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Lena [83]
3 years ago
10

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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The ratio of 'girls' to 'all students' is:  "2: 5 " ;  or, write as:  " 2/5 ".
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The ratio of "girls" to "all students" is:  "2x : 5x "  ;  or " 2x/5x " ; 
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       →  and we have the answer:  " 2 :5 " ;  or, write as:  " 2/5 " . 
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The ratio of 'girls' to 'all students' is:  " 2 :5 " ;  or, write as:  " 2/5 ".
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