Question

g(x) = x³ + 3x² - x - 3
a) state the total # of complex roots
b) use the rational root theorem to list the possible rational roots.
c) solve for the roots of g(x)
d) sketch the graph

Explanation:

Step 1: Analyze Part A (Total Complex Roots)

For a polynomial \( g(x) = x^3 + 3x^2 - x - 3 \), the degree of the polynomial is 3 (since the highest power of \( x \) is 3). By the Fundamental Theorem of Algebra, the total number of complex roots (counting multiplicities) of a polynomial is equal to its degree. So a cubic polynomial (degree 3) has 3 complex roots.

Step 2: Apply Rational Root Theorem (Part B)

The Rational Root Theorem states that any possible rational root \( \frac{p}{q} \) of a polynomial \( a_nx^n + \dots + a_1x + a_0 \) has \( p \) as a factor of the constant term \( a_0 \) and \( q \) as a factor of the leading coefficient \( a_n \). For \( g(x) = x^3 + 3x^2 - x - 3 \), the leading coefficient \( a_n = 1 \) and the constant term \( a_0 = -3 \).

• Factors of \( a_0 = -3 \): \( \pm1, \pm3 \)
• Factors of \( a_n = 1 \): \( \pm1 \)
• So possible rational roots \( \frac{p}{q} \) are \( \pm1, \pm3 \)

Step 3: Solve for Roots (Part C)

We can factor \( g(x) \) by grouping:

$$
LATEXBLOCK0
$$

Setting each factor equal to zero:

• \( x - 1 = 0 \implies x = 1 \)
• \( x + 1 = 0 \implies x = -1 \)
• \( x + 3 = 0 \implies x = -3 \)

Step 4: Sketch the Graph (Part D)

To sketch \( g(x) = (x - 1)(x + 1)(x + 3) \):

• x-intercepts: From the roots, \( x = -3, -1, 1 \). So the graph crosses the x-axis at \( (-3, 0) \), \( (-1, 0) \), and \( (1, 0) \).
• End Behavior: For large \( |x| \), the leading term \( x^3 \) dominates. As \( x \to \infty \), \( x^3 \to \infty \), so the graph goes up to the right. As \( x \to -\infty \), \( x^3 \to -\infty \), so the graph goes down to the left.
• Behavior at Intercepts: The multiplicity of each root is 1 (odd), so the graph crosses the x-axis at each intercept.
• We can also find a few more points: when \( x = 0 \), \( g(0) = -3 \), so the y-intercept is \( (0, -3) \).

Answer:

• Part A: The total number of complex roots of \( g(x) \) is \( \boldsymbol{3} \) (by Fundamental Theorem of Algebra, degree = 3).
• Part B: Possible rational roots are \( \boldsymbol{\pm1, \pm3} \) (from Rational Root Theorem: \( \frac{\text{factors of } -3}{\text{factors of } 1} \)).
• Part C: The roots of \( g(x) \) are \( \boldsymbol{x = -3, x = -1, x = 1} \) (found by factoring \( g(x) = (x + 3)(x + 1)(x - 1) \)).
• Part D: Sketch has x-intercepts at \( (-3, 0) \), \( (-1, 0) \), \( (1, 0) \), y-intercept at \( (0, -3) \), end behavior: down as \( x \to -\infty \), up as \( x \to \infty \), and crosses the x-axis at each intercept.