Question

Determine the zeros of f(x) = x3 – 12x2 + 28x – 9. (1 point)

Answer 1

x²(x-12)+ i wasnt sure how to do the next part because nothing in goes into 9 and 28 . i dont think .

Answer 2

@ganeshie8

Answer 3

since its not working we need to do it lil different

Answer 4

lets split the given polynomial into 6 terms :-

Answer 5

\(f(x) = x^3 – 12x^2 + 28x – 9 \)

\( = x^3 – 9x^2 - 3x^2 + 27x + x – 9 \)

Answer 6

i just wrote,

-12x2 as -9x2 - 3x2

28x as 27x + x

Answer 7

is that fine ?

Answer 8

yes . okk i see what you did there.

Answer 9

next, first two terms pull out x^2

Answer 10

middle two terms pull out -3x

Answer 11

last two terms pull out 1

Answer 12

\(f(x) = x^3 – 12x^2 + 28x – 9 \)

\( = x^3 – 9x^2 - 3x^2 + 27x + x – 9 \)

\( = x^2(x – 9) - 3x(x - 9) + 1(x – 9) \)

Answer 13

now (x-9) is common, pull it out

Answer 14

x3 – 9x²-3x²+27x+x-9
x²(x-9)-3x(x-9)+1(x-9)

Answer 15

okk thats what i got.

Answer 16

okk so (x²-3x+1)(x-9)

Answer 17

\(f(x) = x^3 – 12x^2 + 28x – 9 \)

\( = x^3 – 9x^2 - 3x^2 + 27x + x – 9 \)

\( = x^2(x – 9) - 3x(x - 9) + 1(x – 9) \)

\( =(x – 9)( x^2 - 3x + 1)\)

Answer 18

yes, thats the maximum we can factor. now set it to 0

Answer 19

\((x – 9) (x^2-3x+1) = 0\)

Answer 20

so then x-9=0 and x²-3x+1=0

Answer 21

yes, you need to solve the quadratic

Answer 22

so x=9. im not sure how to do the other one :(

Answer 23

thats one zero, x = 9

Answer 24

x²-3x+1=0

Answer 25

use quadratic formula

Answer 26

\(\large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Answer 27

\(\large x = \frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times 1}}{2 \times 1}\)

Answer 28

\(\large x = \frac{3 \pm \sqrt{9-4}}{2}\)

Answer 29

\(\large x = \frac{3 \pm \sqrt{5}}{2}\)

Answer 30

okk sooo im there so far. then idont what to do from there.

Answer 31

yes thats where we stop :)

Answer 32

so, the zeroes are : \(9, \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\)

Answer 33

ohhhh lol wooww thank you soo much . you are amazing !

Answer 34

ohh ty :) you're amazingly brilliant too :D