Math Problem Statement

A polynomial function P and its graph are given.

P(x) = 3x3 + x2 − 7x − 5

The x y-coordinate plane is given. The curve enters the window in the third quadrant, goes up and right becoming less steep, changes direction at the point (−1, 0), goes down and right becoming more steep, passes through the approximate point (−0.1, −4.2), goes down and right becoming less steep, crosses the y-axis at y = −5, changes direction at the approximate point (0.8, −8.4), goes up and right becoming more steep, crosses the x-axis at approximately x = 1.7, and exits the window in the first quadrant. 

(a)

List all possible rational zeros of P given by the Rational Zeros Theorem. (Enter your answers as a comma-separated list.)

x =     (b)

From the graph, determine which of the possible rational zeros actually turn out to be zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.)

x =    

Solution

Part (a): Possible Rational Zeros

To list all possible rational zeros using the Rational Zeros Theorem, we need to consider the factors of the constant term and the leading coefficient of the polynomial.

Given the polynomial:
P(x)=3x3+x2−7x−5P(x) = 3x^3 + x^2 - 7x - 5

  • The constant term is −5-5, and its factors are ±1,±5\pm 1, \pm 5.
  • The leading coefficient is 33, and its factors are ±1,±3\pm 1, \pm 3.

The Rational Zeros Theorem states that the possible rational zeros are the ratios of the factors of the constant term to the factors of the leading coefficient. So the possible rational zeros are:

Possible Rational Zeros=factors of constant termfactors of leading coefficient\text{Possible Rational Zeros} = \frac{\text{factors of constant term}}{\text{factors of leading coefficient}} Thus, the possible rational zeros are:

x=±1,±13,±5,±53x = \pm 1, \pm \frac{1}{3}, \pm 5, \pm \frac{5}{3}

So, the list of all possible rational zeros is:

x=±1,±13,±5,±53x = \pm 1, \pm \frac{1}{3}, \pm 5, \pm \frac{5}{3}

Part (b): Actual Rational Zeros from the Graph

From the graph description, we know that the curve crosses the xx-axis at two points:

  • One at x=−1x = -1 (this is a zero).
  • Another at approximately x=1.7x = 1.7, which is likely not a rational zero.

Thus, from the possible rational zeros, only x=−1x = -1 turns out to be an actual zero. There is no indication from the graph that any other possible rational zero applies.

Therefore, the actual rational zero is:

x=−1x = -1

Summary:

  • (a) Possible rational zeros: x=±1,±13,±5,±53x = \pm 1, \pm \frac{1}{3}, \pm 5, \pm \frac{5}{3}.
  • (b) Actual rational zero: x=−1x = -1.

Do you want more details or have any questions?

Here are some related questions:

  1. How can we verify that x=−1x = -1 is an actual zero by substitution?
  2. What is the Rational Zeros Theorem, and how is it used in detail?
  3. How can we determine the other non-rational zeros numerically?
  4. What are the properties of cubic polynomials based on their graphs?
  5. How can we use synthetic division to check if a candidate is a zero?

Tip: To find actual zeros, you can always substitute candidate values into the polynomial or use synthetic division.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Graph Analysis

Formulas

P(x) = 3x^3 + x^2 − 7x − 5
Rational Zeros Theorem: Possible Rational Zeros = factors of constant term / factors of leading coefficient

Theorems

Rational Zeros Theorem

Suitable Grade Level

Grades 10-12

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