Identify all zeros for the function f(x)=x^3-29x^2+105x-125.

Answer 1

First, all of the possible zeros are ±1, ±125, ±5, ±25 (all of the factors of your constant divided by all of the factors of your leading coefficient.) Using synthetic division, you can determine whether any of these are a zero.

Let's start with 25, because I know it will work.

x³ - 29x² + 105x -125

25) 1 -29 105 -125

25 -100 125

__________________________

1 -4 5 0

So, 25 is one of the zeros of the function.

Now we have

(x - 25)(x² -4x + 5)

x² - 4x + 5 will not factor without using the quadratic formula.

This is how you can find the other zeros, although they will not be REAL zeros because they will contain imaginary numbers

4 ± √ 16-20

2

4 ± √(-4)

2

4 ±2i

2

2 ± i

And now your zeros are x = 25. x = 2 + i x = 2 - i

Answer 2

Hi,

Graph this function in Desmos Graphing

You will be the function only one time through of x-axis ( at x = 25) that it means it has only one real zero

such as x = 25

I hope it is useful.

Minoo