Question

find all solutions of the polynomial equation:

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

Answer 1

x4−x3−6x2+4x+80 is your answer

Answer 2

you want to use grouping to factor this polynomial

Answer 3

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

Answer 4

how should i group them?

Answer 5

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

Answer 6

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

Answer 7

lets start over

Answer 8

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

Answer 9

use factor and remainder theorem to solve it

Answer 10

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

Answer 11

do you have to use special factors?

Answer 12

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

Answer 13

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

Answer 14

no to special factors, matricked is right also

Answer 15

how'd you get there..

Answer 16

x=-2 and x=2 x=3

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

now set each equation equal to 0

Answer 21

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

Answer 22

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

Answer 23

yes

Answer 24

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

Answer 25

hold on

Answer 26

x^2+4=0

Answer 27

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

Answer 28

you factored it out by grouping

Answer 29

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)

Answer 30

are you learning imaginary numbers?

Answer 31

right, but x^2-x-6 did

Answer 32

you further factored

Answer 33

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

Answer 34

wow, nice job