PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!What are the zeros of the polynomial function?


Select EACH correct answer.

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

Answer:

-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