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The "factoring" play a vital role in mathematics as well in several other fields. Factoring or factorization is the process of finding factors of a number or an expression or an equation. The factors are defined as the integers or expression which when multiplied together result in original number or expression. When something is divided by any of its factors, the remainder is always zero. That means a factor completely divides its original number or expression.

Let us have a look at few examples.

The factors of $15$ are $3\ \times\ 5$ and that of $56$ are $2\ \times\ 2\ \times\ 2\ \times\ 7$.

Similarly, factors of expression $6x\ -\ 9$ are $3(x\ -\ 3)$.

The factoring of $2x^{2}\ +\ 6x^{3}\ +\ 5x^{7}\ +\ 15x^{8}$ can be performed in the following way.

= $x^{2}(2\ +\ 6x\ +5x^{5}\ +\ 15x^{6})$

= $x^{2}[2(1\ +\ 3x)\ +\ 5x^{5}(1\ +\ 3x)]$

= $x^{2}(1\ + 3x)\ (2\ +\ 5x^{5})$

There are different algebraic identities (such as - difference of squares, sum and difference of cubes etc.) which are utilized for factoring expressions. But, these identities generally do not apply on a standard quadratic equation. In this article, we shall learn about methods of factoring of quadratic equations and solved example problems based on those.

We know that an algebraic equation is a mathematical expression involving sign of equality (=) in it. A quadratic equation is kind of equation which can be written in the following form :

$ax^{2}\ +\ bx\ +\ c\ = 0$

Here, $x$ is a variable. Also, a, b and c are constant quantities in which "a" must not be equal to zero, otherwise the equation would become a linear one.

Similarly, a quadratic expression also attains the above form, but without equality sign, i.e.

$ax^{2}\ +\ bx\ +\ c$ with the condition a $\neq\ 0$.

A quadratic equation $ax^{2}\ +\ bx\ +\ c\ =\ 0$ can be factored quite easily using the quadratic formula, according to which the value of variable $x$ can be calculated in the following way :

$x$ = $\frac{-b\ \pm\ \sqrt{b^{2}\ -4ac}}{2a}$

This formula gives two roots of the given equation.

Let these roots are represented by $\alpha$ and $\beta$, then we would have

$\alpha$ = $\frac{-b\ +\ \sqrt{b^{2}\ -4ac}}{2a}$

and

$\beta$ = $\frac{-b\ - \sqrt{b^{2}\ -4ac}}{2a}$

Then, the factors of the above equation would be written as -

$(x\ -\ \alpha)\ (x\ -\ \beta)\ =\ 0$

Similarly, the factors of a quadratic expression $ax^{2}\ +\ bx\ +\ c$ are written as below -

$(x\ -\ \alpha)$ and $(x\ -\ \beta)$

The most common and easy way to factorize a quadratic polynomial is product-sum method. This method is discussed in the following steps.

Remember that a quadratic polynomial takes the form $ax^{2}\ +\ bx\ +\ c$.

Step 1 : First factor out a term that is common (if any) to the whole polynomial.

Step 2 : Determine the values of a, b and c. Also, calculate product of a and c.
For example -

$x^{2}\ +\ 5x\ +\ 6$
here, $a\ =\ 1,\ b\ =\ 5$ and $c\ =\ 6$
$ac\ =\ 1\ \times\ 6\ =\ 6$

Step 3 : Now, choose two numbers whose sum is equal to b and product is equal to ac.
In above example, we can choose $2$ and $3$, since $2\ +\ 3\ =\ 5$ and $2\ \times\ 3\ =\ 6$.

Step 4 : Write these chosen numbers in place of b.
We have
$x^{2}\ +\ 5x\ +\ 6$
= $x^{2}\ +\ (2\ +\ 3)\ x\ +\ 6$
= $x^{2}\ +\ 2x\ +\ 3x\ +\ 6$

Step 5 : Now, factorize by grouping method as we do for linear expression.
$x^{2}\ +\ 2x\ +\ 3x\ +\ 6$
= $x(x\ +\ 2)\ +\ 3(x\ +\ 2)$

Step 6 : Make sure that the terms inside the brackets are same, so that it can be factored out.
$x(x\ +\ 2)\ +\ 3(x\ +\ 2)$
= $(x\ +\ 2)\ (x\ +\ 3)$

In this way, we get the required factors.

Following solved examples of factoring quadratics can be referred for better understanding.
Example 1:

Factorize $x^{2}\ +\ 3x\ - 10$

Solution:

$x^{2}\ +\ 3x\ -\ 10$
= $x^{2}\ +\ (5\ -\ 2)\ x\ -\ 10$
= $x^{2}\ +\ 5x\ -\ 2x\ -\ 10$
= $x(x\ +\ 5)\ -\ 2(x\ +\ 5)$
= $(x\ +\ 5)\ (x\ -\ 2)$
Example 2:

Factorize the following quadratic polynomial : $2x^{2}\ -\ 16x\ +\ 24$

Solution:

$2x^{2}\ -\ 16x\ +\ 24$
$2(x^{2}\ -\ 8x\ +\ 12)$
$2[x^{2}\ -\ (6\ +\ 2)\ x\ +\ 12]$
$2[x^{2}\ -\ 6x\ -\ 2x\ +\ 12]$
$2[x(x\ -\ 6)\ -\ 2(x\ -\ 6)]$
$2(x\ -\ 2)(x\ -\ 6)$
Example 3:

Factorize the following quadratic equation by factoring :

$\frac{1}{4}$ $x^{2}$ - $\frac{3}{4}$ $x\ -7\ = 0$

Solution:

$\frac{1}{4}$ $x^{2}$ - $\frac{3}{4}$ $x\ -7\ =\ 0$

First we have to get rid of fraction. So, multiplying by 4 both sides, we get

$4$[$\frac{1}{4}$ $x^{2}$ - $\frac{3}{4}$ $x\ -7$] = $4.0$

$\Rightarrow\ x^{2}\ -\ 3x\ -\ 28\ =\ 0$

$\Rightarrow\ x^{2}\ -\ (7\ -\ 4)\ x\ -\ 28]\ =\ 0$

$\Rightarrow\ x^{2}\ -\ 7x\ +\ 4x\ -\ 28\ =\ 0$

$\Rightarrow\ x(x\ -\ 7)\ +\ 4(x\ -\ 7)\ =\ 0$

$\Rightarrow\ (x\ -\ 7)(x\ +\ 4)\ = 0$

$x\ =\ 7,\ -4$
Example 4:

Factor the following expression by using quadratic equation :

$t^{2}\ +\ 4t\ +\ 5$

Solution:

$t^{2}\ +\ 4t\ +\ 5$

Here, $a\ =\ 1,\ b\ =\ 4,\ c\ =\ 5$

$t$ = $\frac{-b\ \pm\ \sqrt{b^{2}-4ac}}{2a}$

= $\frac{-4\ \pm\ \sqrt{4^{2}-4.1.5}}{2.1}$

= $\frac{-4\ \pm\ \sqrt{16-20}}{2}$

= $\frac{-4\ \pm\ \sqrt{-4}}{2}$

= $\frac{-4\ \pm\ 2i}{2}$

= $-2\ \pm\ i$

Thus, $t\ =\ -2\ +\ i$ and $t\ =\ -2\ -\ i$

Hence, the factors are $(t\ -\ (-2\ +\ i))$ and $(t\ -\ (-2\ -\ i))$
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