Question

Help please on this

Answer 1

I already did part I. It is x=1,-2,2

Answer 2

@Hero @JustSaiyan @Ultrilliam

Answer 3

it is almost impossible to write synthetic division here, but i can try

Answer 4

on the top line list the coefficients of \(x^3-x^2+8x+10\)
they are \[\large 3~~-1~~8~~10\]

Answer 5

on the left, put a \(\large 1\)

Answer 6

I got it I neeed help withanother problem now

Answer 7

@satellite73

Answer 8

part 1. list the numbers that go in to 4 evenly
there are 6 of them

Answer 9

you know what they are?q

Answer 10

I got 1,-2,2

Answer 11

also \(-1,4,-4\)

Answer 12

OKay got that =)

Answer 13

NOw part II

Answer 14

unless you meant those are the actual roots
part one only asks for the POSSIBLE roots

Answer 15

part two
check the easiest number first, that is \(1\)
\[f(1)=1-1+4-1=0\] got it on the first try

Answer 16

ok typo there, should have been \[F(1)=1-1+4-4=0\]

Answer 17

Okay got that =)

Answer 18

so now you know if factors as \[f(x)=x^3-x^2+4x-4=(x-1)(\text{something})\]

Answer 19

Okay so is this is part of part II as well?

Answer 20

yes, we found one root by checking, we still need to find the others

Answer 21

the easiest way to find the quadratic polynomial is synthetic division, in other words find \[\frac{x^3-x^2+4x-4}{x-1}\]

Answer 22

Okay so we only found 1 we need to find the rest?

Answer 23

yes

Answer 24

Okay for -1 it is...

Answer 25

um I would just put f(-1) and write the rest of the equation right?

Answer 26

no that is not what you should do
you should divide like i wrote above

Answer 27

\[\frac{x^3-x^2+4x-4}{x-1}\]

Answer 28

okay just simply divide

Answer 29

i made another typo, it should be \[\frac{x^3-x^2-4x+4}{x-1}\]

Answer 30

I got x^2-4

Answer 31

i made another typo
it should be \[\frac{x^3-x^2-4x+4}{x-1}\]

Answer 32

yes, you are right

Answer 33

so would this be part III?

Answer 34

so you have \[x^3-x^2-4x+4=(x-1)(x^2-4)\]

Answer 35

no you are still doing part two, find the other zeros

Answer 36

ohhhh okay

Answer 37

since \(x^2-4=(x+2)(x-2)\) the other zeros are \(-2\) and \(2\)
all three zeros are \[\{-2,1,2\}\]

Answer 38

part three is to factor the polynomial completely which you already did
it is \[(x-1)(x-2)(x+2)\]

Answer 39

okay and then step IV

Answer 40

you have to multiply out to show that it is right, which is silly but not hard

Answer 41

\[(x-1)(x-2)(x+2)=(x-1)(x^2-4)\] is a start

Answer 42

then multiply the last two and see that you get \[x^3-x^2-4x+4\] which you will

Answer 43

and then is that it?

Answer 44

yes, that is all it asks for

Answer 45

thank you soooooooo much I really really appreciate it =)

Answer 46

yw