# Thread: The eh.. Distributive property and squares.

1. ## The eh.. Distributive property and squares.

So we have our final exam on monday... so im doin the review.

I tried problems 1-10.. and got 1 right -_-. I fear i am doomed. Lol.

but anyway..

1. (5x+7)^2

Seems ridiculously easy to me..

Heres what i tried...

turned it into

(5x+7)(5x+7)
FOIL
25x^2*5x*7*7*5x*49

25x^2*10X*49*49

it says the answer is 25x^2+70x+49

--------

7. x^2+x-2 / x^3+x^2 * x / x^2+3x+2

Heres what i did:

= x^3 + x^2 - 2x / x^10+3x^7 + 2x^5
= 1 / x^7+3x^5+x^4

x-1 / x(x+1)^2

can someone give me a review on this? XD there's probably some little thing im doing wrong.....

2. Originally Posted by Cait
So we have our final exam on monday... so im doin the review.

I tried problems 1-10.. and got 1 right -_-. I fear i am doomed. Lol.

but anyway..

1. (5x+7)^2

Seems ridiculously easy to me..

Heres what i tried...

turned it into

(5x+7)(5x+7)
FOIL
25x^2*5x*7*7*5x*49

25x^2*10X*49*49

it says the answer is 25x^2+70x+49
we call it foil when we are contracting, that is, putting stuff in brackets. when removing things from brackets, we call it expanding.

anyway, why did you multiply everything together?

let's do the question your way:

(5x + 7)^2 = (5x + 7)(5x + 7)

now we expand this using the distributive law, that is, we take the first term in the first set of brackets and multiply everything in the second set of brackets, then take the second term in the first set of brackets and multiply everything in the second set of brackets. i will color code things so you can follow. here goes:

(5x + 7)(5x + 7) = (5x)(5x) + (5x)(7) + 7(5x) + 7(7)
.......................= 25x^2 + 35x + 35x + 49
.......................= 25x^2 + 70x + 49

An even easier way to do this when expanding squares (IT DOES NOT WORK FOR OTHER POWERS!!!) is by doing the following.

we sqaure the first term
then multiply the first term by the second term and then by 2
then we square the last term

(5x + 7)^2 = (5x)^2 + 2*(5x)*7 + (7)^2
................= 25x^2 + 70x + 49

7. x^2+x-2 / x^3+x^2 * x / x^2+3x+2
please retype this question using brackets

3. uhhh..

(x^2+x-2) / (X^3+x^2) (times) (X) / (X^2+3X+2)

like that>?

"anyway, why did you multiply everything together?"

because i have no idea what im doing >.<...

but anyway.. i tried some more problems from 1-10 and each one ive gotten right.. so thanks!!

still cant figure out the fractions though >.<

4. Originally Posted by Cait
uhhh..

(x^2+x-2) / (X^3+x^2) (times) (X) / (X^2+3X+2)

like that>?
yeah, that's fine. always use parenthesis when expressing complicated expressions, it saves everyone a lot of heartache...ok, let me review what you did now, i'll tell you if you are correct...oh, nevermind, we know you're wrong since the text has a different answer...though texts are wrong once in a while

5. Originally Posted by Cait
uhhh..

(x^2+x-2) / (X^3+x^2) (times) (X) / (X^2+3X+2)

like that>?

"anyway, why did you multiply everything together?"

because i have no idea what im doing >.<...

but anyway.. i tried some more problems from 1-10 and each one ive gotten right.. so thanks!!

still cant figure out the fractions though >.<
With this you want to start making things as simple as possible as soon as possible. don't multiply out at first, try to foil and see if things will cancel

6. Hello, Cait!

. . . x² + x - 2 . . . . . x
7) . ------------ · ---------------
. . . . x³ + x² . . x² + 3x + 2

It says "multiply", but you're supposed to REDUCE the fractions, too.
How do we reduce fractions? . . . Factor and cancel.

. . . . . . . . (x - 1)(x + 2) . . . . . x
We have: . --------------- · ----------------- . .
(Do you see what cancels?)
. . . . . . . . . x²(x + 1) . . .(x + 1)(x + 2)

. . . . . . . . .x - 1
. . . . . . . x(x + 1)²

7. ok so..
i get

(x-1)(x+2) / x^2(x+1) (times) (X) / (X+2)(X+1)

the X+2's cancel and
Then from that i get...
(x-1) / x^2(X+1) * X / (X+1)

like you said..

but then..

theres nothing left to cancel????????

its an x-1 and x+1.. so they cant... or?

i duno

8. Originally Posted by Cait
ok so..
i get

(x-1)(x+2) / x^2(x+1) (times) (X) / (X+2)(X+1)

the X+2's cancel and
Then from that i get...
(x-1) / x^2(X+1) * X / (X+1)

like you said..

but then..

theres nothing left to cancel????????

its an x-1 and x+1.. so they cant... or?

i duno
the x into the x^2 maybe...

9. Ok so if the x^2 and x cancel.. then i'll have..

(x-1) / x(x+1) * 1 / (X+1)

right?

then its ok just combine it.

man this is so hard for me for something so simple >_>...

10. Been practicin all year.. and it only got harder x-x

anyway...

Here i have one..

(x^2-2x) / (x^2-6x+8) * (3x-12) / (2x)

heres what i did..

x^2-2x / (x-2)(x-4) * 3(x-4) / (2x)

=

x^2-1 / x-2 * 3 / 1

=

3x - (3/2)

the answe is supposed to be just 3/2
thanks...

11. Originally Posted by Cait
Been practicin all year.. and it only got harder x-x

anyway...

Here i have one..

(x^2-2x) / (x^2-6x+8) * (3x-12) / (2x)

heres what i did..

x^2-2x / (x-2)(x-4) * 3(x-4) / (2x)

=

x^2-1 / x-2 * 3 / 1

=

3x - (3/2)

the answe is supposed to be just 3/2
thanks...
it seems you canceled the 2x in the previous line to get to the line in bold, you can't do that, it is a part of a sum, it needs to be a part of a product to cancel, try to factorize more

12. ## Re:

RE:

This problem is very easy because you do not even have to get a common denominator. Multiply straight across and simplify at the end if you like...

13. Alrighty.. So I can get it to work for the most part..

Here's another:

( (5u^5v^4) / (25u^6v^7) )2

I got..

= (25u^10v^8) / (625u^12v^14)

= (1) / (25u^-2v^-6)

and we were taught that if theres a negative exponent on the bottom, just put it to the top and leave off the negative..

SO now i get..

= u^2v^6 / 25

1/25u^2v^6

So how'd they get rid of the negative exponents on the bottom?

unless... they put the negatives on the top insted of the bottom

like this: = (u^-2v^-6) / (25)'

and then flipped it to make

1 / 25u^2v^6

That's probably what they did..

but why did they put those exponents on the top and not the bottom????

14. Originally Posted by Cait
Alrighty.. So I can get it to work for the most part..

Here's another:

( (5u^5v^4) / (25u^6v^7) )2

I got..

= (25u^10v^8) / (625u^12v^14)

= (1) / (25u^-2v^-6)

and we were taught that if theres a negative exponent on the bottom, just put it to the top and leave off the negative..

SO now i get..

= u^2v^6 / 25

1/25u^2v^6

So how'd they get rid of the negative exponents on the bottom?

unless... they put the negatives on the top insted of the bottom

like this: = (u^-2v^-6) / (25)'

and then flipped it to make

1 / 25u^2v^6

That's probably what they did..

but why did they put those exponents on the top and not the bottom????
when you say 1/25u^2v^6, do you mean (1/25)u^2v^6 or 1/(25u^2v^6)?? if it's the first one, then it's the same answer, relax, they just factered out the 1/25 and put it at the front

15. 1u^10v^8
__
25u^12v^14

=

u^-2v^-6
__
25

=
1
__
25^2v^6

I guess

They should impliment some sort of.... small drawing board for each post.. that would be cool..
(but hard on bandwidth )

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