Given that
−
2
+
2
𝑖
is a root, which of the follwing options lists all roots of the following polynomial:
𝑃
(
𝑥
)
=
𝑧
3
+
8
𝑧
2
+
24
𝑧
+
32.

Question

Given that 
−
2
+
2
𝑖
 is a root, which of the follwing options lists all roots of the following polynomial:
𝑃
(
𝑥
)
=
𝑧
3
+
8
𝑧
2
+
24
𝑧
+
32.

BlackTom AI · Accepted Answer

Complex roots of polynomials with real coefficients occur in conjugate pairs, so if −2 + 2i is a root then −2 − 2i must also be a root. Form the quadratic factor from that conjugate pair and divide the cubic to find the remaining root.

1) Quadratic factor from the pair:
(z − (−2+2i))(z − (−2−2i)) = (z + 2 − 2i)(z + 2 + 2i)
Use (a−b)(a+b) = a^2 − b^2 with a = z+2, b = 2i:
(z + 2)^2 − (2i)^2 = (z^2 + 4z + 4) − (−4) = z^2 + 4z + 8.

2) Divide P(z) by z^2 + 4z + 8 to get the remaining linear factor:
P(z) = z^3 + 8z^2 + 24z + 32.
Long division steps:
- Leading term: z^3 ÷ z^2 = z. Multiply divisor: z(z^2 + 4z + 8) = z^3 + 4z^2 + 8z.
- Subtract: (z^3 + 8z^2 + 24z + 32) − (z^3 + 4z^2 + 8z) = 4z^2 + 16z + 32.
- Next term: 4z^2 ÷ z^2 = 4. Multiply divisor: 4(z^2 + 4z + 8) = 4z^2 + 16z + 32.
- Subtract remainder = 0. Quotient = z + 4.

Thus P(z) = (z^2 + 4z + 8)(z + 4), so the third root is z = −4.

3) Direct check: P(−4) = (−4)^3 + 8(−4)^2 + 24(−4) + 32 = −64 + 128 − 96 + 32 = 0, confirming −4 is a root.

Why other choices would be wrong: any proposed third root other than −4 would not make the polynomial zero; and omitting the conjugate −2 − 2i would contradict the conjugate-pair property for real-coefficient polynomials.

Conclusion: the roots are −2 + 2i, −2 − 2i, and −4, matching the provided answer.

Given that − 2 + 2 𝑖 is a root, which of the follwing options lists all roots of the following polynomial: 𝑃 ( 𝑥 ) = 𝑧 3 + 8 𝑧 2 + 24 𝑧 + 32.单项选择题

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