# Factoring Quadratic Equations

*See also:*factoring polynomials, ruffiniâ€™s rule*Ref.*quadratic equations

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: \[x=1 \quad x= 3\]