Question

use the quadratic equation to solve for x^3-216=0

Answer 1

factor as the difference of two cubes. you have
\[x^3-6^3\]

Answer 2

factors as
\[(x-6)(x^2+6x+36)\]

Answer 3

so one solution to
\[(x-6)(x^2+6x+36)=0\] is
\[x=6\]

Answer 4

the other two come from the quadratic
\[x^2+6x+36=0\] and the solutions will be complex. do you need these as well?

Answer 5

i used
\[a^3-b^3=(a-b)(a^2+ab+b^2)\] to factor

Answer 6

if you want to solve
\[x^2+6x+36=0\]
\[x^2+6x=-36\]
\[(x+3)^2=-36+3^2=-27\]
\[x+3=\pm\sqrt{-27}\]

Answer 7

the formula is good... but any case u can see dividing and obtain the other part
if we know one factor

Answer 8

and so
\[x=-3\pm3\sqrt{3}i\]

Answer 9

so after you get the x=6 you apply in for the second step?

Answer 10

you could also do it the quick way

\[6e^{i\cdot0},6e^{i\cdot2\pi/3},6e^{i\cdot4\pi/3}\]

Answer 11

;)

Answer 12

yeah sure

Answer 13

not how they wanted it ...but it is quicker

Answer 14

I asked because when i used the a^3 - b^3 i don't get the x^2 + 6x+36

Answer 15

or the real quick way, say it is
\[6\times 1, 6\times \frac{-1}{2}\pm\frac{\sqrt{3}}{2}i\]

Answer 16

ok lets go slow

Answer 17

what you are supposed to recognize is
"the difference of two cubes"
\[a^3-b^3\]

Answer 18

zarkon I feel like you are speaking another language lol I have no idea what those are, i HAVEN'T LEARNED THAT YET.

Answer 19

the difference of two cubes always factors, as does the sum of two cubes. the check is by multipication

Answer 20

Okay I got that part satellite73

Answer 21

that is ok, but it is clear what it means right? "difference" means subtraction
and "cubes" means something cubed

Answer 22

yes

Answer 23

and it is a fact that
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 24

if you multiply out on the right, you will get the expression on the left

Answer 25

now you are given
\[x^3-216\] and since the first term is a cube, figure that the second one must be a cube also

Answer 26

the first side is x-6

Answer 27

and check that
\[6^3=216\] so you have
\[x^3-6^3\] which looks just like
\[a^3-b^3\] right?

Answer 28

ohh okay

Answer 29

so i write
\[x^3-6^3=(x-6)(x^2+6x+6^2)\] just like i would if they were
\[a^3-b^3\]

Answer 30

could it also be written as (x-6)(x+6)(x-6)

Answer 31

absolutely not

Answer 32

there is only one way to factor an expression, just like there is only one way to factor a number

Answer 33

if you multiply out what you wrote, you get
\[x^3-6 x^2-36 x+216\] not
\[x^3-216\]

Answer 34

try it and you will see

Answer 35

what if I used (x+6)(x+6) to get the second part?

Answer 36

or is it that whatevever the cube equals I must apply it in the second? for example x=6. so therefore I could just place the 6 in the second.

Answer 37

ok at the risk of sounding like a math teacher (i apologize in advance) there is only one way to factor the difference of two cubes. so
\[x^3-216=(x-6)(x^2+6x+36\] and no other.

Answer 38

the second part
\[x^2+6x+36\] does not factor using integers

Answer 39

you cannot do it. that is why the zeros are complex numbers.

Answer 40

It's okay the way you sounded but I needed it in a simple form sorry to sound slow but yea my brain shut down on me tonight.

Answer 41

so its just something I must know then? from what you did i noticed the 6 is applied in the b and the c^2. is it like that all the time? even my book did the same thing.

Answer 42

i guessed it because the first term was a cube. it was
\[x^3\]

Answer 43

so the only way this could factor as the difference of two cubes is if the second term was a cube as well

Answer 44

now not being a calculating genius, i took out a calculator and put in
\[\sqrt[3]{216}\] and out popped 6, just like i suspected

Answer 45

that is how i knew it was
\[x^3-6^3\]

Answer 46

so if i see
\[x^3-8\] i think "x cubed minus 2 cubed" and write
\[x^3-8=(x-2)(x^2+2x+4)\]

Answer 47

\[x^3-27=(x-3)(x^2+3x+9)\] and so one

Answer 48

right and the 2 and 3 were applied in the b and c

Answer 49

exactly!

Answer 50

fantastic, you're a life saver!!!! lol

Answer 51

one more thing. if you have "the SUM of two cubes it factors as well
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]

Answer 52

notice the difference in the placement of the minus sign

Answer 53

that equation def came in handy

Answer 54

are you good with math?

Answer 55

fair to middlin'

Answer 56

lol okay, thank you so much, I appreciate it