Question

Find all real solutions for the following:
(x+3)^2(x-1)+(x+3)(x-1)^2=0

Answer 1

factoring is your friend

Answer 2

\[(x+3)^2(x-1)+(x+3)(x-1)^2=0\]

Answer 3

don't divide them

Answer 4

(x+3) 2 (x−1)+(x+3)(x−1) 2
= (x+3)(x-1)((x+3)^2+(x-1))

Answer 5

^Thanks. I was unsure of how I should go about factoring it.

Answer 6

each term has a common factor of \((x+3)(x-1)\) factor it out

Answer 7

you get
\[(x+3)(x-1)(x-3+x+1)\]
\[(x+2)(x-1)(2x-2)\]

Answer 8

Like @funinabox said, factoring is your friend. Factor out (x+3) and (x-1):\[\bf (x+3)^2(x-1)+(x+3)(x-1)^2=0 \rightarrow (x+3)(x-1)[(x+3)+(x-1)]\]Simplifying yields:\[\bf (x+3)(x-1)(2x+2)=2(x+3)(x-1)(x+1)\]

Answer 9

(x+3)^2(x-1)+(x+3)(x-1)^2=0
@satellite73
(x+3)(x−1) ((x+3)^2+x-1)

Answer 10

yeah i screwe, up

Answer 11

@genius12 has it

Answer 12

Thanks everyone!

Answer 13

should be \[(x+2)(x-1)(2x+2)\]

Answer 14

@satellite73 u screwed up again lol.