Find Zeros of f(x) = x^3 - 5x^2 - 8x + 12 Using Factor Theorem

Math Problem Statement

If f(x) = x^3 - 5x^2 - 8x + 12 and x + 2 is a factor of f(x), then find all of the zeros of f(x) algebraically.

Solution

To solve the given problem algebraically, we are tasked with finding all the zeros of the polynomial function $f(x) = x^3 - 5x^2 - 8x + 12$ , given that $x + 2$ is a factor of $f(x)$ .

Step 1: Use the Factor Theorem

Since $x + 2$ is a factor, $f(-2) = 0$ . Let us confirm this: $f(x) = x^3 - 5x^2 - 8x + 12$ Substitute $x = -2$ : $f(-2) = (-2)^3 - 5(-2)^2 - 8(-2) + 12$ $f(-2) = -8 - 20 + 16 + 12 = 0$ Thus, $x + 2$ is indeed a factor.

Step 2: Perform Synthetic or Polynomial Division

We divide $f(x)$ by $x + 2$ to reduce the degree of the polynomial. Using synthetic division:

-2 & 1 & -5 & -8 & 12 \\ & & -2 & 14 & -12 \\ \hline & 1 & -7 & 6 & 0 \\ \end{array}$$ The quotient is $$x^2 - 7x + 6$$, and the remainder is $$0$$. Thus, $$f(x) = (x + 2)(x^2 - 7x + 6)$$ ### Step 3: Factor the Quadratic $$x^2 - 7x + 6$$ We now factorize $$x^2 - 7x + 6$$. Look for two numbers that multiply to 6 and add to -7. These numbers are -6 and -1. Thus, $$x^2 - 7x + 6 = (x - 6)(x - 1)$$ ### Step 4: Combine the Factors The fully factored form of $$f(x)$$ is: $$f(x) = (x + 2)(x - 6)(x - 1)$$ ### Step 5: Solve for the Zeros Set $$f(x) = 0$$ to find the zeros: $$x + 2 = 0 \quad \Rightarrow \quad x = -2$$ $$x - 6 = 0 \quad \Rightarrow \quad x = 6$$ $$x - 1 = 0 \quad \Rightarrow \quad x = 1$$ ### Final Answer The zeros of $$f(x)$$ are: $$x = -2, \, x = 6, \, x = 1$$ Would you like a detailed explanation of any step or additional examples? --- Here are some related questions to explore: 1. How do you verify the roots of a polynomial after solving? 2. What is the significance of the Factor Theorem in polynomial factorization? 3. Can all cubic polynomials be factored completely into linear terms? 4. How would the process change if a complex factor were involved? 5. What are the graphical interpretations of the zeros of $$f(x)$$? **Tip**: Always check your work by substituting the roots back into the original polynomial!

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Math Problem Analysis

Mathematical Concepts

Polynomial Division
Factoring Polynomials
Roots of Polynomials

Formulas

Synthetic Division
Quadratic Factoring

Theorems

Factor Theorem

Suitable Grade Level

Grades 10-12

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