Solution of Polynomials - Algebra-1

1. Fundamental Concepts

Solutions, zeros, roots, and x-intercepts of a polynomial are different terms for the same concept. They all refer to the values of the independent variable x when the value of the polynomial function equals 0, i.e., the x-values that satisfy \(f(x) = 0\).

2. Key Concepts

If a polynomial can be factored into a product of factors (e.g., \(f(x) = (x - a)(x - b)\)), set each factor equal to 0. The resulting x-values ( \(x = a\), \(x = b\)) are the roots of the polynomial.
A linear polynomial (e.g., \(f(x) = mx + n\), where \(m \neq 0\)) has exactly 1 root. A quadratic polynomial (e.g., \(f(x) = ax^2 + bx + c\), where \(a \neq 0\)) has at most 2 roots, and the exact number of roots is determined by the discriminant ( \(b^2 - 4ac\)).

3. Examples

Easy (Linear Polynomial)

Find the root of \(f(x) = 2x + 4\). Solution: Set \(f(x) = 0\), so \(2x + 4 = 0\). Rearrange terms to get \(2x = -4\), and solve for x: \(x = -2\).

Medium (Quadratic Polynomial Factorable Directly)

Find the roots of \(h(x) = x^2 - 13x + 30\). Solution: First, factorize \(h(x)\), which gives \(h(x) = (x - 3)(x - 10)\); Set \((x - 3)(x - 10) = 0\), and solve for x: \(x = 3\) or \(x = 10\).

4. Problem-Solving Techniques

Rearrange the Equation: Transform the equation involving the polynomial into the standard form \(f(x) = 0\) (move all terms to the left side of the equal sign, with the right side being 0).
Factorization (for Quadratic and Higher-Degree Polynomials): If the polynomial can be factored into a product of linear factors, set each factor equal to 0 and solve for x respectively.
Direct Solution (for Linear Polynomials): For a linear polynomial \(f(x) = mx + n\) (where \(m \neq 0\)), solve for x directly by rearranging terms and converting the coefficient of x to 1, using the formula \(x = -\frac{n}{m}\).
Verify the Result: Substitute the obtained x-values back into the original polynomial to check if \(f(x) = 0\), ensuring the correctness of the roots.