BloomLocke367
  • BloomLocke367
Help with factoring?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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BloomLocke367
  • BloomLocke367
\(\Huge x^2-10x+25-36y^2\)
BloomLocke367
  • BloomLocke367
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.
BloomLocke367
  • BloomLocke367
I'm thinking it may be prime... but I'm not sure.

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BloomLocke367
  • BloomLocke367
@ganeshie8
BloomLocke367
  • BloomLocke367
@phi
anonymous
  • anonymous
Think of it like\[\left( x^2- 10x+25\right)-y^2\]Can you factor the trinomial in the brackets?
BloomLocke367
  • BloomLocke367
yes
anonymous
  • anonymous
*36y^2 Sorry
anonymous
  • anonymous
OK. Go ahead. What do you get?
BloomLocke367
  • BloomLocke367
\((x-5)(x-5)\) or\( (x-5)^2\)
anonymous
  • anonymous
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?
BloomLocke367
  • BloomLocke367
I didddd, but it's been like a year since I have. I forgot what to use. :/
BloomLocke367
  • BloomLocke367
I should have kept my notes from Algebra 2 XD
anonymous
  • anonymous
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?
BloomLocke367
  • BloomLocke367
yes, thank you. I'm going to write that down XD
anonymous
  • anonymous
You're welcome.
BloomLocke367
  • BloomLocke367
So I have \((x-5)^2-(6y)^2=(x-5+6y)(x-5-6y)\)
BloomLocke367
  • BloomLocke367
@ospreytriple
anonymous
  • anonymous
Correct. Well done.
BloomLocke367
  • BloomLocke367
I'm not done yet, am I?
anonymous
  • anonymous
Can you simplify these factors any further?
BloomLocke367
  • BloomLocke367
ummmmm, I don't think so
BloomLocke367
  • BloomLocke367
@ospreytriple
anonymous
  • anonymous
Good. Then you're finished. Sorry, I am having connectivity issues.
BloomLocke367
  • BloomLocke367
and while you're here, can you tell me how to find the difference of two cubes? and it's alright :)
anonymous
  • anonymous
Difference of cubes: \[a^3-b^3 = \left( a-b \right)\left( a^2+ab+b^2 \right)\]
BloomLocke367
  • BloomLocke367
ok, thanks!
anonymous
  • anonymous
Oh, and sum of cubes:\[a^3 + b^3 = \left( a+b \right)\left( a^2-ab+b^2 \right)\]
anonymous
  • anonymous
NP.
BloomLocke367
  • BloomLocke367
thank youuuuuuuuuuuuuu
BloomLocke367
  • BloomLocke367
also, is this prime? \(x^2+36\)
anonymous
  • anonymous
Not sure if "prime" is the correct term, but it is unfactorable with real numbers.
BloomLocke367
  • BloomLocke367
prime and unfactorable mean the same thing when it comes to factoring.
anonymous
  • anonymous
Fair enough. I've never come across that term used to describe polynomials.
BloomLocke367
  • BloomLocke367
haha, yeah. and for \(x^2-27\) I got \(x^3-3^3=(x-3)(x^2+3x+9)\). Is that correct?
BloomLocke367
  • BloomLocke367
@Preetha is that correct? ^^
Nnesha
  • Nnesha
:P
BloomLocke367
  • BloomLocke367
you're still here XD is that right?
Nnesha
  • Nnesha
lol do you mean \[\rm x^3-27 \] ?
BloomLocke367
  • BloomLocke367
yesss, my bad XD
BloomLocke367
  • BloomLocke367
I'm pretty sure it's correct because when you distribute it back out, you get \(x^3-27\)
Nnesha
  • Nnesha
yes right :) i know you already know but just want to share SOAP for signs |dw:1440020416579:dw|
BloomLocke367
  • BloomLocke367
soooo I did get it
Nnesha
  • Nnesha
:)
BloomLocke367
  • BloomLocke367
okay, I have one more.
Nnesha
  • Nnesha
ok:)
anonymous
  • anonymous
Sorry. Lost connection again.
BloomLocke367
  • 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.
anonymous
  • anonymous
That's right. What do you think? 2 number that multiply to give -30 and add to -7
Nnesha
  • Nnesha
\[a\cancel{=}1\] i'm not agree if a =1 you can multiply ac
BloomLocke367
  • BloomLocke367
I know. but you could just use c XD
BloomLocke367
  • BloomLocke367
and -10 and 3
Nnesha
  • Nnesha
and when \[a\cancel{=}1\] when you multiply 2 numbers you should get product of AC not just C
BloomLocke367
  • BloomLocke367
I knowwwwwwww. XD lol I just didn't know if -5y^2 was C or not.
Nnesha
  • Nnesha
LOL just saying :P
BloomLocke367
  • BloomLocke367
\(\color{#0cbb34}{\text{Originally Posted by}}\) @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
anonymous
  • anonymous
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.
BloomLocke367
  • BloomLocke367
ohhh okay I got it, thanks! I gotta go, bye! :) thanks so so so so much for the help!
anonymous
  • anonymous
You're welcome

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