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When facing a quartic equation, or any polynomial equation of higher degree, the number of terms and the exponents can make solving it seem daunting. Quadratic substitution is a technique employed to simplify such equations into a more manageable form. The key lies in recognizing patterns within the equation that align with the structure of a quadratic equation: terms that follow the form of \(ax^2 + bx + c = 0\).

In our exercise, \(x^4 - 5x^2 + 4 = 0\), the quadratic substitution involves setting \(x^2\) equal to a new variable, typically \(u\). This transformation changes the quartic equation into a quadratic one in terms of \(u\), as shown by \(u^2 - 5u + 4 = 0\). By doing so, not only does the equation become simpler, but it also utilizes the familiar, established methods for solving quadratic equations.

It's essential to be careful when choosing the right substitution and to clearly track the new variable throughout the solving process. Substitution eases the path to finding the roots, as it converts the problem to one with a more familiar form.

Factoring quadratic equations is an effective root-finding strategy for second-degree polynomials. After reducing a higher-degree polynomial into a quadratic equation using a substitution method, factoring becomes invaluable. A quadratic equation such as \(ax^2 + bx + c = 0\) can often be decomposed into a product of two first-degree binomials, \( (px + q)(rx + s) = 0\).

In our exercise, once \(x^2\) has been substituted by \(u\), the equation \(u^2 - 5u + 4 = 0\) is factored into \( (u - 4)(u - 1) = 0\). This step is pivotal as it lays the groundwork for finding the specific values of \(x\) that satisfy the initial quartic equation. Remember, for the product of two factors to be zero, at least one of them must be zero. This simple principle unlocks the potential solutions for \(u\), and subsequently for \(x\).

Root finding is the process of determining the values for which a polynomial equation equals zero. In essence, it's about pinpointing where the graph of the polynomial intersects the x-axis. These intersection points are the 'roots' or 'solutions.' Polynomial equations can have multiple roots, and quartic equations specifically can have up to four real roots.

In the context of our exercise, after using quadratic substitution and factoring, we arrive at real values for \(u\). Each resultant value of \(u\) corresponds to a pair of potential solutions for \(x\) after performing back substitution. As showcased in the exercise, \(u = 4\) and \(u = 1\) yield four distinct roots for the original quartic equation: \(±2\) and \(±1\), respectively.

Understanding the fundamental theorem of algebra, which states that every non-zero, single-variable polynomial with complex coefficients has as many complex roots as its degree, provides a deeper comprehension of root finding. In practice, finding the roots of a polynomial often involves a mixture of techniques, including substitution, factoring, or even using algorithms for numerical approximation for more complex equations.