In mathematics, we often come across with the concept of factorization or the process of factoring. But what factoring is all about ? It is a process in which a number or a mathematical expression is written in the form of product of other numbers or expressions. We can say that two or more numbers are said to be factors of a given number when they are multiplied together, the given number is obtained.

$6\ =\ 3\ \times\ 2$, so $3$ and $2$ are factors of $6$.

$18\ =\ 3\ \times\ 3 \times\ 2$, thus $3,\ 3$ and $2$ are factors of $18$.

The whole process of finding factors is known as factoring or factorizing or factorization. Smaller numbers are easier to factorize, while bigger numbers are little difficult. It is not complicated to say that $12\ =\ 4\ \times\ 3\ =\ 2\ \times\ 2 \times\ 3$ once we have knowledge of multiplication tables. But, finding factors of 240 would require more work than 12.

There are specific methods for finding factors of both numerical and algebraic expressions. The most simple expressions for factoring are obviously numerical expressions shown above. However in mathematics, we often factorize algebraic expressions too. There are various methods available for factoring different types of expressions and their formats. In this chapter, we are going to learn some common ways to factorize a variety of expressions.

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The binomials are the polynomials having two terms. These are the expressions made up of two terms. The methods of factoring binomials are discussed below.

If there is any number or factor common to both the terms, it is factored out of the expression and each term is divided by this factor.

$7a^{4}\ -\ 21a$

Here, $7a$ is common to both terms.

So, $7a^{4}\ -\ 21a\ =\ 7a(a^{3}\ -\ 3)$

If given binomial is in the form of difference of squares, then the following identity can be applied.

$p^{2}\ -\ q^{2}\ =\ (p\ +\ q)(p\ -\ q)$

$x^{4} - 16$

= $(x^{2})^{2} - 4^{2}$ ____ ( p$^{2}$ - q$^{2}$ form)

= $(x^{2} + 4)(x^{2} - 4)$

= $(x^{2} + 4)(x^{2} - 2^{2})$ ____ again (p$^{2}$ - q$^{2}$ form)

= $(x^{2} + 4)(x + 2)(x - 2)$

When a binomial is in the form of difference of cubes or sum of cubes, the following two identities are applied in order to factorize it.

$p^{3} - q^{3} = (p - q)(p^{2} + q^{2} + pq)$

$p^{3} + q^{3} = (p + q)(p^{2} + q^{2} - pq)$

$t^{3} + 27$

= $t^{3} + 3^{3}$

= $(t + 3)(t^{2} + 3^{2} - t.3)$

= $(t + 3)(t^{2} + 9 - 3t)$

The quadratic equations are the equations having following form px$^{2}$ + qx + r =0, provided that p $\neq$ 0. These equations can be factored using the following methods.

Take out if there is any factor common to all the terms of a quadratic equation.

According to this method, q the coefficient of x is split in two terms in such a way that their sum is equal to q and their product is equal to the product of p and r.

$4x^{2}\ +\ 9x\ +\ 2\ =\ 0$

$\Rightarrow$ $4x^{2}\ +\ 8x\ +\ x\ +\ 2\ =\ 0$

$\Rightarrow$ $4x(x\ +\ 2)\ +\ 1(x\ +\ 2)\ =\ 0$

$\Rightarrow$ $(x\ +\ 2)(4x\ +\ 1)\ =\ 0$

The quadratic formula is given by -

$x$ = $\frac{-q\ \pm\ \sqrt{q^{2}\ -4pr}}{2p}$

Substitute the values of p, q, r and calculate x. It will give two values, say $x_{1}$ and $x_{2}$.

The required factors would be (x - x$_{1}$) and (x - x$_{2}$).

Let us take above example.

Here, p = 4, q = 9, r = 2

Using quadratic formula,

$x$ = $\frac{-9 \pm \sqrt{9^{2}-4.4.2}}{2.4}$

$x$ = $\frac{-9 \pm \sqrt{81-32}}{8}$

$x$ = $\frac{-9 \pm 7}{8}$

$x_{1}$ = $\frac{-9 + 7}{8}$ = $\frac{-1}{4}$

and

$x_{2}$ = $\frac{-9\ - 7}{8}$ = $-2$

Factors are ($x$ + $\frac{1}{4}$) and $(x\ +\ 2)$

When there is nothing that factors of all the terms of an expression, one may divide them in groups and then factor out common terms from each group. One may think of factoring in pairs or even in groups of three or more terms.

$m^{2}\ -\ 4m\ +\ 6m\ -\ 24$

= $m(m\ -\ 4)\ +\ 6(m\ -\ 4)$

= $(m\ -\ 4)(m\ +\ 6)$

We have learnt above how to factorize a linear polynomials and quadratic polynomials. Lets understand here the method of factoring a higher-degree polynomial. Although all the above discussed methods can be applied to a given polynomial, still there is a particular long-division method which is used when other methods are not applicable.

Let us understand this with the help of an example.

$6x^{3}\ -\ 5x^{2}\ -\ 17x\ + 6$

First substitute random values for x (trial and error method) and check whether it gives zero or not. If yes, then that particular value will serve as one of the factors.

Putting $x\ =\ 1$, we get

$6\ -\ 5\ -\ 17\ +\ 6$ which is not zero.

Putting $x\ =\ 2$

$6(2)^{3}\ -\ 5(2)^{2}\ -\ 17(2)\ + 6$

= $48\ -\ 20\ -\ 34\ +\ 6\ =\ 0$

Therefore, x - 2 is one of the factors.

Now, divide the given polynomial by x - 2 using long division method.

$\frac{6x^{3} - 5x^{2} - 17x + 6}{x - 2}$ = $6x^{2} + 7x -3$

$6x^{3} - 5x^{2} - 17x + 6 = (x - 2) (6x^{2} + 7x -3)$

Now, we shall factorize quadratic equation so obtained by a suitable method.

= $(x - 2) (6x^{2} + 9x - 2x - 3)$

= $(x - 2) [3x(2x + 3) - 1(2x + 3)]$

= $(x - 2) (2x + 3) (3x - 1)$

A quadratic polynomial can also be factored using this method. After finding first factor by trial and error, one gets cubic polynomial as a result of long division. This polynomial can be further factored applying same method again. In this way, the higher-degree polynomials can also be factorized.

In mathematics, the reverse factoring is termed as FOIL. Its full form is

It is used to obtain a quadratic expression $(ax^{2} + bx + c)$ from its factored form which is $(x\ -\ a)\ (x\ -\ b)$

Reverse factoring states that multiply first terms, then outer terms, then inner terms and then last terms. Find the sum of all and simplify.

For instance :

(a - 5)(a + 4)

F : a. a = a$^{2}$

O : a . 4 = 4a

I : (-5).a = -5a

L : (-5) .4 = -20

Therefore,

$(a\ -\ 5)(a\ +\ 4)\ =\ a^{2}\ +\ 4a\ -\ 5a\ -\ 20\ =\ a^{2}\ -\ a\ -\ 20$

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