Factoring an Equation Using Its Roots
Any polynomial equation in one variable $$ P(x)=0 $$ can be written in factored form as $$ a \cdot (x - x_n) \cdot ... \cdot (x - x_2) \cdot (x - x_1) $$ Here, a is the leading coefficient (the coefficient of the highest power of x), and x1, ..., xn are the roots or solutions of the equation.
A practical example
Consider the quadratic equation:
$$ 2x^2 - 10x + 2 = -10 $$
To express it in standard form, move all terms to one side:
$$ 2x^2 - 10x + 2 + 10 = 0 $$
$$ 2x^2 - 10x + 12 = 0 $$
The leading coefficient is a = 2.
We can now find the two distinct roots using the quadratic formula:
$$ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 2 \cdot 12}}{2 \cdot 2} $$
$$ x = \frac{10 \pm \sqrt{100 - 96}}{4} $$
$$ x = \frac{10 \pm \sqrt{4}}{4} $$
$$ x = \frac{10 \pm 2}{4} $$
$$ x = \begin{cases} \frac{10 - 2}{4} = \frac{8}{4} = 2 \\ \\ \frac{10 + 2}{4} = \frac{12}{4} = 3 \end{cases} $$
The roots of the equation are x1 = 2 and x2 = 3.
Since a = 2, the equation can be expressed in its equivalent factored form:
$$ 2x^2 - 10x + 12 = 0 $$
$$ a \cdot (x - x_2) \cdot (x - x_1) = 0 $$
$$ 2 \cdot (x - 2) \cdot (x - 3) = 0 $$
Verification. Expanding the factors confirms the result: $$ 2 \cdot (x - 2) \cdot (x - 3) = 0 $$ $$ (2x - 4) \cdot (x - 3) = 0 $$ $$ 2x^2 - 6x - 4x + 12 = 0 $$ $$ 2x^2 - 10x + 12 = 0 $$
This matches the original equation, completing the verification.
