Help with factoring?

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\(\Huge x^2-10x+25-36y^2\)
at first I thought about factoring by grouping, but 25 and -36 don't have a GCF. So, I'm not sure what to do.
I'm thinking it may be prime... but I'm not sure.

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Think of it like\[\left( x^2- 10x+25\right)-y^2\]Can you factor the trinomial in the brackets?
yes
*36y^2 Sorry
OK. Go ahead. What do you get?
\((x-5)(x-5)\) or\( (x-5)^2\)
Excellent. So now you have\[\left( x-5 \right)^2 - 36y^2\]This is a difference of squares. Do you know how to factor these?
I didddd, but it's been like a year since I have. I forgot what to use. :/
I should have kept my notes from Algebra 2 XD
OK. In general, factoring a difference of squares looks like\[a^2-b^2 = \left( a+b \right)\left( a-b \right)\]In your question, you have \(a=x-5\) and \(b=6y\). Can you put it all together now?
yes, thank you. I'm going to write that down XD
You're welcome.
So I have \((x-5)^2-(6y)^2=(x-5+6y)(x-5-6y)\)
Correct. Well done.
I'm not done yet, am I?
Can you simplify these factors any further?
ummmmm, I don't think so
Good. Then you're finished. Sorry, I am having connectivity issues.
and while you're here, can you tell me how to find the difference of two cubes? and it's alright :)
Difference of cubes: \[a^3-b^3 = \left( a-b \right)\left( a^2+ab+b^2 \right)\]
ok, thanks!
Oh, and sum of cubes:\[a^3 + b^3 = \left( a+b \right)\left( a^2-ab+b^2 \right)\]
NP.
thank youuuuuuuuuuuuuu
also, is this prime? \(x^2+36\)
Not sure if "prime" is the correct term, but it is unfactorable with real numbers.
prime and unfactorable mean the same thing when it comes to factoring.
Fair enough. I've never come across that term used to describe polynomials.
haha, yeah. and for \(x^2-27\) I got \(x^3-3^3=(x-3)(x^2+3x+9)\). Is that correct?
@Preetha is that correct? ^^
:P
you're still here XD is that right?
lol do you mean \[\rm x^3-27 \] ?
yesss, my bad XD
I'm pretty sure it's correct because when you distribute it back out, you get \(x^3-27\)
yes right :) i know you already know but just want to share SOAP for signs |dw:1440020416579:dw|
soooo I did get it
:)
okay, I have one more.
ok:)
Sorry. Lost connection again.
\(6x^2-7xy-5y^2\), I'm not sure what to do... does \(5y^2\) count as c? Because I know that when \(a\ne 1\) you take \(ac\) and find the factors of those that multiply to get c but add to get b.
That's right. What do you think? 2 number that multiply to give -30 and add to -7
\[a\cancel{=}1\] i'm not agree if a =1 you can multiply ac
I know. but you could just use c XD
and -10 and 3
and when \[a\cancel{=}1\] when you multiply 2 numbers you should get product of AC not just C
I knowwwwwwww. XD lol I just didn't know if -5y^2 was C or not.
LOL just saying :P
\(\color{#0cbb34}{\text{Originally Posted by}}\) @BloomLocke367 \(6x^2-7xy-5y^2\), I'm not sure what to do... does \(5y^2\) count as c? Because I know that when \(a\ne 1\) you take \(ac\) and find the factors of those that multiply to get c but add to get b. \(\color{#0cbb34}{\text{End of Quote}}\) I meant to say that multiply to get ac.. now I see why you corrected me @Nnesha XD
Perfect. So rewrite as\[6x^2+3xy-10xy-5y^2\]Regroup as two binomials\[\left( 6x^2+3xy \right)-\left( 10xy+5y^2 \right)\]Factor common factor out of each binomial.
ohhh okay I got it, thanks! I gotta go, bye! :) thanks so so so so much for the help!
You're welcome

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