# Factorization

## Basic Concept

##### Factorization of $$Ax^2 + Bx + C$$ with $$A = 1$$

$$x^2 +(p+q)x+pq = (x+p)(x+q)$$

Example 1

Factorize $$x^2 + 5x + 6$$

To factorise this expressions, find two numbers (p and q) that have a product of +6 and a sum of +5.

There are a couple of ways of making +6 by multiplying two numbers.

There are 1 × 6 and 2 × 3.

Only the combination of 2 and 3 will also give a sum of +5, so the two numbers are 2 and 3.

p = 2 and q = 3

\begin{equation*}
\begin{split}
x^2 + 5x + 6& = x^2 + (2 + 3)x + (2 \times 3)\\\\
x^2 + 5x + 6& = (x + 2)(x + 3)
\end{split}
\end{equation*}

Example 2

Factorize $$x^2 -4x - 12$$

To factorise this expressions, find two numbers (p and q) that have a product of −12 and a sum of −4.

There are −(1 × 12),  −(2 × 6), and −(3 × 4)

Only the combination of 2 and −6 will also give a sum of −4, so the two numbers are 2 and −6.

p = 2 and q = −6

\begin{equation*}
\begin{split}
x^2 -4x - 12& =x^2 + (2 -6)x + (2)(-6)\\\\
x^2 -4x - 12&= (x+2)(x-6)
\end{split}
\end{equation*}

##### Factorization of $$Ax^2 + Bx + C$$ with $$A \neq 1$$

Example 1

Factorize $$4x^2 + 23x + 15$$

\begin{equation*}
\begin{split}
4x^2 + \color{blue}23x\color{black} + 15& = 4x^2 + \color{blue} 20x + 3x \color{black} + 15\\\\
4x^2 + \color{blue}23x\color{black} + 15& = 4x\color{purple}(x + 5)\color{black} + 3\color{purple}(x + 5)\\\\
4x^2 + \color{blue}23x\color{black} + 15& = \color{purple}(x + 5)\color{black}(4x + 3)
\end{split}
\end{equation*}

So, $$4x^2 + 23x + 15 = (x + 5)(4x + 3)$$

Example 2

Factorize $$2x^2 +11x -21$$

\begin{equation*}
\begin{split}
2x^2 + \color{blue}11x\color{black} -21& = 2x^2 + \color{blue} 14x - 3x \color{black} -21\\\\
2x^2 + \color{blue}11x\color{black} -21& = 2x\color{purple}(x + 7)\color{black} - 3\color{purple}(x + 7)\\\\
2x^2 + \color{blue}11x\color{black} -21& = \color{purple}(x + 7)\color{black}(2x - 3)
\end{split}
\end{equation*}

So, $$2x^2 +11x -21 = (x + 7)(2x - 3)$$