Question

(x + 7) (x-1) = (x-1)^2

Answer 1

I tried foiling both sides idk what I did wrong but do you think you could do it that way pretty please?

Answer 2

You don't have to FOIL this one, there's a \((x-1)\) on both sides so it's simpler.

Answer 3

okie let's do it...!
as ganeshie8 said

we get equation like this after shifting terms

(x+7) (x-1)- (x-1)^2=0

now you can see we can take x-1 as a common factor..

after taking x-1 as common factor what do you get ?

Answer 4

im confused!

Answer 5

okay just divide both sides of equation

(x+7) (x-1)- (x-1)^2=0 what do you get??

Answer 6

why cant I foil it?

Answer 7

You can. But it makes the problem needlessly complicated.

Answer 8

you can foil it but why go to complicated side..?

Answer 9

okay just divide both sides of equation

(x+7) (x-1)- (x-1)^2=0 what do you get??

Answer 10

\[(x+7)(x-1)-(x-1)^2=0\]Do you see how both terms have \((x-1)\)? You can factor that out.
\[(x-1)\bigl(x+7)-(x-1)\bigr) = 0\]

Answer 11

whait

Answer 12

wait I wrot

Answer 13

i wrote the problem down wrong

Answer 14

it

Answer 15

(x +7) (x -1) = (x +1)^2

Answer 16

You already asked that one.

Answer 17

okie in this you have to expand it...

Answer 18

i did but i still don't get it

Answer 19

Did you expand it?

Answer 20

expand?

Answer 21

FOIL it!

Answer 22

oh yah but i did it wrong

Answer 23

okie tell us what you got.. we will check your answer..

Answer 24

@lovelike_woee

Answer 25

6x-8

Answer 26

you are right what's the matter now? find value of x

Answer 27

alright thank yuo

Answer 28

That isn't what I got...

Answer 29

you are welcome..!

Answer 30

@SACAPUNTAS check you answer..!

Answer 31

Ok.
\[(x+7)(x-1) - (x+1)^2\]\[(x^2 + 7x - x -7) - (x^2 + x + x + 1)\]\[x^2-x^2+7x-3x+7-1\]\[4x-8\]
You messed up a minus sign somewhere I guess.

Answer 32

And I messed up copying. -7 - 1. But yeah.

Answer 33

@SACAPUNTAS yep you are right... thanks for correcting bro... you are right...! keep up the good work..! :)

Answer 34

@lovelike_woee ans must be 4x-8

Answer 35

yah i figured it out too thanks though!