anonymous
  • anonymous
(y+3)^3+8 ? how to factor completely ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UsukiDoll
  • UsukiDoll
sum of cubes formula \[(a+b)^3 =(a+b)(a^2-ab+b^2) \] rewrite \[(y+3)^3=8\] as \[(y+3)^3=2^3\] so let a = y+3 and b = 2
anonymous
  • anonymous
hmm... then ?:(
UsukiDoll
  • UsukiDoll
well the b portion is easy because if b = 2 then b^2 is ?

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UsukiDoll
  • UsukiDoll
OH MAN! I GOT MY LATEX WRONG NUGH! \[\large (y+3)^3+ 8 \rightarrow = (y+3)^3+2^3 \]
UsukiDoll
  • UsukiDoll
but the sum of cubes formula is correct thank goodness.
UsukiDoll
  • UsukiDoll
so if b = 2 what's b^2 = ?
anonymous
  • anonymous
4?
UsukiDoll
  • UsukiDoll
\[\large (y+3)^3+ 8 \rightarrow (y+3)^3+2^3\] idk what's with me in the middle of the night
UsukiDoll
  • UsukiDoll
but yes b^2 = 4 because b = 2 2^2 = 4 woo.
UsukiDoll
  • UsukiDoll
\[((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4)\]
anonymous
  • anonymous
(y+3+2) = (y+5) (_______)
UsukiDoll
  • UsukiDoll
so we need to expand (y+3)^2 and distribute the -2 on y+3
anonymous
  • anonymous
(y+5)(y^2+9-(y+3+2)+4) ?
UsukiDoll
  • UsukiDoll
wait.. try expand this in parts expand \[(y+3)^2 \] and then distribute -2(y+3)
imqwerty
  • imqwerty
i didn't even say something :(
anonymous
  • anonymous
y^2+3y+9 then?
imqwerty
  • imqwerty
i ws here to help k..sry bye i won't interrupt u frm nw .
UsukiDoll
  • UsukiDoll
then distribute the -2 for -2(y+3)
anonymous
  • anonymous
-2y-6
UsukiDoll
  • UsukiDoll
yes so now we can place them into our equation \[((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4) \] \[((y+3)+(2))^3 =(y+5)(y^2+3y+9-2y-6+4) \]
UsukiDoll
  • UsukiDoll
then we can rearrange and combine like terms...
UsukiDoll
  • UsukiDoll
\[(y+5)(y^2+3y-2y-6+4+9)\]
UsukiDoll
  • UsukiDoll
so combine like terms in the super long equation what's 3y-2y what's -6+4+9
anonymous
  • anonymous
(y+5)(y^2+y+7) ?
UsukiDoll
  • UsukiDoll
cool so now we need to ... wow distribute \[(y+5)(y^2+y+7)\] so distribute the y throughout \[y^2+y+7\] like \[y(y^2+y+7)\]
UsukiDoll
  • UsukiDoll
\[(y^2(y)+y(y)+7(y))\]
UsukiDoll
  • UsukiDoll
do you know the exponent laws ?
anonymous
  • anonymous
yup
UsukiDoll
  • UsukiDoll
alrighty then :)
anonymous
  • anonymous
y^3+y^2+7y
UsukiDoll
  • UsukiDoll
excellent! now we do the same thing for the 5 \[(y+5)(y^2+y+7) \] so distribute 5 throughout \[y^2+y+7\] \[5(y^2+y+7)\]
UsukiDoll
  • UsukiDoll
so it's like 5 x y^2 5 x y 5 x 7
anonymous
  • anonymous
5y^2 +5y+35
UsukiDoll
  • UsukiDoll
yeah! so now we combine like terms \[y^3+y^2+7y+5y^2+5y+35\]
UsukiDoll
  • UsukiDoll
-_- something is wrong ... our middle terms are off.
anonymous
  • anonymous
y^3+5y^2+12y+35
UsukiDoll
  • UsukiDoll
yeah but I'm double checking through a calculator and it claims that it's 9y^2+27y unless ... hey @Haseeb96 could you come over here and check this?
UsukiDoll
  • UsukiDoll
oh nevermind.. wait let me think... our first term and last term is correct..
anonymous
  • anonymous
i hate factoring completely :(
anonymous
  • anonymous
bye :) i have lot to do :/
UsukiDoll
  • UsukiDoll
ok. I'll try and figure out what's up before I go sleep. Maybe we needed pascal's triangle/binomial theorem.
anonymous
  • anonymous
thanks for your time dude :)
UsukiDoll
  • UsukiDoll
you're welcome. I'm gonna post the pascal's triangle version.. I see it clearer on how they got the middle term... another method is to exapand (y+3)^3 manually and add 8 but who wants to do that ? that's time consuming.
UsukiDoll
  • UsukiDoll
|dw:1437311220374:dw|
UsukiDoll
  • UsukiDoll
so from k = 0 to k = 3 ...
UsukiDoll
  • UsukiDoll
|dw:1437311295892:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311393568:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311428990:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311489980:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311527971:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311547143:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311614971:dw|
UsukiDoll
  • UsukiDoll
@freckles wait don't go! I was using sum of cubes and the middle terms were wonky . What gives?
UsukiDoll
  • UsukiDoll
|dw:1437311655716:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311666239:dw|
UsukiDoll
  • UsukiDoll
ugh this thing is a troll x.x
UsukiDoll
  • UsukiDoll
y^3+9y^2+27y+35 ... should be it but the middle terms splash out when I did sum of cubes. So I'm doing Pascal's triangle / Binomial Theorem and that's working so far
UsukiDoll
  • UsukiDoll
|dw:1437311833243:dw|
freckles
  • freckles
\[(y+3)^3+8 \\ a^3+b^3=(a+b)(a^2-ab+b^2) \\ a=y+3 \\ b=2 \\ a^2=(y+3)^2 \\ b^2=4 \\ a^3=(y+3)^3 \\ b^3=8 \\ \\ \text{ input into formula } \\ (y+3)^3+8=(y+3+2)([y+3]^2-[y+3](2)+4) \\ (y+3)^3+8=(y+5)(y^2+6y+9-2y-6+4) \\ \text{ combine like terms inside the second () }\]
UsukiDoll
  • UsukiDoll
|dw:1437311848473:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311880098:dw|
UsukiDoll
  • UsukiDoll
|dw:1437311906118:dw| that should be 1 since everything cancels.
UsukiDoll
  • UsukiDoll
|dw:1437311933179:dw|
freckles
  • freckles
I think I see what you did wrong
freckles
  • freckles
\[(y+3)^2=y^2+6y+9 \\ (y+3)^2 \neq y^2+3y+9\]
UsukiDoll
  • UsukiDoll
|dw:1437311941673:dw|
UsukiDoll
  • UsukiDoll
OMg x.x Not my night
freckles
  • freckles
it was probably just a type-o and you just never fixed and you forgot about it or whatever happens to me too
UsukiDoll
  • UsukiDoll
|dw:1437312099694:dw|
UsukiDoll
  • UsukiDoll
|dw:1437312142306:dw|
UsukiDoll
  • UsukiDoll
oh yay
UsukiDoll
  • UsukiDoll
\[ (y+5)(y^2+4y+7)\] \[y^3+4y^2+7y+5y^2+20y+35 = y^3+9y^2+27y+35\]
UsukiDoll
  • UsukiDoll
there we go
UsukiDoll
  • UsukiDoll
because (y+3)(y+3) = y^2+3x+3x+9 = y^2+6x+9 I must have overlooked it. so sum of cubes can be used after all.
UsukiDoll
  • UsukiDoll
@freckles I got it to work :)

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