Question

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Answer 1

no i dont think that is right

Answer 2

ohh...

Answer 3

i am going to guess that it is what you get when you multiply out \((x-5)^2(x-1)^2\) lets see

Answer 4

no it isn't damn

Answer 5

one way is plug in 1 for x, and see if you get 0

that reduces the choices to the last 2 equations

Answer 6

you can do the same for x=5, or you can "guess" that the constant should be a multiple of 5

Answer 7

well then that makes it easy, because the constant must be 5 or -5

Answer 8

yeah that is the one it has to be E

Answer 9

How do you get B?
y = x4 + 6x3 – 6x2 + 6x – 5
replace all the x^n with 1 (because 1 to any power is 1)
y= 1+6-6+6-5
that is not 0

Answer 10

i think you were supposed to check \(x=1\) then divide, check \(x=5\) then divide again
but really this is a lot of work

Answer 11

I'am really confused now -_-

Answer 12

The "zeros" are the x values that make the expression 0
checking if 1 is a zero is (relatively) easy.
notice that the 1st equation (with x replaced with 1) is

–x4 – 6x3 + 6x2 – 6x + 5
-1 -6 +6 -6 +5
that adds up to -2

so the first equation is out. Same for the 2nd equation
the last 2 work.

now check if 5 is a root. synthetic division is one way. replace x with 5 is another (harder)
a third is knowing that the constant (last term) should be a multiple of 5 and you can toss the 3rd equation. That leaves the 4th as the answer

Answer 13

ok cool :D I get it lol Thanks ^_^