x^3+2x^2-16x-32=0?

Question

can you please breakdown how you come up with the answers..THANKS....I am 36yrs old...was an Algebra wiz 25 yrs ago...cant remember how to breakdown factoring an equation..i do know the answers are: x= -2, x= -4 and x = 4

Answer 1

f ( 4 ) = 64 + 32 - 64 - 32 = 0
Thus x - 4 is a factor
Find other factors by synthetic division.
"""""""|1"""""2""""- 16"""""-32
"""""4|"""""""4""""""24""""""32
"""""""|1"""""6"" """""8""""""0
( x - 4 ) ( x ² + 6x + 8 ) = 0
( x - 4 ) ( x + 4 ) ( x + 2 ) = 0
x = - 2 , x = - 4, x = 4

Answer 2

x³ + 2x² - 16x - 32 =0
Regroup
(x³ -16x) + (2x² - 32) = 0
Factor both terms
x(x² - 16) + 2(x² -16) = 0
Now Factor:
(x + 2)(x² -16) = 0
Factor x² -16:
(x + 2 )(x - 4)( x+ 4) =0
Now x +2 = 0, x - 4 = 0, and x +4 = 0
Solving these 3 equations gives you:
x = -2, 4, -4

Answer 3

(x^3+2x^2)-(16x+32) = 0, regrouping
x^2(x+2)-16(x+2) = 0, factoring each group
(x^2-16)(x+2) = 0, since (x+2) is a common factor
(x+4)(x-4)(x+2) = 0, since x^2-16 = (x+4)(x-4)
Solve x+4=0, x-4=0 and x+2=0 for x,
x = -4, 4, -2

Answer 4

x^3+2x^2-16x-32=0
x^2(x+2)-16(x-2)=0
(x^2-16)(x+2) = 0
(x+4)(x-4)(x+2) =0
Therefore x =-4; 4; -2

Answer 5

x^3 + 2x^2 - 16x - 32
x^2(x + 2) - 16(x + 2) = 0
(x + 2)(x^2 - 16) = 0
(x + 2)(x + 4)(x - 4)= 0 (since x^2 - 16 = x^2 - 4^2 = (x+4)(x-4))
x = -2, + or - 4

Answer 6

x^3+2x^2-16x-32=0

x^2(x+2)-16(x+2)

because you have two common things (x+2), you group the remaining two (x^2-16)---factor this one, set the other one to zero.

x=-2
x=-4,4