Help with factoring?

- BloomLocke367

Help with factoring?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- BloomLocke367

\(\Huge x^2-10x+25-36y^2\)

- 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

I'm thinking it may be prime... but I'm not sure.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- BloomLocke367

@ganeshie8

- BloomLocke367

@phi

- anonymous

Think of it like\[\left( x^2- 10x+25\right)-y^2\]Can you factor the trinomial in the brackets?

- BloomLocke367

yes

- anonymous

*36y^2 Sorry

- anonymous

OK. Go ahead. What do you get?

- BloomLocke367

\((x-5)(x-5)\) or\( (x-5)^2\)

- 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

I didddd, but it's been like a year since I have. I forgot what to use. :/

- BloomLocke367

I should have kept my notes from Algebra 2 XD

- 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

yes, thank you. I'm going to write that down XD

- anonymous

You're welcome.

- BloomLocke367

So I have \((x-5)^2-(6y)^2=(x-5+6y)(x-5-6y)\)

- BloomLocke367

@ospreytriple

- anonymous

Correct. Well done.

- BloomLocke367

I'm not done yet, am I?

- anonymous

Can you simplify these factors any further?

- BloomLocke367

ummmmm, I don't think so

- BloomLocke367

@ospreytriple

- anonymous

Good. Then you're finished. Sorry, I am having connectivity issues.

- BloomLocke367

and while you're here, can you tell me how to find the difference of two cubes? and it's alright :)

- anonymous

Difference of cubes:
\[a^3-b^3 = \left( a-b \right)\left( a^2+ab+b^2 \right)\]

- BloomLocke367

ok, thanks!

- anonymous

Oh, and sum of cubes:\[a^3 + b^3 = \left( a+b \right)\left( a^2-ab+b^2 \right)\]

- anonymous

NP.

- BloomLocke367

thank youuuuuuuuuuuuuu

- BloomLocke367

also, is this prime? \(x^2+36\)

- anonymous

Not sure if "prime" is the correct term, but it is unfactorable with real numbers.

- BloomLocke367

prime and unfactorable mean the same thing when it comes to factoring.

- anonymous

Fair enough. I've never come across that term used to describe polynomials.

- BloomLocke367

haha, yeah. and for \(x^2-27\) I got \(x^3-3^3=(x-3)(x^2+3x+9)\). Is that correct?

- BloomLocke367

@Preetha is that correct? ^^

- Nnesha

:P

- BloomLocke367

you're still here XD is that right?

- Nnesha

lol
do you mean \[\rm x^3-27 \] ?

- BloomLocke367

yesss, my bad XD

- BloomLocke367

I'm pretty sure it's correct because when you distribute it back out, you get \(x^3-27\)

- Nnesha

yes right :)
i know you already know but just want to share SOAP for signs |dw:1440020416579:dw|

- BloomLocke367

soooo I did get it

- Nnesha

:)

- BloomLocke367

okay, I have one more.

- Nnesha

ok:)

- anonymous

Sorry. Lost connection again.

- 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

That's right. What do you think? 2 number that multiply to give -30 and add to -7

- Nnesha

\[a\cancel{=}1\] i'm not agree
if a =1 you can multiply ac

- BloomLocke367

I know. but you could just use c XD

- BloomLocke367

and -10 and 3

- Nnesha

and when \[a\cancel{=}1\] when you multiply 2 numbers you should get product of AC not just C

- BloomLocke367

I knowwwwwwww. XD lol I just didn't know if -5y^2 was C or not.

- Nnesha

LOL just saying :P

- 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

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

ohhh okay I got it, thanks! I gotta go, bye! :) thanks so so so so much for the help!

- anonymous

You're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.