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 polynomials involves expressing it as a multiplication of irreducible factors, usually of lower degree.

A quadratic equation in the form \(ax^{2}+bx+c=0 \) 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.

\[ P(x) = ax^2 + bx + cx\]


We can factor out the coefficient \(a\) and rewrite the expression:

\[P(x) = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) \]


The expression becomes clearer due to the relationships between the solutions of a second-degree equation and its coefficients.

\[\begin{align} P(x) &= a[x^2-(x_1+x_2)x+x_1x_2] \\[1ex] &= a(x^2-x_1x-x_2x+x_1x_2) \\[1ex] &= a[x(x-x_1)-x_2(x-x_1) ]\\[1ex] &= a(x-x_1) (x-x_2) \end{align}\]

This demonstrates the expression \((1)\).

\[P(x)= ax^{2}+bx+c = a(x-x_1)(x-x_2)\]

In general, we must find two numbers \(x_1, x_2\) that meet the following constraints:

\[x_1 \cdot x_2=a \cdot c \] \[x_1+x_2=b \]

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:

\[(x-x_1)=0 \to x = x_1\] \[(x-x_2)=0 \to x = x_2\]

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 \(x^2-4x+3\) can be factorized in the form \((x-1)(x-3)\). In fact we have to find two numbers, \(x_1, x_2\) whose product \(P\) is \(3\) and sum \(S\) is \(-4\).

We can use this simple scheme to find the numbers that satisfy our constraints. \begin{array}{rrrr} & x_1 & x_2 & P & S \\ \hline & 1 & 3 & 3 & 4 \\ & -1 & -3 & 3 & -4 \\ \end{array}


The second row satisfies our constraints, so the equation associated with the polynomial becomes: \[x^2-4x+3 = (x-1)(x- 3) = 0\]

The solutions to the associated equation are: \[x -1=0 \to x = 1\] \[x -3=0 \to x = 3\]

The solution to the equation is: \[x=1 \quad x= 3\]