Question

Help with factoring?

Answer 1

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

Answer 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.

Answer 3

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

Answer 4

@ganeshie8

Answer 5

@phi

Answer 6

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

Answer 7

yes

Answer 8

*36y^2 Sorry

Answer 9

OK. Go ahead. What do you get?

Answer 10

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

Answer 11

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?

Answer 12

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

Answer 13

I should have kept my notes from Algebra 2 XD

Answer 14

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?

Answer 15

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

Answer 16

You're welcome.

Answer 17

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

Answer 18

@ospreytriple

Answer 19

Correct. Well done.

Answer 20

I'm not done yet, am I?

Answer 21

Can you simplify these factors any further?

Answer 22

ummmmm, I don't think so

Answer 23

@ospreytriple

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

ok, thanks!

Answer 28

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

Answer 29

NP.

Answer 30

thank youuuuuuuuuuuuuu

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

@Preetha is that correct? ^^

Answer 37

:P

Answer 38

you're still here XD is that right?

Answer 39

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

Answer 40

yesss, my bad XD

Answer 41

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

Answer 42

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