Algebra

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