Question

Which answer is equal to the quotient in the expression below?

(2x3 - x2 - 3x) (x2 + x)
A.2x - 3
B.2x2 - 3x
C.2x2 + x
D.2x - 3x

Answer 1

i dont see a quotient

Answer 2

it's divided by

Answer 3

are you doing (2x^3-x^2-3x)/(x^2+x)

Answer 4

never mind that one.. how about this
Which answer is equal to the difference in the expression below?

(2x3 - 6x2 + 2x - 2) - (-2x3 - 3x2 + 3x + 3)
A.4x3 - 3x2 + x - 5
B.4x3 - 3x2 - 5
C.4x3 + 9x2 + x - 5
D.4x3 - 3x2 - x - 5

Answer 5

\[\frac{2x^3-x^2-3x}{x^2+x}\]

Answer 6

is this what you meant?

Answer 7

you don't want to do this one?

Answer 8

ok for this next one distribute you the negative in front of the parenthesis and then put like terms together

Answer 9

no i justed guessed on it, this was my next one. but i forgot how to set up subtraction

Answer 10

if you have
(2y-3x)-(-5y+7x)
=2y-3x-(-5y)+(-7x)
=2y-3x+5y-7x
=2y+5y-3x-7x
=7y-10x

see i distribute the negative 1 to each term in the second paranthese
then i grouped my like terms together

let me see you try and i can correct where you went wrong

Answer 11

where are the ys coming from?

Answer 12

i have (2x^3-6x^2+2x-2)+(2x^3-3x^2+3x+3)

Answer 13

that was an example

Answer 14

oh haa

Answer 15

see that i got rid the grouping symbols after distributing the negative 1
you have (2x3 - 6x2 + 2x - 2) - (-2x3 - 3x2 + 3x + 3)
=2x^3-6x^2+2x-2+2x^3+3x^2-3x-3

Answer 16

not put your like terms togther

Answer 17

now*

Answer 18

okay, now next step?

Answer 19

ohh alright.. so that's... my two 2x^3

Answer 20

do i add them and get 4x^3?

Answer 21

ok!
what about your x^2's?

Answer 22

well -6x^2+3x^2

Answer 23

right

Answer 24

-6+3=-3
so
-6x^2+3x^2=

Answer 25

-3x^2?

Answer 26

:)
ok
x's?

Answer 27

-x?

Answer 28

:)
and finally the constants?

Answer 29

+1?

Answer 30

-2-3 is not 1
2+3=5
so -(2+3)=-(5)=-5
-(2+3)=-2+(-3)=-2-3=-5

Answer 31

-5?

Answer 32

yes
so D is the answer

Answer 33

woop woop. haha thank youuu. :)

Answer 34

np