## anonymous one year ago Part A: Factor 3x2y2 − 2xy2 − 8y2. Show your work. (4 points) Part B: Factor x2 + 10x + 25. Show your work. (3 points) Part C: Factor x2 − 36. Show your work. (3 points)

1. anonymous

i need steps i have answers! Part A.) y^2(3x+4)(x-2) Part B.) (x-5)^2 Part C.) (x+6)(x-6)

2. anonymous

@zepdrix

3. anonymous

sorry I cant help u on this but I love the name :)

4. anonymous

ty :}

5. anonymous

yw and good luck on ur question!

6. anonymous

ty

7. anonymous

yw

8. zepdrix

$\large\rm 3x^2y^2 − 2xy^2 − 8y^2$What are we doing? :U Factoring this thing?

9. zepdrix

$\large\rm 3x^2\color{orangered}{y^2} − 2x\color{orangered}{y^2} − 8\color{orangered}{y^2}$Hmm there isn't a whole lot that we can do with this first one. Notice that the last term has no x's, so we can't pull out any factor out any amount of x. The last two terms have a 2 in them (8 is 4 times 2) but the first term does not, so we can't pull a 2 out. You can do something with the y's though, ya?

10. anonymous

Yes.

11. zepdrix

Oh I guess we can factor the x's after that though :3 hah

12. zepdrix

$\large\rm \color{orangered}{y^2}(3x^2−2x−8)$So you ok with that first step? :o

13. anonymous

Um (y2)(3x2+−2x+−8) =(y2)(3x2)+(y2)(−2x)+(y2)(−8) =3x2y2−2xy2−8y2 =3x2y2−2xy2−8y2

14. zepdrix

You just un-did the first step :) Not sure why hehe

15. anonymous

Opps >_<

16. anonymous

I am confused on this..

17. zepdrix

|dw:1440970256523:dw|So we're taking y^2 out of each term.

18. zepdrix

And from there, we're going to factor our x's. That first step giving you trouble though? :d

19. anonymous

y^2(3x+4)(x-2)

20. zepdrix

Yes, that's the answer. But we're not through the steps quite yet :) lol

21. anonymous

Ok lol :]

22. zepdrix

Since the x^2 has a coefficient, we'll have to try factoring by grouping for this one:$\large\rm 3x^2-2x-8\qquad\to\qquad ax^2+bx+c$Multiplying a and c gives us -24, ya? So we need factors of -24 that will add to -2. Example: $$\large\rm -12 * 2 = -24$$ But $$\large\rm -12 + 2\ne-2$$ Hmm so -12 and 2 are NOT the factors we're looking for.

23. zepdrix

$$\large\rm 3\cdot-8=-24$$ But $$\large\rm 3+-8\ne-2$$ Hmm looks like those factors don't work either :[

24. zepdrix

Any other ones you can try? :)

25. anonymous

Um... 2*-8=(-16)

26. anonymous

$2\times-8= -16$

27. zepdrix

No no, we want numbers that multiply to -24.

28. anonymous

OH OK.

29. zepdrix

3 and 8 didn't work. 12 and 2 didn't work. Any other combinations we can try? :o

30. anonymous

$(-12) *2= -24$ OR $12*(-2)=-24$

31. zepdrix

No we already tried 12 and 2. Gotta try some other factors of 24

32. anonymous

k

33. anonymous

$6•4$

34. zepdrix

$$\large\rm 6\cdot-4=-24$$ Ok those work for multiplication. Let's try adding them. $$\large\rm 6+-4=2$$ Hmm I got a 2 instead of a -2, that simply means I Put the negative on the wrong number. So the factors we want are -6 and 4. $$\large\rm -6\cdot4=-24$$ $$\large\rm -6+4=-2$$

35. anonymous

^0^

36. zepdrix

$\large\rm 3x^2\color{orangered}{−2x}−8$So what we want to do is, split up this -2x into -6x and 4x.$\large\rm 3x^2\color{orangered}{−6x+4x}−8$

37. zepdrix

And from there, we can do some grouping.$\large\rm (3x^2−6x)+(4x−8)$

38. anonymous

=)

39. zepdrix

So uh, factor some stuff :O what can you pull out of each bracket?

40. anonymous

3x2−2x−8

41. anonymous

$3x^2−2x−8$

42. zepdrix

...?

43. zepdrix

factor some stuff out of this: (3x^2-6x)

44. anonymous

3x(x−2)

45. zepdrix

$\large\rm (3x^2−6x)+(4x−8)$$\large\rm 3x(x−2)+(4x−8)$Good good good :O Do something similar with the other set of brackets.

46. anonymous

$(3x+4)(x−2)$

47. zepdrix

Mmm k, so you get a 4 from the other brackets, ya?$\large\rm 3x(x−2)+(4x−8)$$\large\rm 3x(x−2)+4(x−2)$And you go a step further and pull the (x-2) from each term,$\large\rm (x−2)(3x+4)$Mmm k good, and ya, you get that. With the y^2 still in front.$\large\rm y^2(3x+4)(x-2)$

48. anonymous

k

49. zepdrix

For Part C recall your difference of squares formula:$\large\rm a^2-b^2=(a-b)(a+b)$Notice that in the formula, both numbers are squared on the left side of the equation. So in order to apply this rule to our equation,$\large\rm x^2-36$We'll first need to rewrite 36 as a square somehow.

50. anonymous

ok maybe $6^2$

51. zepdrix

ok great!$\large\rm x^2-6^2$Good. Now you can apply the formula:$\large\rm x^2-6^2=(x-6)(x+6)$Nice easy one step problem. It's really just about remembering your formula for that one.

52. anonymous

Can you write like Part A Part B And Part C I am really confused what is what >_< lol

53. zepdrix

We skipped Part B, maybe that's the confusion. I gotta go make a sammich, so I was trying to rush through it >.< lol sorry The big long explanation was Part A. And then these last two comments I posted were Part C.

54. anonymous

I really need part B too!

55. zepdrix

you -_- you and your ways lol

56. zepdrix

ok sec :3 lemme read it again

57. anonymous

Thank yyou!

58. zepdrix

$\large\rm x^2 + 10x + 25$So this is our typical factoring problem. Don't have a good grasp on these types yet? :o You look for factors of 25 that will sum to 10. Lemme try a one and see what happens. $$\large\rm 25\cdot1=25$$ $$\large\rm 25+1=26$$ hmm, nope. they didn't add to 10.

59. zepdrix

Any other factors we can try?

60. anonymous

$25+(-15)$

61. zepdrix

factors of 25 is what you're looking for. $$\large\rm 25\cdot(-15)\ne25$$ Those are not factors of 25.

62. zepdrix

Example: If I was looking for factors of 24, i would list them like this maybe: 8 * 3 = 24 6 * 4 = 24 12 * 2 = 24 24 * 1 = 24 So those are all of the combinations I would want to try.

63. anonymous

There is not one.

64. zepdrix

25 ends in a 5, so it's 5 times something

65. anonymous

Yes 5*5...

66. zepdrix

$$\large\rm 5\cdot5=25$$ $$\large\rm 5+5=10$$ Ah there we go! That's the one we were looking for.

67. anonymous

Is that part B?

68. anonymous

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

69. anonymous

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

70. zepdrix

So for part B, you can either do factoring by grouping again:$\large\rm x^2 + 10x + 25$$\large\rm x^2 + 5x+5x + 25$$\large\rm x(x + 5)+5(x + 5)$$\large\rm (x+5)(x + 5)$$\large\rm (x+5)^2$Or just remember your shortcut for factoring. Yes. but not -5, +5 silly :)

71. zepdrix

The solution you listed at the start is incorrect. That's strange :P

72. anonymous

I am confused whats part a and wahts part c?

73. anonymous

*whats