- brodybarnes35

Which of the following represents the factored form of f(x) = x^3 − 64x?
f(x) = x(x + 8)(x − 8)
f(x) = (x − 8)(x + 8)
f(x) = x(x − 8)^2
f(x) = x(x^2 − 8)

- chestercat

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- brodybarnes35

- brodybarnes35

FAN AND MEDAL

- brodybarnes35

idk :/ thats embarrassing wow

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## More answers

- brodybarnes35

but really idk could you explain

- brodybarnes35

ohhhh im so stupid sorry X

- brodybarnes35

-64 = x? what

- brodybarnes35

9 and 7

- brodybarnes35

no thats 63 my b

- brodybarnes35

OMG WTH IM DUMB

- brodybarnes35

8*8 UGHHH

- brodybarnes35

-8 * 8

- brodybarnes35

???

- Hero

@brodybarnes35, what happens to the expression after you factor out the x?

- brodybarnes35

so its either a or b right

- brodybarnes35

idk @hero

- brodybarnes35

@Mehek14 ???

- brodybarnes35

o ok :/

- brodybarnes35

@Nnesha can you help?

- Nnesha

first find GCF(greatest common factor )
so there is x common in both terms right ?
now take out the x from x^3 and -64x
or in other words divide both terms by common factor
\[\rm x^3-64x=x(???-???)\]

- Nnesha

|dw:1439937064914:dw|
divide both terms by common factor and write your answer in the parentheses.

- brodybarnes35

if you divide by x the outcome would be the same?

- brodybarnes35

- Nnesha

not yet.there is one more step.
just divide x^3 by x and then -64x by x

- Nnesha

x^3/x= ??
-64x/x= ?

- brodybarnes35

= x^3
=-64x
the outcome is the same when you divide...or am i doing it wrong

- brodybarnes35

wait does the x disappear?

- Nnesha

i don't get it. what do you mean?
x^3 over x isn't equal to x^3

- Nnesha

x^3 is same as x times x times x \[\huge\rm \frac{ x \times x \times x }{ x }\]can you divide now?

- brodybarnes35

its 1x or x?

- Nnesha

x^3 is same as x times x times x \[\huge\rm \frac{ x \times x \times \cancel{x} }{\cancel{ x }}\]how many x's wre there ?

- brodybarnes35

2x? now

- Nnesha

x^3 is same as x times x times x \[\huge\rm \frac{\color{Red}{ x} \times\color{Red}{ x} \times \cancel{\color{Red}{x}} }{\cancel{ x }}\]how many x's wre there ?

- brodybarnes35

2x

- Nnesha

no when you multiply same bases you should add their exponents
\[\huge\rm x^m \times x^n = x^{m+n}\]
when you ADD same bases then you should add their coefficient
1x+1x=(1+1))x=2x

- Nnesha

and x is same as x to the one power \[\rm x^1 \times x^1 =x^{??}\]

- brodybarnes35

x^2

- Nnesha

right so x^3 over x = x^2|dw:1439937679559:dw|
now divide -64x over x ?

- Hero

Another way to understand factoring is to think of it in terms of the distributive property.
ab + ac = a(b + c)
In other words, a is common to both terms on the left so we factor it out to get the expression on the right.

- brodybarnes35

ummm that would be -64x^2?

- Nnesha

\[\frac{ -64x }{ x} = ?\] here you have to divide by x
what would get when you divide same bases ?
like 2/2 = ?

- brodybarnes35

2/2 = 1 right so -64x would be -64x?

- Nnesha

you are right.
-64x would be -64x
but what is \[\huge\rm \frac{ -64\color{Red}{x} }{ \color{Red}{x} } = ?\]

- brodybarnes35

x^1????

- Nnesha

Take out x from -64x. that's it.

- brodybarnes35

omg wow that was simple -.- im dumb

- Nnesha

so when you take out x from -64x what will u have left with ?

- brodybarnes35

-64?

- Nnesha

right!!

- brodybarnes35

ok so what happens now lol?

- Nnesha

|dw:1439938027333:dw|
\[\huge\rm x(x^2-64)\] now apply the difference of square method \[\huge\rm a^2-b^2=(a+b)(a-b)\]

- Nnesha

\[\huge\rm x(\color{red}{x^2-64})\] now apply the difference of square method \[\huge\rm a^2-b^2=(a+b)(a-b)\]
take square of both terms in the parentheses write ur answer as
(sqrt of 1st term + sqrt of 2nd term)(sqrt of 1st term - sqrt of 2nd term )

- brodybarnes35

x^3 - 64x

- Nnesha

that's the original equation what about it ?

- brodybarnes35

oh nvm

- brodybarnes35

what a and what is b

- Nnesha

a and b are variables
a = first term
b= 2nd term

- brodybarnes35

yeah but what are they like number wise..

- Nnesha

|dw:1439938335253:dw|

- brodybarnes35

x^2 - 64^2 = ( x2 + 64) ( x2 - 64)?

- Nnesha

no take square root of both terms look at the equation i gave you
it's not \[\huge\rm a^2-b^2\cancel{=}(a^2-b^2)(a^2+b^2)\]

- brodybarnes35

B

- Nnesha

how did you get b? what about common that's outside the parentheses?

- Nnesha

common factor*

- brodybarnes35

because just they look alike ( - )( + ) the signs match... am i right

- Nnesha

|dw:1439938542139:dw|
take square root of both terms write your answer in the parentheses

- brodybarnes35

it is B!!

- anonymous

its A x(x-8)(x+8)

- Nnesha

don't look at the answer choices

- brodybarnes35

oh its a yikes

- Nnesha

why it's a ?

- brodybarnes35

she said it

- Nnesha

she is not gonna be there when you have to take final exam.

- brodybarnes35

(x - 8)(x + 8) right?

- anonymous

x(x^2-64)= x^3-64x

- Nnesha

that should go in the parentheses|dw:1439938908541:dw|
common should stay there outside the parentheses.

- brodybarnes35

ohhhh so A

- brodybarnes35

- anonymous

yes

- Nnesha

yes|dw:1439938990977:dw|

- brodybarnes35

ok thanks can you help with one more?

- brodybarnes35

please

- brodybarnes35

???

- Nnesha

@ana123456 don't give out direct answer
especially when someone spnd one HOUR to teach.

- Nnesha

and sorry i have to go :) good luck!!

- brodybarnes35

ok

- brodybarnes35

ana can you help lol

- brodybarnes35

- brodybarnes35

one more question pls

- anonymous

sure

- brodybarnes35

nvm i got it thanks

- anonymous

you're welcome

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