Question

Factor the trinomial...
w18 – 9w9y5 + 14y10
Walk me through it please?

Answer 1

\[w^{18}-9w ^{9}y ^{5}+14y ^{10}\]

Answer 2

is that it?

Answer 3

Yes ma'am

Answer 4

thats a tricky..

Answer 5

Lol I haven't done these in a few weeks so my mind is baffled

Answer 6

i would start by figuring out how you can factor the w such as how it is normally
(x+ )(x+ )
i would figure out a
(w+ )(w+ )
does that make sense?

Answer 7

Yes it does

Answer 8

Would I have to do that for y too?

Answer 9

mhmm, but once you figure out the w it will make the y a lot easier..at least in my opinion.

Answer 10

Okay then, uhmmm

Answer 11

so how would you factor the w?

Answer 12

Do I have to involve the exponents?

Answer 13

yes.

Answer 14

I've never factored with that so I'm a bit lost

Answer 15

since the leading term is w^18 i would try using any two factors of 18 as your exponents. Honestly this one is extremely hard and its gonna be a lot of guessing and reasonable deduction :/ unless there is something im missing.

Answer 16

looking at your question, I would let
\[a = w ^{9}\] and \[b=y ^{5}\]

so your question will look like this
\[a ^{2} -9ab +14 b ^{2}\]

Answer 17

you have a negative 9ab and positive 14b^2, which mean in your factor, both terms have to be (a- )(a- ). Because two negative numbers multiply together will give you a positive 14b^2

Answer 18

I'M SO LOST omgeeeeez, hold on please

Answer 19

because by factor out the power of 18 is difficult. so you need to simplified your equation by substitute valuable without change your equation. what I did is let
a=w9
and
b=y5
so your original question will become
\[(w^{9})^{2}-9w ^{9}y ^{5}+14(y ^{5})^{2}\]

Answer 20

by substitute a and b
your equation now become
\[a ^{2}-9ab+14b ^{2}\]
are you follow?

Answer 21

note \[(w ^{9})^{2}=w ^{18}\]
so I am not changing any value of your original equation, I just write it differently.

Answer 22

after you factor in terms of a and b, then just put back w^9 and y^5 where a and b are respectively.