## BloomLocke367 one year ago Help with factoring?

1. BloomLocke367

$$\Huge x^2-10x+25-36y^2$$

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.

3. BloomLocke367

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

4. BloomLocke367

@ganeshie8

5. BloomLocke367

@phi

6. anonymous

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

7. BloomLocke367

yes

8. anonymous

*36y^2 Sorry

9. anonymous

OK. Go ahead. What do you get?

10. BloomLocke367

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

11. 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?

12. BloomLocke367

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

13. BloomLocke367

I should have kept my notes from Algebra 2 XD

14. 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?

15. BloomLocke367

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

16. anonymous

You're welcome.

17. BloomLocke367

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

18. BloomLocke367

@ospreytriple

19. anonymous

Correct. Well done.

20. BloomLocke367

I'm not done yet, am I?

21. anonymous

Can you simplify these factors any further?

22. BloomLocke367

ummmmm, I don't think so

23. BloomLocke367

@ospreytriple

24. anonymous

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

25. BloomLocke367

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

26. anonymous

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

27. BloomLocke367

ok, thanks!

28. anonymous

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

29. anonymous

NP.

30. BloomLocke367

thank youuuuuuuuuuuuuu

31. BloomLocke367

also, is this prime? $$x^2+36$$

32. anonymous

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

33. BloomLocke367

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

34. anonymous

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

35. BloomLocke367

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

36. BloomLocke367

@Preetha is that correct? ^^

37. Nnesha

:P

38. BloomLocke367

you're still here XD is that right?

39. Nnesha

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

40. BloomLocke367

41. BloomLocke367

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

42. Nnesha

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

43. BloomLocke367

soooo I did get it

44. Nnesha

:)

45. BloomLocke367

okay, I have one more.

46. Nnesha

ok:)

47. anonymous

Sorry. Lost connection again.

48. 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.

49. anonymous

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

50. Nnesha

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

51. BloomLocke367

I know. but you could just use c XD

52. BloomLocke367

and -10 and 3

53. Nnesha

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

54. BloomLocke367

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

55. Nnesha

LOL just saying :P

56. 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

57. 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.

58. BloomLocke367

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

59. anonymous

You're welcome