Math Problem Statement

What are the zeros, holes, or intercepts of f(x) = (x^2 - 1) / (x-1)^2, when analyzing the polynomial and rational functions? What is the remainder of P(x) divided by 4x - 1, and how many turning points are there in the polynomial f(x) = 5(x-a)(x-b)(x-c)^2?

Solution

Here is the breakdown of the provided math problems and explanations for the questions visible in the image:


Question 6:

The zeros of g(x)=x21g(x) = x^2 - 1 are the _________ of f(x)=x21(x1)2f(x) = \frac{x^2 - 1}{(x-1)^2}.

  1. First, factorize x21x^2 - 1:
    x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1)
    So, g(x)g(x) has zeros at x=1x = 1 and x=1x = -1.

  2. Now analyze f(x)f(x):
    f(x)=x21(x1)2=(x1)(x+1)(x1)2f(x) = \frac{x^2 - 1}{(x-1)^2} = \frac{(x-1)(x+1)}{(x-1)^2}

    Simplify f(x)f(x):
    f(x) = \frac{x+1}{x-1}, \quad \text{(as long as x \neq 1)}.

  3. At x=1x = 1:

    • The denominator becomes zero, but the numerator does not eliminate the (x1)(x-1) term entirely.
    • Therefore, there is a hole at x=1x = 1.
  4. At x=1x = -1:

    • The numerator equals zero, and the denominator is not zero.
    • Therefore, there is an xx-intercept at x=1x = -1.

Correct Answer:

  • B. hole at x=1x = 1.
  • A. x-intercepts at x=1x = -1.

Question 7:

When a polynomial P(x)P(x) is divided by 4x14x - 1, the remainder is 10-10.

This is based on the Remainder Theorem:

  • If P(x)P(x) is divided by axbax - b, then the remainder is P(ba)P\left(\frac{b}{a}\right).
  • Here axb=4x1x=14ax - b = 4x - 1 \Rightarrow x = \frac{1}{4}.

Thus, the remainder is given as:
P(14)=10.P\left(\frac{1}{4}\right) = -10.

Correct Answer:

  • D. -10

Question 8:

What is the number of turning points in the polynomial f(x)=5(xa)(xb)(xc)2f(x) = 5(x - a)(x - b)(x - c)^2, where c>b>ac > b > a?

  1. The number of turning points is determined by the degree of the polynomial:

    • (xa)(x - a) and (xb)(x - b) are linear factors (degree 11).
    • (xc)2(x - c)^2 is a quadratic factor (degree 22).
    • Therefore, the degree of f(x)f(x) is 1+1+2=41 + 1 + 2 = 4.
  2. The maximum number of turning points in a polynomial of degree nn is n1n - 1:

    • Here n=4n = 4, so the maximum number of turning points is 41=34 - 1 = 3.

Correct Answer:

  • A. 3

Recap of Answers:

  1. 6: B. hole
  2. 7: D. -10
  3. 8: A. 3

Follow-Up Questions:

  1. How do you determine holes and asymptotes in rational functions?
  2. What is the Remainder Theorem, and how is it applied?
  3. Why does a polynomial of degree nn have at most n1n-1 turning points?
  4. Can a polynomial have fewer turning points than the maximum possible? Why?
  5. How do you factorize and analyze rational functions like f(x)f(x)?

Tip:

To identify holes in a rational function, look for factors in both the numerator and denominator that cancel out—this is where the function becomes undefined (a "hole").

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Math Problem Analysis

Mathematical Concepts

Polynomial Zeros
Rational Functions
Holes and Asymptotes
Remainder Theorem
Turning Points of Polynomials

Formulas

Simplification of Rational Functions
Remainder Theorem: If P(x) is divided by (ax - b), the remainder is P(b/a)
Turning Points = Degree of Polynomial - 1

Theorems

Remainder Theorem
Degree-Turning Points Relationship

Suitable Grade Level

Grades 10-12