brodybarnes35
  • 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)
Mathematics
chestercat
  • chestercat
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brodybarnes35
  • brodybarnes35
brodybarnes35
  • brodybarnes35
FAN AND MEDAL
brodybarnes35
  • brodybarnes35
idk :/ thats embarrassing wow

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brodybarnes35
  • brodybarnes35
but really idk could you explain
brodybarnes35
  • brodybarnes35
ohhhh im so stupid sorry X
brodybarnes35
  • brodybarnes35
-64 = x? what
brodybarnes35
  • brodybarnes35
9 and 7
brodybarnes35
  • brodybarnes35
no thats 63 my b
brodybarnes35
  • brodybarnes35
OMG WTH IM DUMB
brodybarnes35
  • brodybarnes35
8*8 UGHHH
brodybarnes35
  • brodybarnes35
-8 * 8
brodybarnes35
  • brodybarnes35
???
Hero
  • Hero
@brodybarnes35, what happens to the expression after you factor out the x?
brodybarnes35
  • brodybarnes35
so its either a or b right
brodybarnes35
  • brodybarnes35
idk @hero
brodybarnes35
  • brodybarnes35
a or b right @mehek14 @hero ?
brodybarnes35
  • brodybarnes35
brodybarnes35
  • brodybarnes35
o ok :/
brodybarnes35
  • brodybarnes35
@Nnesha can you help?
Nnesha
  • 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
  • Nnesha
|dw:1439937064914:dw| divide both terms by common factor and write your answer in the parentheses.
brodybarnes35
  • brodybarnes35
if you divide by x the outcome would be the same?
brodybarnes35
  • brodybarnes35
Nnesha
  • Nnesha
not yet.there is one more step. just divide x^3 by x and then -64x by x
Nnesha
  • Nnesha
x^3/x= ?? -64x/x= ?
brodybarnes35
  • brodybarnes35
= x^3 =-64x the outcome is the same when you divide...or am i doing it wrong
brodybarnes35
  • brodybarnes35
wait does the x disappear?
Nnesha
  • Nnesha
i don't get it. what do you mean? x^3 over x isn't equal to x^3
Nnesha
  • 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
  • brodybarnes35
its 1x or x?
Nnesha
  • 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
  • brodybarnes35
2x? now
Nnesha
  • 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
  • brodybarnes35
2x
Nnesha
  • 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
  • Nnesha
and x is same as x to the one power \[\rm x^1 \times x^1 =x^{??}\]
brodybarnes35
  • brodybarnes35
x^2
Nnesha
  • Nnesha
right so x^3 over x = x^2|dw:1439937679559:dw| now divide -64x over x ?
Hero
  • 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
  • brodybarnes35
ummm that would be -64x^2?
Nnesha
  • Nnesha
\[\frac{ -64x }{ x} = ?\] here you have to divide by x what would get when you divide same bases ? like 2/2 = ?
brodybarnes35
  • brodybarnes35
2/2 = 1 right so -64x would be -64x?
Nnesha
  • Nnesha
you are right. -64x would be -64x but what is \[\huge\rm \frac{ -64\color{Red}{x} }{ \color{Red}{x} } = ?\]
brodybarnes35
  • brodybarnes35
x^1????
Nnesha
  • Nnesha
Take out x from -64x. that's it.
brodybarnes35
  • brodybarnes35
omg wow that was simple -.- im dumb
Nnesha
  • Nnesha
so when you take out x from -64x what will u have left with ?
brodybarnes35
  • brodybarnes35
-64?
Nnesha
  • Nnesha
right!!
brodybarnes35
  • brodybarnes35
ok so what happens now lol?
Nnesha
  • 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
  • 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
  • brodybarnes35
x^3 - 64x
Nnesha
  • Nnesha
that's the original equation what about it ?
brodybarnes35
  • brodybarnes35
oh nvm
brodybarnes35
  • brodybarnes35
what a and what is b
Nnesha
  • Nnesha
a and b are variables a = first term b= 2nd term
brodybarnes35
  • brodybarnes35
yeah but what are they like number wise..
Nnesha
  • Nnesha
|dw:1439938335253:dw|
brodybarnes35
  • brodybarnes35
x^2 - 64^2 = ( x2 + 64) ( x2 - 64)?
Nnesha
  • 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
  • brodybarnes35
B
Nnesha
  • Nnesha
how did you get b? what about common that's outside the parentheses?
Nnesha
  • Nnesha
common factor*
brodybarnes35
  • brodybarnes35
because just they look alike ( - )( + ) the signs match... am i right
Nnesha
  • Nnesha
|dw:1439938542139:dw| take square root of both terms write your answer in the parentheses
brodybarnes35
  • brodybarnes35
it is B!!
anonymous
  • anonymous
its A x(x-8)(x+8)
Nnesha
  • Nnesha
don't look at the answer choices
brodybarnes35
  • brodybarnes35
oh its a yikes
Nnesha
  • Nnesha
why it's a ?
brodybarnes35
  • brodybarnes35
she said it
Nnesha
  • Nnesha
she is not gonna be there when you have to take final exam.
brodybarnes35
  • brodybarnes35
(x - 8)(x + 8) right?
anonymous
  • anonymous
x(x^2-64)= x^3-64x
Nnesha
  • Nnesha
that should go in the parentheses|dw:1439938908541:dw| common should stay there outside the parentheses.
brodybarnes35
  • brodybarnes35
ohhhh so A
brodybarnes35
  • brodybarnes35
anonymous
  • anonymous
yes
Nnesha
  • Nnesha
yes|dw:1439938990977:dw|
brodybarnes35
  • brodybarnes35
ok thanks can you help with one more?
brodybarnes35
  • brodybarnes35
please
brodybarnes35
  • brodybarnes35
???
Nnesha
  • Nnesha
@ana123456 don't give out direct answer especially when someone spnd one HOUR to teach.
Nnesha
  • Nnesha
and sorry i have to go :) good luck!!
brodybarnes35
  • brodybarnes35
ok
brodybarnes35
  • brodybarnes35
ana can you help lol
brodybarnes35
  • brodybarnes35
brodybarnes35
  • brodybarnes35
one more question pls
anonymous
  • anonymous
sure
brodybarnes35
  • brodybarnes35
nvm i got it thanks
anonymous
  • anonymous
you're welcome

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