Question

find the zeros of the polynomial p(x), one has been provided for you.

1. p(x) = x³ - 6x² + 11x - 6; x = 2
2. p(x) = x³ - 3x² - 4x + 12; x = 2
3. p(x) = x³ - 4x² - 5x; x = -1
4. p(x) = 3x³ + 4x² - 45x + 50; x = -5

Explanation:

Problem 7: \( p(x) = x^3 - 6x^2 + 11x - 6 \); \( x = 2 \)

Step 1: Factor using the given zero

Since \( x = 2 \) is a zero, \( (x - 2) \) is a factor. We perform polynomial division or use synthetic division to divide \( p(x) \) by \( (x - 2) \). Using synthetic division with root 2:

$$
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$$

So, \( p(x) = (x - 2)(x^2 - 4x + 3) \).

Step 2: Factor the quadratic

Factor \( x^2 - 4x + 3 \). We need two numbers that multiply to 3 and add to -4. Those numbers are -1 and -3. So, \( x^2 - 4x + 3 = (x - 1)(x - 3) \).

Step 3: Find all zeros

Set each factor equal to zero:

• \( x - 2 = 0 \) gives \( x = 2 \) (given)
• \( x - 1 = 0 \) gives \( x = 1 \)
• \( x - 3 = 0 \) gives \( x = 3 \)

Step 1: Factor using the given zero

Since \( x = 2 \) is a zero, \( (x - 2) \) is a factor. Use synthetic division with root 2:

$$
LATEXBLOCK0
$$

So, \( p(x) = (x - 2)(x^2 - x - 6) \).

Step 2: Factor the quadratic

Factor \( x^2 - x - 6 \). We need two numbers that multiply to -6 and add to -1. Those numbers are -3 and 2. So, \( x^2 - x - 6 = (x - 3)(x + 2) \).

Step 3: Find all zeros

Set each factor equal to zero:

• \( x - 2 = 0 \) gives \( x = 2 \) (given)
• \( x - 3 = 0 \) gives \( x = 3 \)
• \( x + 2 = 0 \) gives \( x = -2 \)

Step 1: Factor out the GCF

First, factor out \( x \) from \( p(x) \): \( p(x) = x(x^2 - 4x - 5) \).

Step 2: Check the given zero

Check if \( x = -1 \) is a zero: \( p(-1) = (-1)((-1)^2 - 4(-1) - 5) = (-1)(1 + 4 - 5) = (-1)(0) = 0 \), so \( (x + 1) \) is a factor. Factor \( x^2 - 4x - 5 \) using \( x = -1 \) (or find two numbers that multiply to -5 and add to -4: -5 and 1). So, \( x^2 - 4x - 5 = (x - 5)(x + 1) \).

Step 3: Find all zeros

Set each factor equal to zero:

• \( x = 0 \)
• \( x - 5 = 0 \) gives \( x = 5 \)
• \( x + 1 = 0 \) gives \( x = -1 \) (given)

Answer:

The zeros are \( x = 1 \), \( x = 2 \), and \( x = 3 \).

Problem 8: \( p(x) = x^3 - 3x^2 - 4x + 12 \); \( x = 2 \)