Factoring Quadratic Equations
Introduction
A quadratic equation is a type of polynomial expression that consists of one or more terms, where each term is a product of a constant coefficient, a variable raised to a power, and a possible constant term. Factoring a polynomial involves expressing it as a multiplication of irreducible factors, usually of lower degree.
A quadratic equation in the form $a x^{2} + b x + c = 0$ can be factored into an equivalent equation if its discriminant is not negative:
\[a ( x - x_{1} ) ( x - x_{2} ) = 0\]where $x_{1}$ and $x_{2}$ are the roots of the equation.
Demonstration
The proof of the expression above is straightforward. We include it here for completeness.
\[P ( x ) = a x^{2} + b x + c\]We can factor out the coefficient $a$ and rewrite the expression:
\[P ( x ) = a ( x^{2} + \frac{b}{a} x + \frac{c}{a} )\]The expression becomes clearer due to the relationships between the solutions of a second-degree equation and its coefficients, in particular, from what are defined as Viète’s relations:
\(x_{1} \cdot x_{2} = \frac{c}{a}\) \(x_{1} + x_{2} = - \frac{b}{a}\)
Replacing the values, we obtain:
\[P ( x ) & = a [ x^{2} - ( x_{1} + x_{2} ) x + x_{1} x_{2} ] \\ & = a ( x^{2} - x_{1} x - x_{2} x + x_{1} x_{2} ) \\ & = a [ x ( x - x_{1} ) - x_{2} ( x - x_{1} ) ] \\ & = a ( x - x_{1} ) ( x - x_{2} )\]This demonstrates the expression $( 1 )$.
\[P ( x ) = a x^{2} + b x + c = a ( x - x_{1} ) ( x - x_{2} )\]This method of factoring polynomials is known as the AC method. The solutions to the equation can be found by setting each factor equal to zero:
\(( x - x_{1} ) = 0 \rightarrow x = x_{1}\) \(( x - x_{2} ) = 0 \rightarrow x = x_{2}\)
This method is effective for simple polynomial equations. However, it becomes impractical for more complex equations. In these cases, it is better to use the quadratic formula.
Example
The polynomial $x^{2} - 4 x + 3$ can be factorized in the form $( x - 1 ) ( x - 3 )$. We have to find two numbers, $m , n$ whose product $P$ is $3$ $( a \cdot c = 1 \cdot 3 )$ and sum $S$ is $- 4$ $( b )$.
We can use this simple scheme to find the numbers that satisfy our constraints. \(& m & n & P & S \\ & 1 & 3 & 3 & 4 \\ & - 1 & - 3 & 3 & - 4\)
The second row satisfies our constraints, so the equation associated with the polynomial becomes: \(x^{2} - 4 x + 3 = ( x - 1 ) ( x - 3 ) = 0\)
The solutions to the associated equation are: \(x - 1 = 0 \rightarrow x = 1\) \(x - 3 = 0 \rightarrow x = 3\)
The solution to the equation is: \(x_{1} = 1 \text{and} x_{2} = 3\)