Question

divide (y^4-81)/ (y-3) @SalvadorV

Answer 1

that one is a little tricky, do you know how to divide functions?

Answer 2

no not really

Answer 3

Factorise (y^4-81) and then cancel out the common factor i.e,(y-3).

Answer 4

what you can see there is that you have Y^4+3^4, (3^4=81)

Answer 5

sorry y^4-3^4

Answer 6

@Onlymaths can you draw that out for me @SalvadorV yes

Answer 7

\[y^4-3^4\]

Answer 8

@SalvadorV where did you get 3^4 =81 from and why

Answer 9

3^4= 3*3*3*3= 9*3*3=27*3=81

Answer 10

y^4-3^4=(y^2-9)(y^2+9)

Answer 11

so, we know that \[(a+b)(a-b)=a^2+ab-ab-b^2=a^2-b^2\]

Answer 12

okay yes

Answer 13

so then on the same logic \[(a^2-b^2)(a^2+b^2)=a^4+a^2b^2-a^2b^2-b^4=a^4-b^4\]

Answer 14

do you see why?

Answer 15

no not really

Answer 16

it is the same thing, just that they are now squared

Answer 17

ookay yea go on

Answer 18

so we have \[a^4-b^4=(a^2-b^2)(a^2+b^2)=((a+b)(a-b))(a^2+b^2)\]

Answer 19

yea what is next

Answer 20

you see, that is the problem we are looking for

Answer 21

your answer is there, can you see how?

Answer 22

umm no can you break it down more draw it please

Answer 23

@SalvadorV

Answer 24

say we have our\[a^4-b^4=((a-b)(a+b))(a^2+b^2)\] but I say a=y and b=3
then it would be \[y^4-3^4=((y-3)(y+3))(y^2+3^2)\]

Answer 25

can you add the numbers in please

Answer 26

so, can you see now, what we want to do is \[((y-3)(y+3)(y^2+3^2))/(y-3)\]

Answer 27

it is \[y^4-81=(y-3)(y+3)(y^2-9)\]

Answer 28

do you know what you have to do here?