Question

I can across this problem on my sat prep book can someone teach me how to do it? I don't want the answer lol if I wanted I would of went to the back of the book:D

Answer 1

Here

Answer 2

@Seratul

Answer 3

Can you check the back of the book real quick or no? :)
I don't want to explain this if I did it incorrectly.
Is the answer c=3?

Answer 4

yes

Answer 5

it is c=3

Answer 6

Since the polynomial is divisible by y-2,
it means we can divide y-2 out of the polynomial,\[\large\rm y^5-2y^4-cxy+6x\quad=\quad (y-2)(stuff)\]And it will leave us with some other polynomial stuff, but no remainder.
What happens if we plug y=2 into this new thing?\[\large\rm (y-2)(stuff)=(2-2)(stuff)\]The whole thing ends up giving us zero, right?
That's what our remainder theorem tells us.

Our function evaluated at y=2 will give us zero because the expression is divisible by y-2.

Answer 7

\[\large\rm y^5-2y^4-cxy+6x\]So when we plug in y=2,
we can set this expression equal to zero,\[\large\rm 2^5-2(2)^4-cx(2)+6x=0\]

Answer 8

okay so basically for these problems we set the problem to 0 and plug in the numbers for y?

Answer 9

For this specific type of problem, where we have divisibility,
yes it looks like that will always be the case! :)

Answer 10

okay so if its y+2 it would be y=-2?

Answer 11

Yes

Answer 12

oh okay so any type of problems that says polynomial divisible by we just make it opposite and plug it into the letter and equal to zero.

Answer 13

just wanted to make sure and thank you

Answer 14

Yes.

It's helpful if you understand `why` this is true though. That's what I was trying to do with the (y-2)(stuff) explanation. But as long as you have some type of shortcut memorized, that will work out nicely. :) np

Answer 15

i was worrying so much about it a shortcut lol we didn't solve it @zepdrix i tried to simplify it but i ended up something else then 3

Answer 16

if u could help me with that, i would appreciate it thx

Answer 17

\[\large\rm 2^5-2(2)^4-cx(2)+6x=0\]2*2^4 is the same as 2^5, right?
So the first two terms subtract off,\[\large\rm -cx(2)+6x=0\]

Answer 18

yes

Answer 19

Let's rewrite our equation like this:\[\large\rm 6x-2cx=0\]From here we should try to factor.
I guess we can pull an x out of each term,
and maybe a 2 also, ya?

Answer 20

oh so u canceled out the x