Math Problem Statement

Find the real zeros of f. Use the real zeros to factor f. f left parenthesis x right parenthesis equals x cubed minus 10 x squared plus 28 x minus 16 Question content area bottom Part 1 The real​ zero(s) of f​ is/are    enter your response here ​(Simplify your answer. Type an exact​ answer, using radicals as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as​ needed.) Part 2 Factor f. ​f(x)equals    enter your response here ​(Factor completely. Type an exact​ answer, using radicals as needed. Use integers or fractions for any numbers in the​ expression.)

Solution

Let's find the real zeros and factor the polynomial function f(x)=x310x2+28x16f(x) = x^3 - 10x^2 + 28x - 16.

Part 1: Finding the Real Zeros

To find the real zeros of the function, we can attempt to use the Rational Root Theorem. The Rational Root Theorem suggests that any rational root of the polynomial, if it exists, must be a factor of the constant term (in this case, 16-16) divided by a factor of the leading coefficient (in this case, 11).

The constant term is 16-16, and the leading coefficient is 11. The factors of 16-16 are: ±1,±2,±4,±8,±16\pm 1, \pm 2, \pm 4, \pm 8, \pm 16

We will now test these potential rational roots using synthetic division or direct substitution into f(x)f(x).

Testing potential roots:

  1. Let's test x=2x = 2: f(2)=2310(22)+28(2)16=840+5616=8f(2) = 2^3 - 10(2^2) + 28(2) - 16 = 8 - 40 + 56 - 16 = 8 So, x=2x = 2 is not a root.

  2. Next, let's try x=4x = 4: f(4)=4310(42)+28(4)16=64160+11216=0f(4) = 4^3 - 10(4^2) + 28(4) - 16 = 64 - 160 + 112 - 16 = 0 So, x=4x = 4 is a root.

Part 2: Factoring the Polynomial

Now that we know x=4x = 4 is a root, we can factor f(x)f(x) as (x4)(x - 4) times a quadratic polynomial. To find this quadratic factor, we perform synthetic division on f(x)f(x) by (x4)(x - 4).

Using synthetic division:

4 & 1 & -10 & 28 & -16 \\ & & 4 & -24 & 16 \\ \hline & 1 & -6 & 4 & 0 \\ \end{array}$$ The quotient is $$x^2 - 6x + 4$$, so we have: $$f(x) = (x - 4)(x^2 - 6x + 4)$$ Next, let's see if the quadratic $$x^2 - 6x + 4$$ can be factored further. The discriminant is: $$\Delta = (-6)^2 - 4(1)(4) = 36 - 16 = 20$$ Since the discriminant is positive but not a perfect square, the quadratic does not factor nicely into rational numbers. However, we can find the roots using the quadratic formula: $$x = \frac{-(-6) \pm \sqrt{20}}{2(1)} = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$$ So, the real zeros are: $$x = 4, x = 3 + \sqrt{5}, \text{ and } x = 3 - \sqrt{5}$$ ### Final Answer: 1. The real zeros of $$f(x)$$ are: $$x = 4, \, 3 + \sqrt{5}, \, 3 - \sqrt{5}$$ 2. The factored form of $$f(x)$$ is: $$f(x) = (x - 4)(x - (3 + \sqrt{5}))(x - (3 - \sqrt{5}))$$ Let me know if you need further clarification or have any other questions! --- ### Further Exploration: 1. How does the Rational Root Theorem work for polynomials? 2. What happens if the discriminant of the quadratic is negative? 3. Can we find complex roots for quadratics with a negative discriminant? 4. How do you check your factorization of polynomials? 5. What role does synthetic division play in factoring polynomials? ### Tip: When solving cubic equations, always start by checking for simple rational roots first (using the Rational Root Theorem) and use synthetic division to simplify the polynomial.

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Math Problem Analysis

Mathematical Concepts

Algebra
Cubic Equations
Synthetic Division
Rational Root Theorem
Quadratic Equations

Formulas

Quadratic Formula
Rational Root Theorem

Theorems

Rational Root Theorem

Suitable Grade Level

Grades 9-11