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BloomLocke367
 one year ago
Help with factoring?
BloomLocke367
 one year ago
Help with factoring?

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BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(\Huge x^210x+2536y^2\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0at 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
 one year ago
Best ResponseYou've already chosen the best response.0I'm thinking it may be prime... but I'm not sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Think of it like\[\left( x^2 10x+25\right)y^2\]Can you factor the trinomial in the brackets?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK. Go ahead. What do you get?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\((x5)(x5)\) or\( (x5)^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Excellent. So now you have\[\left( x5 \right)^2  36y^2\]This is a difference of squares. Do you know how to factor these?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I didddd, but it's been like a year since I have. I forgot what to use. :/

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I should have kept my notes from Algebra 2 XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK. In general, factoring a difference of squares looks like\[a^2b^2 = \left( a+b \right)\left( ab \right)\]In your question, you have \(a=x5\) and \(b=6y\). Can you put it all together now?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0yes, thank you. I'm going to write that down XD

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0So I have \((x5)^2(6y)^2=(x5+6y)(x56y)\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I'm not done yet, am I?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you simplify these factors any further?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0ummmmm, I don't think so

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good. Then you're finished. Sorry, I am having connectivity issues.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0and while you're here, can you tell me how to find the difference of two cubes? and it's alright :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Difference of cubes: \[a^3b^3 = \left( ab \right)\left( a^2+ab+b^2 \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, and sum of cubes:\[a^3 + b^3 = \left( a+b \right)\left( a^2ab+b^2 \right)\]

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0thank youuuuuuuuuuuuuu

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0also, is this prime? \(x^2+36\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not sure if "prime" is the correct term, but it is unfactorable with real numbers.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0prime and unfactorable mean the same thing when it comes to factoring.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Fair enough. I've never come across that term used to describe polynomials.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0haha, yeah. and for \(x^227\) I got \(x^33^3=(x3)(x^2+3x+9)\). Is that correct?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0@Preetha is that correct? ^^

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0you're still here XD is that right?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0lol do you mean \[\rm x^327 \] ?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0yesss, my bad XD

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I'm pretty sure it's correct because when you distribute it back out, you get \(x^327\)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes right :) i know you already know but just want to share SOAP for signs dw:1440020416579:dw

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0soooo I did get it

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0okay, I have one more.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry. Lost connection again.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(6x^27xy5y^2\), 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
 one year ago
Best ResponseYou've already chosen the best response.0That's right. What do you think? 2 number that multiply to give 30 and add to 7

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[a\cancel{=}1\] i'm not agree if a =1 you can multiply ac

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I know. but you could just use c XD

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0and when \[a\cancel{=}1\] when you multiply 2 numbers you should get product of AC not just C

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I knowwwwwwww. XD lol I just didn't know if 5y^2 was C or not.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @BloomLocke367 \(6x^27xy5y^2\), 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
 one year ago
Best ResponseYou've already chosen the best response.0Perfect. So rewrite as\[6x^2+3xy10xy5y^2\]Regroup as two binomials\[\left( 6x^2+3xy \right)\left( 10xy+5y^2 \right)\]Factor common factor out of each binomial.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0ohhh okay I got it, thanks! I gotta go, bye! :) thanks so so so so much for the help!
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