### Theory:

Factorisation of polynomials are done in the following two ways:
• Using algebraic identities.
• By splitting the middle term of the polynomial.
Factorising a polynomial using algebraic identities

This a direct method to find the factors of the polynomial. The polynomial is simply resolved into factors by comparing it with the algebraic identityies similar to its form.

The following basic identities are used in factorizing a polynomial.
• ${a}^{2}+2\mathit{ab}+{b}^{2}={\left(a+b\right)}^{2}$
• ${a}^{2}-2\mathit{ab}+{b}^{2}={\left(a-b\right)}^{2}$
• ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
• ${a}^{2}+{b}^{2}+{c}^{2}+2\mathit{ab}+2\mathit{bc}+2\mathit{ca}={\left(a+b+c\right)}^{2}$
• ${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+\mathit{ab}+{b}^{2}\right)$
• ${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-\mathit{ab}+{b}^{2}\right)$
• ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ca}\right)$
Important!
The following are some special cases derived from the identities.

$\begin{array}{l}{\left(a+b\right)}^{2}+{\left(a-b\right)}^{2}=\left({a}^{2}+2\mathit{ab}+{b}^{2}\right)+\left({a}^{2}-2\mathit{ab}+{b}^{2}\right)\\ =2{a}^{2}+2{b}^{2}\\ =2\left({a}^{2}+{b}^{2}\right)\end{array}$
• ${\left(a+b\right)}^{2}+{\left(a-b\right)}^{2}=2\left({a}^{2}+{b}^{2}\right)$
$\begin{array}{l}{a}^{4}-{b}^{4}={\left({a}^{2}\right)}^{2}-{\left({b}^{2}\right)}^{2}\\ =\left({a}^{2}+{b}^{2}\right)\left({a}^{2}-{b}^{2}\right)\\ =\left({a}^{2}+{b}^{2}\right)\left(a+b\right)\left(a-b\right)\end{array}$
• ${a}^{4}-{b}^{4}=\left({a}^{2}+{b}^{2}\right)\left(a+b\right)\left(a-b\right)$
$\begin{array}{l}{\left(a+b\right)}^{2}-{\left(a-b\right)}^{2}=\left({a}^{2}+2\mathit{ab}+{b}^{2}\right)-\left({a}^{2}-2\mathit{ab}+{b}^{2}\right)\\ =2\mathit{ab}+2\mathit{ab}\\ =4\mathit{ab}\end{array}$
• ${\left(a+b\right)}^{2}-{\left(a-b\right)}^{2}=4\mathit{ab}$
$\begin{array}{l}{a}^{6}-{b}^{6}={\left({a}^{3}\right)}^{2}-{\left({b}^{3}\right)}^{2}\\ =\left({a}^{3}+{b}^{3}\right)\left({a}^{3}-{b}^{3}\right)\\ =\left(a+b\right)\left({a}^{2}-\mathit{ab}+{b}^{2}\right)\left(a-b\right)\left({a}^{2}+\mathit{ab}+{b}^{2}\right)\end{array}$
• ${a}^{6}-{b}^{6}=\left(a+b\right)\left({a}^{2}-\mathit{ab}+{b}^{2}\right)\left(a-b\right)\left({a}^{2}+\mathit{ab}+{b}^{2}\right)$

Factorising a polynomial by splitting the middle term

A quadratic polynomial  of the form $$ax^2 + bx +c$$ where $$a \neq 0$$ is factorized by splitting the middle term.

Working rule:

Step 1:   Multiply the co-efficient of $$x^2$$ and the constant term in the quadratic polynomial.

Step 2: Split the product into two factors such that their sum is equal to the coefficient of $$x$$ and its product is equal to the product of the co-efficient of $$x^2$$ and the constant term.

Step 3: Group the terms into two pairs and factorise.
Example:
Factorize .

Solution:

Compare  with .

Here ($$a = 6$$, $$b = 19$$ and $$c = 10$$).

Step 1: Find the product of the first and last term $$(a × c)$$.

.

Step 2: Split the product $$60$$ into two factors in such a way that the addition or subtraction of the factors is equal to $$19$$ and its product is equal to $$60$$.

and .

Step 3: Write the middle term using the sum of the $$2$$ new factors, including the correct signs.

Step 4: Group the terms to make pairs using the first two terms and the last two terms. Factor each pair by finding the common factors.

Step 5: Factor out the common factors from the term.

The factors of  is .