Question

Find all of the zeros of the function f(x) = x3 – 23x2 + 161x – 303.

Answer 1

Answer:

• x = 3
• x = 10 ± i

Step-by-step explanation:

A graph shows the only real zero to be at x = 3.

Factoring that out gives the quadratic whose vertex form is ...

  y = (x -10)² +1

The roots of this quadratic are the complex numbers x = 10 ± i.

_____

For y = (x -10)² +1, the zeros are ...

  (x -10)² +1 = 0

  (x -10)² = -1 . . . . . . . . . . subtract 1

  x -10 = ±√(-1) = ±i . . . . .take the square root

  x = 10 ± i . . . . . . . . . . . . add 10

Answer 2

Answer:

3, 10±i

Step-by-step explanation:

Given is a function $f(x) = x^3 - 23x^2 + 161x -303.$

By rational roots theorem, this can have zeroes as ±1, ±3,±101

By trial and error checking we find f(3) =0

Hence x-3 is a factor

f(x) = $(x-3)(x^2-20x+101)$

II being a quadratic equation we find zeroes using formula

$x=\frac{20±\sqrt{400-404} }{2} =10+i, 10-i$

zeroes are 3, 10±i