Solving Polynomial Equations
Overview
Solving polynomial equations involves finding the values of the variable that make the equation true. Polynomials can have multiple solutions, and these solutions are called the "roots" of the equation. The fundamental theorem of algebra tells us that a polynomial equation of degree \( n \) has \( n \) solutions (real or complex).
There are several methods for solving polynomial equations, including:
- Factoring: Factoring the polynomial into simpler expressions and solving each factor separately.
- Using the Zero Product Property: If a product of terms equals zero, one or more of the terms must be zero.
- Graphing: Plotting the polynomial and finding the x-intercepts, which represent the solutions.
- Using the Quadratic Formula: If the polynomial is a quadratic, use the quadratic formula to find the solutions.
Example 1: Solving by Factoring
Consider the equation \( x^2 - 5x + 6 = 0 \). First, factor the quadratic polynomial:
\( x^2 - 5x + 6 = (x - 2)(x - 3) \)
Now, use the Zero Product Property. Set each factor equal to zero:
\( x - 2 = 0 \quad \rightarrow \quad x = 2 \)
\( x - 3 = 0 \quad \rightarrow \quad x = 3 \)
The solutions are \( x = 2 \) and \( x = 3 \).
Example 2: Solving by Graphing
Consider the equation \( x^2 - 4 = 0 \). To solve this graphically, plot the polynomial and find the x-intercepts:
\( x^2 - 4 = (x + 2)(x - 2) \)
The graph crosses the x-axis at \( x = -2 \) and \( x = 2 \), so the solutions are \( x = -2 \) and \( x = 2 \).
Example 3: Solving a Cubic Equation
Consider the cubic equation \( x^3 - 3x^2 - 4x + 12 = 0 \). First, factor the polynomial:
\( x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6) \)
Next, factor the quadratic term:
\( x^2 - x - 6 = (x - 3)(x + 2) \)
So, the complete factorization is:
\( (x - 2)(x - 3)(x + 2) = 0 \)
Using the Zero Product Property, set each factor equal to zero:
\( x - 2 = 0 \quad \rightarrow \quad x = 2 \)
\( x - 3 = 0 \quad \rightarrow \quad x = 3 \)
\( x + 2 = 0 \quad \rightarrow \quad x = -2 \)
The solutions are \( x = 2 \), \( x = 3 \), and \( x = -2 \).
Practice Questions
- Question 1: Solve the equation \( x^2 - 7x + 10 = 0 \).
Solution
Factor the quadratic equation:
\( x^2 - 7x + 10 = (x - 2)(x - 5) \)
Now, use the Zero Product Property:
\( x - 2 = 0 \quad \rightarrow \quad x = 2 \)
\( x - 5 = 0 \quad \rightarrow \quad x = 5 \)
The solutions are \( x = 2 \) and \( x = 5 \).
- Question 2: Solve the equation \( x^2 + 4x + 4 = 0 \).
Solution
Factor the quadratic equation:
\( x^2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)^2 \)
Set the factor equal to zero:
\( x + 2 = 0 \quad \rightarrow \quad x = -2 \)
The solution is \( x = -2 \).
- Question 3: Solve the equation \( x^3 - 6x^2 + 11x - 6 = 0 \).
Solution
First, factor the cubic polynomial:
\( x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \)
Next, factor the quadratic part:
\( x^2 - 5x + 6 = (x - 2)(x - 3) \)
The complete factorization is:
\( (x - 1)(x - 2)(x - 3) = 0 \)
Set each factor equal to zero:
\( x - 1 = 0 \quad \rightarrow \quad x = 1 \)
\( x - 2 = 0 \quad \rightarrow \quad x = 2 \)
\( x - 3 = 0 \quad \rightarrow \quad x = 3 \)
The solutions are \( x = 1 \), \( x = 2 \), and \( x = 3 \).
- Question 4: Solve the equation \( x^2 + 3x - 4 = 0 \).
Solution
Factor the quadratic equation:
\( x^2 + 3x - 4 = (x - 1)(x + 4) \)
Now, use the Zero Product Property:
\( x - 1 = 0 \quad \rightarrow \quad x = 1 \)
\( x + 4 = 0 \quad \rightarrow \quad x = -4 \)
The solutions are \( x = 1 \) and \( x = -4 \).
- Question 5: Solve the equation \( x^3 + 5x^2 + 6x = 0 \).
Solution
Factor out the greatest common factor (GCF):
\( x(x^2 + 5x + 6) = 0 \)
Factor the quadratic expression:
\( x(x + 2)(x + 3) = 0 \)
Now, use the Zero Product Property:
\( x = 0 \)
\( x + 2 = 0 \quad \rightarrow \quad x = -2 \)
\( x + 3 = 0 \quad \rightarrow \quad x = -3 \)
The solutions are \( x = 0 \), \( x = -2 \), and \( x = -3 \).