Solve by Using Factors - Algebra-2

1. Fundamental Concepts

Definition: A polynomial equation is an equation that can be written in the form $$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$$, where $$a_n, a_{n-1}, \ldots, a_0$$ are constants and $$x$$ is the variable.
Factors: Factors of a polynomial are expressions that, when multiplied together, give the original polynomial. For example, if $$p(x) = (x - r)(x - s)$$, then $$r$$ and $$s$$ are roots of the polynomial.
Zero Product Property: If the product of two or more factors is zero, then at least one of the factors must be zero. This property is used to solve polynomial equations by setting each factor equal to zero.

2. Key Concepts

Factoring Quadratics: $$ax^2 + bx + c = 0$$ can be factored into $$(px + q)(rx + s) = 0$$, where $$p, q, r, s$$ are constants.

Roots and Zeros: The solutions to the equation $$p(x) = 0$$ are the roots or zeros of the polynomial $$p(x)$$.

Application: Solving polynomial equations by factoring is used in various fields such as physics, engineering, and economics to find critical points, equilibrium states, and more.

3. Examples

Example 1 (Basic)

Problem: Solve the quadratic equation $$x^2 - 5x + 6 = 0$$.

Step-by-Step Solution:

Factor the quadratic: $$x^2 - 5x + 6 = (x - 2)(x - 3)$$
Set each factor equal to zero:
- $$x - 2 = 0 \Rightarrow x = 2$$
- $$x - 3 = 0 \Rightarrow x = 3$$

Validation: Substitute $$x = 2$$ and $$x = 3$$ into the original equation:

For $$x = 2$$: $$2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$$ ✓
For $$x = 3$$: $$3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$$ ✓

Example 2 (Intermediate)

Problem: Solve the cubic equation $$x^3 - 4x^2 - 7x + 10 = 0$$.

Step-by-Step Solution:

Use the Rational Root Theorem to test possible rational roots: $$\pm 1, \pm 2, \pm 5, \pm 10$$
Test $$x = 1$$: $$1^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0$$, so $$x = 1$$ is a root.
Divide the polynomial by $$x - 1$$ using synthetic division or polynomial long division to get: $$x^3 - 4x^2 - 7x + 10 = (x - 1)(x^2 - 3x - 10)$$
Factor the quadratic: $$x^2 - 3x - 10 = (x - 5)(x + 2)$$
Set each factor equal to zero:
- $$x - 1 = 0 \Rightarrow x = 1$$
- $$x - 5 = 0 \Rightarrow x = 5$$
- $$x + 2 = 0 \Rightarrow x = -2$$

Validation: Substitute $$x = 1, 5, -2$$ into the original equation:

For $$x = 1$$: $$1^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0$$ ✓
For $$x = 5$$: $$5^3 - 4(5)^2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0$$ ✓
For $$x = -2$$: $$(-2)^3 - 4(-2)^2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0$$ ✓

4. Problem-Solving Techniques

Rational Root Theorem: Use this theorem to list possible rational roots and test them to find actual roots.
Synthetic Division: Use synthetic division to divide the polynomial by a known factor to reduce the degree of the polynomial.
Factoring by Grouping: For polynomials with four or more terms, group terms and factor out common factors to simplify the expression.
Graphical Analysis: Use graphing tools to visualize the polynomial and estimate the roots, which can then be verified algebraically.