Question

(2y^2 +4y-8)-(4y^2+3y-5)
MEDAL REWARDED!!
Use the rules of polynomials.

Answer 1

@satellite73

Answer 2

\[(2y^2 +4y-8)-(4y^2+3y-5)\] distribute the minus sign first and start with
\[2y^2+4y-8-4y^2-2y+5\]

Answer 3

Okay

Answer 4

then we can group the like terms together as before

Answer 5

Which one do we group first?

Answer 6

\[2y^2-4y^2+4y-3y-8+5\]

Answer 7

What you can do is combine like terms. What are the like terms in this problem? 2y^2 and 4y^2, 4y and 3y, and -8 and -5.

You'd want to apply the negative sign to all of the terms in the second part. So, 2y^2 + 4y - 8 - 4y^2 - 3y + 5. Combine like terms to get -2y^2 + y - 3.

Answer 8

i put the \(y^2\) next to each other
then the \(y\) terms,
then the numbers

Answer 9

That's why, I was confused at first.

Answer 10

Thanks aurorablue01!

Answer 11

then combine as before
what is \(2y^2-4y^2\) or what is \(2-4\) ?

Answer 12

-2

Answer 13

No problem at all but I'd give satellite the medal.

Answer 14

k good so first term is \(-2y^2\)

Answer 15

Yea

Answer 16

Okay

Answer 17

1y

Answer 18

oops i meant \(4y-3y\)

Answer 19

yeah, get rid of the 1 though and just write \(y\)

Answer 20

That's okay, it turns out to be the same solution. ;)

Answer 21

It's still y though lol

Answer 22

and finally \(-8+5\)

Answer 23

Okay

Answer 24

Next one be -2?

Answer 25

hmmm no i don't think so

Answer 26

\(-8+5\) or \(5-8\)

Answer 27

I think I made a mistake

Answer 28

Somehow

Answer 29

And @satellite, you've answered almost 30k questions?!?! Talk about incredible :)

Answer 30

Yes!

Answer 31

takes a lickin and keeps on ticken

Answer 32

Talk about genius!

Answer 33

All too true :)

Answer 34

Lol, that rhymes!

Answer 35

First 100 I've seen here. A lot of 99s but never a 100 before.

Answer 36

True

Answer 37

The first real one. ;)

Answer 38

I've gotta go now. See you guys around.

Answer 39

Or gals, whatever. lol