Solving Quadratic Equations by Factorisation
The process of writing an expression as a product of two or more common factors is called the method of factorization. For example:
 $${x^2} + 5x + 6 = \left( {x + 2} \right)\left( {x + 3} \right)$$
 $$5{x^2} + 8x = x\left( {5x + 8} \right)$$
 $$30 = 2 \times 3 \times 5$$
In the above examples, $$\left( {x + 2} \right)\left( {x + 3} \right)$$ are the factors of expression $${x^2} + 5x + 6$$, $$x\left( {5x + 8} \right)$$ are the factors of $$5{x^2} + 8x$$ and $$2 \times 3 \times 5$$ are the factors of $$30$$.
While solving the quadratic equation by the method of factorization, we have the following steps:

Convert the quadratic equation into standard form if necessary, i.e. $$a{x^2} + bx + c = 0$$, where $$a \ne 0$$.

Multiply the coefficient of $${x^2}$$ by constant terms, and we get $$a \times c = ac$$.

Now try to find two numbers whose product is $$ac$$ and sum or difference is equal to $$b$$ (coefficient of $$x$$).

Factorise the given expression on L.H.S.

Equate each factor equal to zero.

We get the required roots, say $${x_1}$$, $${x_2}$$.
Example:
Solve the equation by factorization method.
\[5{x^2} – 11x + 6 = 0\]
Solution:
The given equation in standard form is $$5{x^2} – 11x + 6 = 0$$
Multiplying the coefficient of $${x^2}$$ and the constant term, we get $$5 \times 6 = 30$$
Divide $$30$$ into two parts such that their difference or sum is $$11$$
Possible factors of $$30$$

Sum or difference of factors

$$30 \times 1 = 30$$

$$30 – 1 = 29,{\text{ }}30 + 1 = 31$$ (not possible)

$$15 \times 2 = 30$$

$$15 – 2 = 13,{\text{ 15}} + 2 = 17$$ (not possible)

$$10 \times 3 = 30$$

$$10 – 3 = 7,{\text{ 1}}0 + 3 = 13$$ (not possible)

$$6 \times 5 = 30$$

$$6 – 5 = 1,{\text{ 6}} + 5 = 11$$ (possible)

Therefore, $$5{x^2} – 11x + 6 = 0$$
$$5{x^2} – \left( {5 + 6} \right)x + 6 = 0$$
$$5{x^2} – 5x – 6x + 6 = 0$$
$$5x\left( {x – 1} \right) – 6\left( {x – 1} \right) = 0$$
$$\left( {x – 1} \right)\left( {5x – 6} \right) = 0$$
Either $$x – 1 = 0$$ or $$5x – 6 = 0$$
$$x = 1$$ or $$x = \frac{6}{5}$$
Example:
Solve the equation by factorization method.
\[8x + 4{x^2} = 0\]
Solution:
The given equation in standard form is $$4{x^2} + 8x = 0$$
Here, the constant term is absent; its factorization is very simple.
Taking common $$4x$$, we get
$$4x\left( {x + 2} \right) = 0$$
Either $$4x = 0$$ or $$x + 2 = 0$$
$$x = 0$$ or $$x = – 2$$