Correct answer only ! i cannot retake what are the zeros of the polynomial function? select each correct answer.

x = {-3, -1, 1, 7}

Step-by-step explanation:

f(x) = x⁴ - 4x³ - 22x² + 4x + 21

possible roots are: +/- 1, 3, 7, 21

Let's try x = 1

1  |  1     -4     -22       4       21

  |  ↓      1       -3     -25     -21

     1     -3      -25    -21       0     remainder is 0 so (x - 1) is a factor

(x - 1)(x³ - 3x² - 25x - 21)

Next, let's try x = -1

-1  |  1     -3     -25      -21

   |  ↓     -1        4        21

      1     -4      -21       0     remainder is 0 so (x + 1) is a factor

(x - 1)(x + 1)(x² - 4x - 21)

Then, factor the third polynomial

(x - 1)(x + 1)(x + 3)(x - 7)

Now, set each factor equal to zero and solve each one.

(x - 1) = 0    ⇒     x = 1

(x + 1) = 0   ⇒     x = -1

(x + 3) = 0   ⇒    x = -3

(x - 7) = 0   ⇒     x = 7

-3, -1, 1, 7

Step-by-step explanation:

Evaluate the function at each x value in the choices. If the polynomial evaluates to zero, then that x value is a solution.

f(-3) = (-3)^4 - 4(-3)^3 - 22(-3)^2 + 4(-3) + 21 = 0

x = -3 is a solution

f(-1) = (-1)^4 - 4(-1)^3 - 22(-1)^2 + 4(-1) + 21 = 0

x = -1 is a solution

f(0) = (0)^4 - 4(0)^3 - 22(0)^2 + 4(0) + 21 = 21

x = 0 is not a solution

f(1) = (1)^4 - 4(1)^3 - 22(1)^2 + 4(1) + 21 = 0

x = 1 is a solution

f(3) = (3)^4 - 4(3)^3 - 22(3)^2 + 4(3) + 21 = -192

x = 3 is not a solution

f(7) = (7)^4 - 4(7)^3 - 22(7)^2 + 4(7) + 21 = 0

x = 7 is a solution

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.