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x4−x3−6x2+4x+80 is your answer

you want to use grouping to factor this polynomial

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

how should i group them?

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

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

lets start over

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

use factor and remainder theorem to solve it

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

do you have to use special factors?

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

no to special factors, matricked is right also

how'd you get there..

x=-2 and x=2 x=3

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

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

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

now set each equation equal to 0

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

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

yes

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

hold on

x^2+4=0

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

you factored it out by grouping

are you learning imaginary numbers?

right, but x^2-x-6 did

you further factored

wow, nice job