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My name is Ann [436]
3 years ago
6

In Problems 23–30, use the given zero to find the remaining zeros of each function

Mathematics
1 answer:
Talja [164]3 years ago
4 0

Answer:

x =  2i, x = -2i and x = 4 are the roots of given polynomial.

Step-by-step explanation:

We are given the following expression in the question:

f(x) = x^3 - 4x^2+ 4x - 16

One of the zeroes of the above polynomial is 2i, that is :

f(x) = x^3 - 4x^2+ 4x - 16\\f(2i) = (2i)^3 - 4(2i)^2+ 4(2i) - 16\\= -8i+ 16+8i-16 = 0

Thus, we can write

(x-2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Now, we check if -2i is a root of the given polynomial:

f(x) = x^3 - 4x^2+ 4x - 16\\f(-2i) = (-2i)^3 - 4(-2i)^2+ 4(-2i) - 16\\= 8i+ 16-8i-16 = 0

Thus, we can write

(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Therefore,

(x-2i)(x+2i)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16\\(x^2 + 4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

Dividing the given polynomial:

\displaystyle\frac{x^3 - 4x^2 + 4x - 16}{x^2+4} = x -4

Thus,

(x-4)\text{ is a factor of polynomial }x^3 - 4x^2 + 4x - 16

X = 4 is a root of the given polynomial.

f(x) = x^3 - 4x^2+ 4x - 16\\f(4) = (4)^3 - 4(4)^2+ 4(4) - 16\\= 64-64+16-16 = 0

Thus, 2i, -2i and 4 are the roots of given polynomial.

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Answer:

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Step-by-step explanation:

The general equation for an arithmetic progression is:

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Let's solve this by elimination by addition and subtraction.  Subtract the first equation from the second.  We get:

                                                             

Substituting 1/12 for d in the first equation, we get:

a(1) + 14(1/12) = 11/2 or 66/12 (using the LCD 12)

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