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Sophie [7]
3 years ago
8

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

Mathematics
1 answer:
dlinn [17]3 years ago
5 0

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}


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