Question

Same type of question

Answer 1

okay so for part I I got +-1,+-,2,+-3,+-4,+-6,+-12

Answer 2

then for part II i got 1

Answer 3

am i on the right track so far?

Answer 4

the root 1 does not produce a remainder of 0 after synthetic division so you will have to ry another root

Answer 5

2

Answer 6

not quite, check your calculations again

if you plug in the root into the equation it should equal 0

Answer 7

to speed this process up you can use a calculator to start plugging in x values and see which ones give you 0, starting with 1 or -1, then 2 and -2, etc. until you get 0

Answer 8

silly if you're reading this is there an easier way to narrow down the possible roots w/o using calculus

Answer 9

okay so zero for part II

Answer 10

@sillybilly123

Answer 11

0 is not a possible root (it is not in your list of rational roots)

Answer 12

\(\color{#0cbb34}{\text{Originally Posted by}}\) @zarkam21
okay so for part I I got +-1,+-,2,+-3,+-4,+-6,+-12
\(\color{#0cbb34}{\text{End of Quote}}\)

^ your root must be one of these

Answer 13

4

Answer 14

when you plug in x = 4, your result is not 0 so keep trying

Answer 15

i got it

Answer 16

its 3

Answer 17

good

for part III) divide the function by (x-3) and write the resulting function

Answer 18

x^3+4x^2+6x+4

Answer 19

awesome, then for part IV try to find another root by testing the possible roots from part I

Answer 20

good, now try testing these roots to see if you can find another solution

Answer 21

so just substitute for x right

Answer 22

yes

Answer 23

into the cubic function right not the original

Answer 24

yes (technically both will work but cubic is easier)

Answer 25

I'm getting 15,40,85

Answer 26

Am i on the right track

Answer 27

what did you do to obtain these values?

Answer 28

oh i have to set them equal to zero

Answer 29

if you are plugging in roots, keep going until the result is 0

Answer 30

oh okY

Answer 31

-2?

Answer 32

good, so -2 is the other root so that's your answer for IV

Answer 33

OKAY SO for part V it would just be x=-1,+-i,-2

Answer 34

part V is not in the original screenshot

Answer 35

PArtV use the quadratic formula to find the final 2 roots

Answer 36

after you divide the cubic polynomial by (x+2) what polynomial do you get?

Answer 37

x^2+2x+2

Answer 38

good, now apply the quadratic formula to get the last two roots

Answer 39

x=-1+i

Answer 40

good but don't forget the other one x = -1 - i (remember the quadratic formula has a +/- sign)

and that's it