Question

(y+3)^3+8 ? how to factor completely ?

Answer 1

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

Answer 2

hmm... then ?:(

Answer 3

well the b portion is easy because if b = 2 then b^2 is ?

Answer 4

OH MAN! I GOT MY LATEX WRONG NUGH!
\[\large (y+3)^3+ 8 \rightarrow = (y+3)^3+2^3 \]

Answer 5

but the sum of cubes formula is correct thank goodness.

Answer 6

so if b = 2 what's b^2 = ?

Answer 7

4?

Answer 8

\[\large (y+3)^3+ 8 \rightarrow (y+3)^3+2^3\] idk what's with me in the middle of the night

Answer 9

but yes b^2 = 4 because b = 2
2^2 = 4 woo.

Answer 10

\[((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4)\]

Answer 11

(y+3+2) = (y+5) (_______)

Answer 12

so we need to expand (y+3)^2 and distribute the -2 on y+3

Answer 13

(y+5)(y^2+9-(y+3+2)+4) ?

Answer 14

wait.. try expand this in parts
expand \[(y+3)^2 \]
and then distribute
-2(y+3)

Answer 15

i didn't even say something :(

Answer 16

y^2+3y+9 then?

Answer 17

i ws here to help k..sry bye i won't interrupt u frm nw .

Answer 18

then distribute the -2 for -2(y+3)

Answer 19

-2y-6

Answer 20

yes so now we can place them into our equation
\[((y+3)+(2))^3 =((y+3)+2)((y+3)^2-2(y+3)+4) \]
\[((y+3)+(2))^3 =(y+5)(y^2+3y+9-2y-6+4) \]

Answer 21

then we can rearrange and combine like terms...

Answer 22

\[(y+5)(y^2+3y-2y-6+4+9)\]

Answer 23

so combine like terms in the super long equation
what's 3y-2y
what's -6+4+9

Answer 24

(y+5)(y^2+y+7) ?

Answer 25

cool so now we need to ... wow distribute
\[(y+5)(y^2+y+7)\]
so distribute the y throughout \[y^2+y+7\]

like \[y(y^2+y+7)\]

Answer 26

\[(y^2(y)+y(y)+7(y))\]

Answer 27

do you know the exponent laws ?

Answer 28

yup

Answer 29

alrighty then :)

Answer 30

y^3+y^2+7y

Answer 31

excellent!
now we do the same thing for the 5
\[(y+5)(y^2+y+7) \]

so
distribute 5 throughout \[y^2+y+7\]
\[5(y^2+y+7)\]

Answer 32

so it's like
5 x y^2
5 x y
5 x 7

Answer 33

5y^2 +5y+35

Answer 34

yeah!
so now we combine like terms \[y^3+y^2+7y+5y^2+5y+35\]

Answer 35

-_- something is wrong ... our middle terms are off.

Answer 36

y^3+5y^2+12y+35

Answer 37

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?

Answer 38

oh nevermind.. wait let me think... our first term and last term is correct..

Answer 39

i hate factoring completely :(

Answer 40

bye :) i have lot to do :/

Answer 41

ok. I'll try and figure out what's up before I go sleep. Maybe we needed pascal's triangle/binomial theorem.

Answer 42

thanks for your time dude :)

Answer 43

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.