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Mathematics 29 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

factor as the difference of two cubes. you have $x^3-6^3$

OpenStudy (anonymous):

factors as $(x-6)(x^2+6x+36)$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i used $a^3-b^3=(a-b)(a^2+ab+b^2)$ to factor

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (zarkon):

you could also do it the quick way $6e^{i\cdot0},6e^{i\cdot2\pi/3},6e^{i\cdot4\pi/3}$

OpenStudy (zarkon):

;)

OpenStudy (anonymous):

yeah sure

OpenStudy (zarkon):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok lets go slow

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Okay I got that part satellite73

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the first side is x-6

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohh okay

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

absolutely not

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

try it and you will see

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

the second part $x^2+6x+36$ does not factor using integers

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

i guessed it because the first term was a cube. it was $x^3$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

that is how i knew it was $x^3-6^3$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

$x^3-27=(x-3)(x^2+3x+9)$ and so one

OpenStudy (anonymous):

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

OpenStudy (anonymous):

exactly!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

notice the difference in the placement of the minus sign

OpenStudy (anonymous):

that equation def came in handy

OpenStudy (anonymous):

are you good with math?

OpenStudy (anonymous):

fair to middlin'

OpenStudy (anonymous):

lol okay, thank you so much, I appreciate it

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