162x^5y^2-32xy^2
What are you trying to do?
Factor completely. I'm not sure how to do these types of problems and my HW is due in 2 hours. :/
Tell me the coefficients you see in your question
162, -32
okay so what is the greatest common denominator of both coefficients?
Sorry for the slow reply there... Well we see that they are factorable by 2 and they can also be factorable by one x and y^2... Thus 2xy^2(81x^4-16) is the answer
that is the answer^
So 2 is the GCD?
Yes.
Thank you both so much! :)
No problem at all ^_^
I typed that answer in and it's saying it's wrong...
hmm something else should be factored then
There is nothing that can be factored left...
81 is uneven... 16 is even... they share no common factor
OrthodoxMan, did you get a different answer?
well that i know but there has to be something wrong with the answer and no i got the same as him
Super weird... Guess I'll leave it blank and ask my professor tomorrow. Thank you! :)
are you sure? im known to not give up
haha if you want to try and figure it out you are welcome and I'll keep putting whatever answers you get.
I copied and pasted exactly as it is on my HW: 162x^{5}y^{2}-32xy^{2}
alright show me the screenshot saying it's wrong with the answer
K. hang on... Gotta get my phone.
The answer is wrong, it could be reduced further
It can? Are you going to make me figure it out?
well what do you want me to do? Tell you the answer and then explain after?
sure... lol
lol alright you better not run away after i tell you :P
I definitely won't.
the answer is 2xy^2(9x^2 +4)(3x-2)(3x + 2)
:) you're awesome! Now tell me how you did that.
awesomee alright so you already have your greatest common denominator factored out along with the highest powers right?
Right...
Do you get what i mean by highest powers?
5?
nope If i say the highest power that can be factored out I mean like the lowest exponent that exists in the equation So what we had was: 162x^5y^2-32xy^2 The lowest exponent for x is 1 and that can be seen in 32x^1y^2 Notice we dont show x^1 but we write it as x What is the lowest exponent for y?
:/ this gives me a headache. Lowest exponent for y, 1?
2?
there you go :)
:) lol thank you! Can you come do my final on Wednesday for me?
i hope that headache has disappeared for now We're not done yet
Oh man! OK..
what we have after factoring the highest power and the greatest common denominator is 2xy^2(81x^4 - 16)
what do you notice about the two coefficients in the bracket?
I do not know. I'm telling you I'm horrible at math.
Haha dont worry, we'll work on that together Do you know what perfect squares are?
No. haha
lol well that is what we're going to do next I'll make this real short What is the answer to 3^2?
9
that my friend is a perfect square What about 4^2?
16
;)
and 9^2?
81
ah I got it!
yeahh now what do you think happens to this: 2xy^2(81x^4-16)
I think... So we just had to figure out what goes into 81 and 16?
which was 9 for 81 and... now I'm lost again...
dont confuse yourself now this is what we want to reduce: (81x^4-16) In algebra there is a way to reduce them are you ready for the magical formula?
Yes!
This formula is called the difference of squares Before I step further, let me break it down for you To reduce squares, say (16x^2 -4) you would have to square root the coefficients first the square coefficients would be 4 and 2 square rooting x^2 would give you x Whenever you have a subtraction in the brackets you would have to multiply the addition and subtraction versions
Now comes the formula you will consider (a^2 - b^2) = (a+b)(a-b)
do you get this so far? let me know if something confuses you
I think I understand...
okay, ill give you an example right after this to make sure SO looking back at our example: (16x^2 -4) a^2 = 16 a = 4 b^2 = 4 b = 2 The square root of x^2 is simply x now you make the factor using the formula (16x^2 -4) = (4x -2)(4x+2) you can reduce it to become (16x^2 -4) = 2(2x -1)(2x+1)
you get that?
why is b^2=4? and x^2=x? I am sooooo confused. I don't think there's any hope for me getting this tonight! lol
lol there is Im staying up to make sure
I understand that in my original problem the answer could be reduced further.
let's put this aside and go back to our question Maybe it's the numbers that are the same Could probably be confusing you
So we have this: (81x^4-16) What would be a^2?
Use the formula i gave you: (a^2 - b^2) = (a+b)(a-b)
a? I have no idea.
16
Look at this side of the formula: (a^2 - b^2) Look at this now: (81x^4-16) Match it up. Hint: Include the x as well.
a^2 and b^2 are perfect squares
All you have to do is tell me what a^2 and b^2 are :)
16?
what is that? a^2 or b^2 It's one of them :)
haha you're so mean!
a
haha what did i say?
b^2=4 ;)
and nopeeee b^2 = 16
OMG
(81x^4-16) look at that
to get b you have to take the square root of b^2
you know how to square root?
Yes. But how do you square root a letter with a power?
you subtract it's exponent But I know what you're thinking b^2 is just a variable squared we want to find b okay :) Want me to run a small example?
Sure, can't guarantee I'll understand.
I can :) Say you have an expression as follows: x^2 = 16 Your goal is to square root both sides and find x what would the square root of 16 be?
4
Good!! Now what is the square root of x^2? Remember divide the exponents by 2
16?
nope i just want the square root of x^2 dont look at 16 look at x^2 you have two sides of the equation x^2 is your left side 16 is on the right side correct?
Looking at this would help x^2 = 16 Left right
Tell me. haha
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