Question

Find all the real zeros of the polynomial
P(x)=x^4 - 3 x^3 - 6 x^2 + 8 x .

Answer 1

Can you factor the polynomial?

Once you've factored it into the form
\[P(x) = (x-a)(x-b)(x-c)(x-d)\]a-d are the roots.

Answer 2

thank you

Answer 3

Post your answers and I'll check them for you if you want.

Answer 4

this is as far as i was able to get x(x^2-3x^2-6x+8) can someone please help me?

Answer 5

have you tried (x-1) as a root?

Answer 6

no because it was all tied up in a poly and I thought i had to factor first to break it down

Answer 7

Divide \(x(x^3-3x^2-6x+8)\) with \((x-1)\) then tell me what you got :)

Answer 8

I don't even know how to compute that.

Answer 9

Use synthetic division; have you learned that yet?

Answer 10

yes. i'm just confuse with that x in the front.

Answer 11

you can ignore that when you do the synthetic division, and then multiply it back afterwards. just divide \((x^3−3x^2−6^x+8)\) with \((x-1)\)

Answer 12

so when when I did the division my numbers were 1, 4, 10, 2 there wasn't a zero in there. I tried it with the other factors of 8 and 4 seemed to work it gave me 1, 1, -2, 0 but i'm not sure what to do with that now

Answer 13

it's ok.when you divide \(x^3-3x^2-6^x+8\) with \(x-1\) you should get \(x^2-2x-8\), so the whold equation becomes \[x(x-1)(x^2-2x-8)=0\]right?

Answer 14

I didn't get those numbers but i probably did it wrong

Answer 15

okay, yes i just did

Answer 16

do i need to take the 1 and create another division with the new numbers of 1, -2, 8 and 0 now?

Answer 17

well no zero bc there wasn't one when i did it

Answer 18

1| 1 -3 -6 8
| 1 -2 -8
-----------
1 -2 -8 | 0
That's my synth. div. result

Answer 19

when you look for zeros, it means to look for x's that would make this equation zero, which means solve the equation

Answer 20

okay, yes see. i accidentally didn't computer my -6 and -2 right

Answer 21

it's ok. then just solve \(x^2-2x-8=0\) as you would, as in factor it.

Answer 22

so would my set up next be:
x-1 1 -2 -8

Answer 23

oh. okay.

Answer 24

i think i need to do the quadratic formula since this isn't factoring right

Answer 25

yup

Answer 26

so for my answer I got x= +/- sqrt 19

Answer 27

maybe i did something wrong, that isn't looking right either

Answer 28

\(x^2-2x-8=(x-4)(x+2)\) ok?

Answer 29

so i didn't need to do the formula?

Answer 30

you need to factor \(x^2-2x-8\) do you know how to do that?

Answer 31

so now that I have those zeros, 4 and -2 what do i do with them?

Answer 32

for your (x-4)(x-2) when i just went to check it i got x^2-8x+8

Answer 33

(x-4)(x\(+\)2) not MINUS

Answer 34

x+2

Answer 35

check again

Answer 36

okay, sorry. yes i got it right this time

Answer 37

so this is what you got \[x^4-3x^3-6x^2+8x=x(x-1)(x+2)(x-4)\]right?

Answer 38

no, but that's oaky

Answer 39

where did that x(x-1) pop up from?

Answer 40

from the beginning...?

Answer 41

oh, yeah i forgot that we took it out

Answer 42

\[x^4-3x^3-6x^2+8x\]\[=x(x^3-3x^2-6x+8)\]\[=(x)(x-1)(x^2-2x-8)\]\[=(x)(x-1)(x-4)(x+2)\]ok?

Answer 43

yes

Answer 44

Then set \[x(x-1)(x+2)(x-4)=0\]and you should be able to find all of the solutions aka zeros.

Answer 45

so my zeros would be 1, 4, -2 ?

Answer 46

do you have the solutions?

Answer 47

yes, x=0 x=1 x=-2 and x=4

Answer 48

yes! YAY YOU GOT IT!

Answer 49

do i need to do something with those? like put them in another synthetic division thing? when I plugged those answers into my computer they came back wrong

Answer 50

no... your answers should be 0,1,-2,4...

Answer 51

well thank you very much for your help sorry it took me so long to get it.

Answer 52

it's ok and anytime!