Question

find all the zeros of the following polynomial:
$x^3 + 6x^2 + 11x + 6$
-1, 2, 3
-1, -2, -3
1, 2
1, 2, 3

Explanation:

Step1: Test x = -1

Substitute \( x = -1 \) into \( x^3 + 6x^2 + 11x + 6 \):
\( (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \). So -1 is a zero.

Step2: Factor the polynomial

Since \( x = -1 \) is a zero, \( (x + 1) \) is a factor. Use polynomial division or synthetic division to factor \( x^3 + 6x^2 + 11x + 6 \) by \( (x + 1) \).
Using synthetic division:
-1 | 1   6   11   6
|    -1   -5   -6
----------------
1   5    6   0
So the polynomial factors as \( (x + 1)(x^2 + 5x + 6) \).

Step3: Factor the quadratic

Factor \( x^2 + 5x + 6 \): \( x^2 + 5x + 6 = (x + 2)(x + 3) \).

Step4: Find all zeros

The polynomial is \( (x + 1)(x + 2)(x + 3) \), so the zeros are \( x = -1, -2, -3 \).

Answer:

B. -1, -2, -3