Math Problem Statement

Write the polynomial as the product of linear factors and list all the zeros of the function: f(x) = x^4 - 4x^3 + 8x^2 - 16x - 16.

Solution

The given polynomial is:

f(x)=x4−4x3+8x2−16x−16f(x) = x^4 - 4x^3 + 8x^2 - 16x - 16

Step 1: Apply Rational Root Theorem

To find the roots, we can apply the Rational Root Theorem, which gives possible rational roots as the factors of the constant term (−16) divided by the factors of the leading coefficient (1). The factors of −16 are:

±1,±2,±4,±8,±16\pm 1, \pm 2, \pm 4, \pm 8, \pm 16

Thus, the possible rational roots are ±1,±2,±4,±8,±16\pm 1, \pm 2, \pm 4, \pm 8, \pm 16.

Step 2: Synthetic Division or Polynomial Division

Let's test x=2x = 2 as a potential root using synthetic division:

2 & 1 & -4 & 8 & -16 & -16 \\ & & 2 & -4 & 8 & -16 \\ \hline & 1 & -2 & 4 & -8 & 0 \\ \end{array}$$ Since the remainder is 0, $$x = 2$$ is a root. Thus, we can factor out $$(x - 2)$$, and the quotient is: $$x^3 - 2x^2 + 4x - 8$$ ### Step 3: Factor the Remaining Cubic Now we need to factor $$x^3 - 2x^2 + 4x - 8$$. Let's again use synthetic division to test $$x = 2$$: $$\begin{array}{r|rrrr} 2 & 1 & -2 & 4 & -8 \\ & & 2 & 0 & 8 \\ \hline & 1 & 0 & 4 & 0 \\ \end{array}$$ The remainder is 0, so $$x = 2$$ is a root again. The quotient is: $$x^2 + 4$$ ### Step 4: Factor Completely Thus, we have factored the polynomial as: $$f(x) = (x - 2)^2 (x^2 + 4)$$ We can now factor $$x^2 + 4$$ as: $$x^2 + 4 = (x - 2i)(x + 2i)$$ ### Final Factorization The complete factorization of the polynomial is: $$f(x) = (x - 2)^2 (x - 2i)(x + 2i)$$ ### Zeros of the Polynomial The zeros of the polynomial are: - $$x = 2$$ (with multiplicity 2), - $$x = 2i$$, - $$x = -2i$$. Would you like more details or explanations on any part of this process? Here are some additional questions you can explore: 1. What is the multiplicity of the root $$x = 2$$? 2. How does the Rational Root Theorem help in factoring polynomials? 3. Can synthetic division be used with any polynomial? 4. Why are $$2i$$ and $$-2i$$ considered complex roots? 5. How does factoring help in finding the zeros of a polynomial? **Tip:** When a polynomial has complex roots, they always occur in conjugate pairs if the coefficients are real.

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Math Problem Analysis

Mathematical Concepts

Polynomial Factorization
Complex Numbers
Rational Root Theorem

Formulas

Rational Root Theorem
Synthetic Division

Theorems

Rational Root Theorem

Suitable Grade Level

Grades 10-12

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