + 14$ also is not in the form $ax^2 + bx + c$

Now, let’s replace $x^2$ with $u$

$x^{4} + 9x^2 + 14 = \left(x^2 \right)^2 + 9x^2 + 14 = u^2 + 9u + 14$

$= u^2 + 2u + 7u + 14 = \left(u^2 + 2u \right) + \left(7u + 14 \right)$

$u\left(u + 2 \right) + 7\left(u + 2 \right) = (u + 2)(u + 7)$

$= \left(x^2 + 2 \right) \left(x^2 + 7 \right)$

Therefore, $2x^{4} + 9x^2 + 14 = \left(x^2 + 2 \right) \left(x^2 + 7 \right)$.

#### 3.3 Factoring of Trinomials Which are Not Quadratic Polynomials

Let’s consider some examples to understand the factorization of these types of polynomials.

#### Examples

**Ex 1:** If $x^2 + px + q = (x + a) (x + b)$, then factorize $x^2 + pxy + qy^2$.

We have $x^2 + px + q = (x + a) (x + b)$

$=> x^2 + px + q = x^2 + x(a + b) + ab$

On equating the coefficients of like powers of $x$, we get

$p = a + b$ and $q = ab$

Therefore, $x^2 + pxy + qy^2 = x^2 + (a + b)xy + aby^2$

$= \left(x^2 + axy \right) + \left(bxy + aby^2 \right)$

$= x(x + ay) + by(x + ay)$

$= (x + ay) (x + by)$

**Ex 2:** Factorize the following expression $x^2y^2 – xy – 72$

In order to factorize $x^2y^2 – xy – 72$, we have to find two numbers $p$ and $q$ such that $p + q = -1$ and $pq = -72$

These numbers are $-9$ and $8$.

$-9 + 8 = -1$ and $-9 \times 8 = -72$.

So, we write the middle term $-xy$ of $x^2y^2 – xy – 72$ as $-9xy + 8xy$, so that we have

$x^2y^2 – xy – 72 = x^2y^2 – 9xy + 8xy – 72$

$= \left(x^2y^2 – 9xy \right) + \left(8xy – 72 \right)$

$= xy (xy – 9) + 8 (xy – 9)$

$= (xy – 9) (xy + 8)$

Therefore, factorization of $x^2y^2 – xy – 72$ is $= (xy – 9) (xy + 8)$.

### 4. Factoring Using Algebraic Identities

The process of factoring polynomials can be easily performed using algebraic identities. When the given polynomial represents any algebraic identities, we can use that identity to factorize the polynomial. The identities that are used in factoring polynomials are

- $a^2 + 2ab + b^2 = (a + b)^2$
- $a^2 – 2ab + b^2 = (a – b)^2$
- $a^2 – b^2 = (a + b)(a – b)$
- $a^3 – b^3 = \left(a – b \right) \left(a^2 + ab + b^2 \right)$
- $a^3 + b^3 = \left(a + b \right) \left(a^2 – ab + b^2 \right)$

### Examples

**Ex 1:** Factorize $4x^2 + 12x + 9$

$4x^2 + 12x + 9 = \left(2x \right)^2 + 2 \times 2x \times 3 + 3^2$, which is of the form $a^2 + 2ab + b^2$, where $a = 2x$ and $b = 3$.

Therefore, $4x^2 + 12x + 9$ is written as $(2x + 3)^{2} = (2x + 3)(2x + 3)$

Thus, factorization of $4x^2 + 12x + 9$ is $(2x + 3)(2x + 3)$

**Ex 2:** Factorize $25y^2 – 20xy + 4x^2$

$25y^2 – 20xy + 4x^2 = \left(5y \right)^2 – 2 \times 5y \times 2x + \left(2x \right)^2$, which is of the form $a^2 – 2ab + b^2$, where $a = 5y$ and $b = 2x$.

Therefore, $25y^2 – 20xy + 4x^2 = (5y – 2x)^2 = (5y – 2x)(5y – 2x)$.

Thus, factorization of $25y^2 – 20xy + 4x^2$ is $(5y – 2x)(5y – 2x)$.

**Ex 3:** Factorize $4x^2 – 9y^2$

$4x^2 – 9y^2 = \left(2x \right)^2 – \left(3y \right)^2$, which is of the form $a^2 – b^2$, where $a = 2x$ and $b = 3y$.

Therefore, $4x^2 – 9y^2 = (2x – 3y)(2x + 3y)$

**Ex 4:** Factorize $\frac{a^4}{9} – \frac{b^4}{16}$

$\frac{a^4}{9} – \frac{b^4}{16} = \left(\frac{a^2}{3} \right)^2 – \left(\frac{b^2}{4} \right)^2$

$\left(\frac{a^2}{3} \right)^2 – \left(\frac{b^2}{4} \right)^2$ is of the form $x^2 – y^2$, which can be written as $(x – y)(x + y)$, where $x = \frac{a^2}{3}$ and $y = \frac{b^2}{4}$

Therefore, $\left(\frac{a^2}{3} \right)^2 – \left(\frac{b^2}{4} \right)^2 = \left(\frac{a^2}{3} – \frac{b^2}{4} \right) \left(\frac{a^2}{3} + \frac{b^2}{4} \right)$

$\frac{a^2}{3} – \frac{b^2}{4}$ can again be written as $\left(\frac{a}{\sqrt{3}} \right)^2 – \left(\frac{b}{2} \right)^2 = \left(\frac{a}{\sqrt{3}} – \frac{b}{2} \right) \left(\frac{a}{\sqrt{3}} + \frac{b}{2}\right)$

Therefore, $\frac{a^4}{9} – \frac{b^4}{16} = \left(\frac{a}{\sqrt{3}} – \frac{b}{2} \right) \left(\frac{a}{\sqrt{3}} + \frac{b}{2}\right) \left(\frac{a^2}{3} + \frac{b^2}{4} \right)$

**Ex 5:** Factorize $8x^3 – 27$

$8x^3 – 27 = (2x)^3 – 3^3$ which is of the form $a^3 – b^3$, where $a = 2x$ and $b = 3$

Therefore, $8x^3 – 27 = (2x)^3 – 3^3 = \left(2x – 3 \right) \left(\left(2x \right)^2 + 2x \times 3b + 3^2 \right) = \left(2x – 3 \right) \left(4x^2 + 6bx + 9 \right)$

## Practice Problems

Factorize the following polynomials

- $54x^2 + 42x^3 – 30x^4$
- $2x^2yz + 2xy^2z + 4xyz$
- $30xy – 12x + 10y – 4$
- $z – 19 + 19xy – xyz$
- $100x^2 – 80xy + 16y^2$
- $16x^4 – y^4$
- $x^2 + 6x + 8$
- $49y^2 – 1$

## FAQs

### What is factoring polynomials?

Factoring polynomials means decomposing the given polynomial into a product of two or more polynomials. For example, $bx + ay + xy + ab$ can be written as $(x + a)(y + b)$ which is its factored form, and $(x + a)$ and $(y + b)$ are called the prime factors of $bx + ay + xy + ab$.

### What is the meaning of factoring polynomials by grouping?

Factoring polynomials by grouping means factoring the polynomial by the method of grouping that allows us to rearrange the terms of the expression, to easily identify and find factors of the polynomial expression.

### What are the four methods of factoring polynomials?

The four methods of factoring polynomials are

a) Method of Common Factors

b) Grouping Method

c) Factoring by splitting terms

d) Factoring Using Algebraic Identities

### How is the factor theorem useful in factoring polynomials?

The factor theorem is used to find the factors of an $n$-degree polynomial without actual division. If a value $x = a$ satisfies an $n$-degree polynomial $f(x)$, and $f(a) = 0$, then $(x – a)$ is a factor of the polynomial expression. Further, we can find a few factors using the factor theorem and the remaining can be found using the factorization of a quadratic equation.

## Conclusion

A factor is an entity(a number, a polynomial, a function, a matrix, etc) that divides another entity, leaving no remainder. Factorization or factoring is defined as the breaking or decomposition of an entity. The four most commonly used methods of factoring polynomials are a method of common factors, grouping method, factoring by splitting terms, and factoring using algebraic identities.

## Recommended Reading

- Factor Theorem of Polynomials – Definition, Proof & Examples
- Remainder Theorem of Polynomials – Definition, Proof & Examples
- What are Polynomials? (Definition, Types & Examples)
- Nature of Roots of Quadratic Equation(With Methods & Examples)
- Solving Quadratic Equations – Formulas, Tricks & Examples
- Quadratic Equation Definition (With Different Forms & Examples)
- Pair of Linear Equations in Two Variables(With Methods & Examples)
- Linear Equations in Two Variables – Definition, Types, and Graphs
- Linear Equations in One Variable – Graph & Method of Solving
- What are Algebraic Identities(With Definition, Types & Derivations)
- What is the Meaning of Equation – Definition, Types & Examples
- Division of Algebraic Expressions(With Methods & Examples)
- Multiplication of Algebraic Expressions(With Methods & Examples)
- Subtraction of Algebraic Expressions(With Methods & Examples)
- Addition of Algebraic Expressions(With Methods & Examples)
- What is Algebraic Expression(Definition, Formulas & Examples)
- What is Algebra – Definition, Basics & Examples
- What is Pattern in Math (Definition, Types & Examples)