Question

Find all zeros x^3+2x^2+4x+21

Answer 1

Answer:

see explanation

Step-by-step explanation:

note when x = - 3

(- 3)³ + 2(- 3)² + 4(- 3) + 21 = - 27 + 18 - 12 + 21 = 0

hence x = - 3 is a zero and (x + 3) is a factor and dividing gives

$\frac{x^3+2x^2+4x+21}{x+3}$ = (x + 3)(x² - x + 7)

For zeros equate to zero

(x + 3)(x² - x + 7) = 0

equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x² - x + 7 = 0 ← solve using quadratic formula

x = (1 ± $\sqrt{1-28}$ ) / 2 = (1 ± 3i$\sqrt{3}$ ) / 2

x = $\frac{1}{2}$ ± $\frac{3i\sqrt{3} }{2}$

zeros are x = - 3, x = $\frac{1}{2}$ ±$\frac{3i\sqrt{3} }{2}$