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3 years ago

Find a third-degree polynomial equation with rational coefficients that has the roots 2 and -3 + i.

Mathematics

1 answer:

NeTakaya3 years ago

5 0

Answer:

Option B

Step-by-step explanation:

Complex roots occur as conjugate pairs so the third root is -3 - i ( note that the sign changes from + to -).

So in factor form we have:-

(x - 2)(x - (-3 + i))(x - (-3 - i)) = 0 Let's expand the last 2 factors first:-

(x - (-3 + i))(x - (-3 - i))

= (x + 3 - i)(x + 3 + i)

= x^2 + 3x +ix + 3x + 9 + 3i - ix - 3i - i^2

= x^2 + 6x + 9 - (-1)

= x^2 + 6x + 10

Now multiplying by (x - 2):-

(x - 2)(x^2 + 6x + 10) = 0

x^3 + 6x^2 + 10x - 2x^2 - 12x - 20 = 0

x^3 + 4x^2 - 2x - 20 = 0 (answer)

Option B

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Step-by-step explanation:

The recursive rule is a term defined in terms of other terms in the sequence.

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The common ratio can be found by dividing a term by previous term.

For example, all of these are equal:

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They are all equal to $\frac{-1}{2}$ .

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