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The function f(x) = x3 – 8x2 x 42 has zeros located at 7, –2, 3. verify the zeros of f(x) and explain how you verified them. describe the end behavior of the function.

Sagot :

The function [tex]f(x)=x^3 - 8x^2 + 42[/tex] contains zeros located at 7, –2, 3.

where the other two zeros exist X = 5, and x = -2.

How to determine the zeros of the function

[tex]f(x)=x^3 - 8x^2 + 42[/tex]?

Given: [tex]f(x)=x^3 - 8x^2 + 42[/tex]

To turn roots into factors and then multiply then you have

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

[tex]f(x) = a(x + 2)(x^2 - 10x + 25)[/tex]

[tex]f(x) = a(x^3 - 8x^2 + 5x + 50)[/tex]

By using synthetical or the factor theorem,

f(-2) = a(-8 - 8(4) + 5(-2) + 50) = a(0) = 0.

f = -2

€(5) = a(125 - 8(25) + 5(5) + 50) = a(0) = 0

f = 5 works

Divide it into the second multiplying

And you should get [tex]$x^2 - 3x - 10 = (x - 5)(x + 2)[/tex],

where the other two zeros exist X = 5, and x = -2.

To learn more about factors refer to:

https://brainly.com/question/11930302

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