Question

Can someone help?

(x+7)(x-1)=(x+1)^2
(x+7)(x-1)=(x+1)(x+1)
2x+6=2x+2

Is this break down correct so far?

Answer 1

Now, how did you get (x+7)(x-1)=(x+1)^2?

Answer 2

Or was that the original question?

Answer 3

It is the original equation
I am suppose to solve it

Answer 4

Well the correct answer is 2

Answer 5

your breakdown is wrong

Answer 6

Asnaseer were am I messing up at?

Answer 7

for example:

(x+1)(x+1) = x*(x+1) + 1*(x+1)
= x^2 + x + x + 1
= x^2 + 2x + 1

Answer 8

try applying the same rules to: (x+7)(x-1)

Answer 9

My book shows the answer... but we are suppose to show the work.

Answer 10

well that is 2x+6 right?

Answer 11

no - look at the way I expanded (x+1)(x+1) above

Answer 12

(x+7)(x-1) = x*(x-1) + 7*(x-1)

try completing it from there

Answer 13

do you understand?

Answer 14

No.

Answer 15

I am trying to see where you are coming up with that

Answer 16

ok, lets try it with actual numbers to help you understand this better...

Answer 17

8*5 = 40

so we can write this as:

(1+7)(6-1) = 40

agreed?

Answer 18

Agree

Answer 19

ok, now lets try to expand this:

(1+7)(6-1) = 1*(6-1) + 7*(6-1)
= 1*5 + 7*5
= 5 + 35
= 40

does that help?

Answer 20

I am lost with this section = 1*(6-1) + 7*(6-1)

Answer 21

So are we suppose to only multiply by the numbers in the first set of parenthesis

Answer 22

If so then it makes sense but I didn't think that was how it was done.

Answer 23

ok, lets try another method. suppose we had:

x + 7x

do you agree this can be written as:

(1 + 7)*x

Answer 24

I don't see how that would be the same

Answer 25

well, now if we replace the x with (6-1) we get:

(6-1) + 7(6-1) = (1+7)(6-1)

Answer 26

in general:

(a+b)(c+d) = a(c+d) + b(c+d)
= ac + ad + bc + bd

Answer 27

okay..

Answer 28

if you are still stuck then try looking at this site: http://www.algebrahelp.com/lessons/simplifying/foilmethod/

it might help you understand better

Answer 29

do you think you can now expand this correctly?

(x+7)(x-1) =

Answer 30

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

Answer 31

=(x+1)^2 is 2x + 2 right?

Answer 32

no :(

Answer 33

is it 2x+1

Answer 34

I would advise you to take a look at that site above first to get a better understanding

Answer 35

I'm sorry I'm being so dumb. My class hasn't started yet but I have some health issues and I'm trying to get ahead start and my professor went ahead and sent me my assignments so I'm trying to learn this without having the class... =/

Answer 36

don't worry - you are certainly not dumb :)
but if you really want to learn this I would spend some time on that site above, then come back here and we can continue.

Answer 37

okay so the first part of the equation should be x^2-1x+7x+-7
???

Answer 38

perfect!

Answer 39

so then what would I need to do to do (x+1)^2

Answer 40

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

Answer 41

x^2+x+x+1

Answer 42

perfect again!
now you should first be able to simplify both expressions.

Answer 43

you know that +- turns into a minus correct?

Answer 44

and you can collect "like" terms

Answer 45

yes

Answer 46

so, for example:

x^2-1x+7x+-7 = x^2 - x + 7x - 7
= x^2 + 6x - 7

Answer 47

so that breaks down to

x^2+6x-7=x^2+2x+1

Answer 48

thats right so far

Answer 49

now bring all the terms involving x from the right hand side to the left hand side - remember to change their signs when you do this

Answer 50

x^2+4x=x^2+8

Answer 51

also bring over the x^2 term from the RHS

Answer 52

I'm not sure how to handle x^2???

Answer 53

4x=8
x=2

Answer 54

bingo! see you CAN do it! :)

Answer 55

okay
so now what would you do if only one side had x^2

Answer 56

how would you break that down

Answer 57

before tackling that - one thing to remember is that you can check your answer by substituting x=2 into your original expressions

Answer 58

yes.. I learned that with my other equations

Answer 59

good :)

Answer 60

now, if you did end up with some x^2 terms, then you would have what is known as a quadratic equation.
have you studied these yet?

Answer 61

if not, this site should help you understand: http://www.mathsisfun.com/algebra/quadratic-equation.html

Answer 62

wonderful! Thanks so much

Answer 63

are you on here a lot?

Answer 64

there is also another very good site that is filled with lots of short video tutorials on various subjects here: http://www.khanacademy.org/

Answer 65

I am on this site quite a lot - but as I work full time, its usually in the evenings (UK time)

Answer 66

Well would you mind if I needed help which I know I will if I messaged you?

Answer 67

there are many many helpful people on this site.
I would suggest you just post your questions as usual and, only if you find no one is helping, then by all means send me a message and I'll do what I can to help out.

it's great to find students like yourself that are eager to learn - you should be very proud of yourself :)

Answer 68

Well thank you.. I get discouraged sometimes... I don't seem to get a lot of support from my family and often wonder if it will all pay off

Answer 69

have faith in your abilities - you are certainly a quick learner!

Answer 70

if x=0 is that a no solution problem?

Answer 71

no - x=0 could well be a valid solution to some equation. e.g. x=0 is a solution to:

2x = 3x

Answer 72

x(2x-3)=(2x+1)(x-4)
2x-3x=2x-8x+1x-4
-1x=-5x-4
4x=-4
x=-1

Answer 73

Were am I messing up with this equation?

Answer 74

where*

Answer 75

you made a mistake up there - remember that x * 2x = 2x^2

Answer 76

I need to go now - but good luck with the rest.
remember to post each new question separately in the list to the left
<----

Answer 77

thanks so much!

Answer 78

yw :)