Given a polynomial that has zeros of −5, 4i, and −4i and has a value of 40 when x=3 write the polynomial equation

Answer:

[tex]f(x) = 0.2(x^3 + 5x^2 + 16x + 80)[/tex]

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots [tex]x_{1}, x_{2}, x_{n}[/tex] such that it can be written as: [tex]a(x - x_{1})*(x - x_{2})*...*(x-x_n)[/tex], in which a is the leading coefficient.

Zeros of −5, 4i, and −4i

This means that [tex]x_1 = -5, x_2 = 4i, x_3 = -4i[/tex]. So

[tex]f(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)[/tex]

[tex]f(x) = a(x - (-5))*(x - 4i)*(x - (-4i))[/tex]

[tex]f(x) = a(x + 5)(x - 4i)(x + 4i)[/tex]

[tex]f(x) = a(x + 5)(x^2 - 16i^2)[/tex]

Since [tex]i^2 = -1[/tex]

[tex]f(x) = a(x + 5)(x^2 + 16)[/tex]

[tex]f(x) = a(x^3 + 5x^2 + 16x + 80)[/tex]

Has a value of 40 when x=3

We use this to find a, which means that we have [tex]x = 3, f(x) = 40[/tex]. So

[tex]a(3^3 + 5*3^2 + 16*3 + 80) = 40[/tex]

[tex]200a = 40[/tex]

[tex]a = \frac{40}{200} = \frac{1}{5} = 0.2[/tex]

So

[tex]f(x) = 0.2(x^3 + 5x^2 + 16x + 80)[/tex]

13. Using figure 1, angles 5 and 6 form a linear pair, If mz6=95", find m25.
a. 5°
b. 85"
c. 90°
d.95"
14. In figure 1, 25 and 28 are vertical angles. If m2s 2x + 45 and m28= 7x - 15
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