find all solutions of the polynomial equation:
x^4 - x^3 - 6x^2 + 4x + 8 = 0

- anonymous

find all solutions of the polynomial equation:
x^4 - x^3 - 6x^2 + 4x + 8 = 0

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

x4−x3−6x2+4x+80 is your answer

- anonymous

you want to use grouping to factor this polynomial

- anonymous

for the first 3 terms, you can factor out an X^2

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

how should i group them?

- anonymous

i got x^2(x^2-x-6)+4(x+2)

- anonymous

x^2(x-3)(x+2)+4(x+2)
(x+2)(x^2(x-3)+4) :/

- anonymous

lets start over

- anonymous

x^4 - x^3 - 6x^2 + 4x + 8 = 0

- anonymous

use factor and remainder theorem to solve it

- anonymous

x^2(x^2 - x - 6) + 4(x + 2) = 0

- anonymous

do you have to use special factors?

- anonymous

see by plugging x=-2 the expression x^4 - x^3 - 6x^2 + 4x + 8 becomes 0
hence x+2 is one of the factor for getting other factors divide x^4 - x^3 - 6x^2 + 4x + 8 by x+2

- anonymous

no... (x^2+4)(x-3)(x+2)^2=0 now solve each equal to zero

- anonymous

no to special factors, matricked is right also

- anonymous

how'd you get there..

- anonymous

x=-2 and x=2 x=3

- anonymous

you skipped a couple steps from x^2(x^2 - x - 6) + 4(x + 2) = 0

- anonymous

i further factored (x^2-x-6) and grouped (x^2+4)

- anonymous

can you show me how, from there, you got (x^2+4)(x-3)(x+2)^2=0?

- anonymous

now set each equation equal to 0

- anonymous

the x^2 and +4 you pulled out go in their own equation

- anonymous

wait so from x^2(x-3)(x+2) + 4(x+2) = (x^2+4)(x+2)^2 (x-3)?

- anonymous

yes

- anonymous

now set each equal to zero by themselves and solve for x

- anonymous

hold on

- anonymous

x^2+4=0

- anonymous

the rest i know how, but h\where's the logic in that move, where you can take x^2+4

- anonymous

you factored it out by grouping

- anonymous

it is my understanding that in order to factor by grouping both pieces need to have common factors like, "4(x+2)" doesn't have a (x-3)

- anonymous

are you learning imaginary numbers?

- anonymous

right, but x^2-x-6 did

- anonymous

you further factored

- anonymous

x^4 - x^3 - 6x^2 + 4x + 8=0
x^4 +2x^3 -3x^3 - 6x^2 + 4x + 8=0
X^3(x+2) -3x^2(x+2) +4(x+2)=0
(x+2)(x^3-3x^2+4)=0
(x+2)(x^3-2x^2-x^2+4)=0
(x+2)(x^2(x-2) - (x+2)(x-2))=0
(x+2)(x-2)(x^2-x-2)=0
(x+2)(x-2)(x^2-2x+x-2)=0
(x+2)(x-2)(x(x-2)+1(x-2))=0
or(x+2)(x-2)(x-2)(x+1)=0
now set each equal to 0
thus the roots are -1,-2,2,2

- anonymous

wow, nice job

Looking for something else?

Not the answer you are looking for? Search for more explanations.