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Hunter-Best [27]
3 years ago
14

If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

Mathematics
1 answer:
Naddik [55]3 years ago
5 0

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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1) Hundreadth

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Answer:

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Step-by-step explanation:

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3 years ago
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3 years ago
Please answer this in steps....​
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3 years ago
Maria bought a blouse on sale for 30% off. The sale price was $28.76. What was the original price? Write an equation to model th
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Answer:

The equation is <u>sale price</u>=p and the original price is $41.09.

                              .70

Step-by-step explanation:

sale price= (1-.30)p      

<u>sale price=.70</u>p

.70              .70

<u>sale price</u>=p                  

.70

<u>28.76</u>=p

.70

$41.09=p

7 0
3 years ago
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