## YanaSidlinskiy one year ago Answers need to be checked. DO NOT SPAM! I'M NOT JOKING!

1. YanaSidlinskiy

1.) Factor out the GCF. $16x^7-5x^5+5x^4$

2. YanaSidlinskiy

My answer: $x^4(16x^3-5x+5)$

3. YanaSidlinskiy

2.) Factor out the GCF:$60x^2-12xy+28x$ My answer:$4x(15x-3y+7)$

4. YanaSidlinskiy

3.) Factor by grouping. $4x^2-8xy-3x+6y$ My answer: $(4x-3)(x-2y)$

5. YanaSidlinskiy

4.) Factor:$x^2-x-42$ My answer:$-6,7$ >>>> I'm thinking I have this down wrong or something.. <<<<

6. Nnesha

you don't need to solve for x ^^^ -6 and 7 are zero (x-intercept )

7. anonymous

8. YanaSidlinskiy

5.) Which of the following trinomials has a greatest common factor that can be factored out first? $100x^2+140x+49$ $9x^2+24x+16$ $x^2+4x+4$ $8x^2+16x+8$

9. YanaSidlinskiy

My answer is the last one and @mathway no. My answer for number for as you can see above...

10. YanaSidlinskiy

6.) Factor completely. $1-4x^2$ My answer: $(1-2x)^2$

11. anonymous

Then you're wrong. (x-6)(x+7) is not equal to the polynomial. You might want to check your signs.

12. YanaSidlinskiy

Ok, thought so too!

13. Nnesha

6)hint apply difference of squares method

14. YanaSidlinskiy

6.) Isn't that what I did??

15. Nnesha

difference of squares $\huge\rm a^2-b^2 =(a\color{Red}{-}b)(a\color{reD}{+}b)$ one parentheses with negative sign and other one should be positive

16. Nnesha

$\huge\rm -4x^2+1$ first take out the negative sign $\huge\rm -(4x^2-1)$

17. YanaSidlinskiy

So it'd be: $2x^2-1$?

18. Nnesha

that is one of the factor (a-b)

19. YanaSidlinskiy

Ohh! Wait. I know what I did wrong.. It'd be: $(1-2x)(1+2x)$

20. YanaSidlinskiy

Is that right? John I see you spying ;)

21. Nnesha

well that's right or you can leave it as $-(2x-1)(2x+1)$ remember we already factor out the negative sign :=)

22. Nnesha

thanks o^_^o

23. YanaSidlinskiy

So the one is a negative then?

24. Nnesha

i would leave it as -(2x-1)(2x+1) which is same as (-2x+1)(2x+1) |dw:1438901505115:dw|

25. YanaSidlinskiy

Ahh, ok. What about number 4? What's up with it??

26. YanaSidlinskiy

Would it be (x+6)(x-7)?

27. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @YanaSidlinskiy 6.) Factor completely. $1-4x^2$ My answer: $(1-2x)^2$ $$\color{blue}{\text{End of Quote}}$$ (a+b)(a+d) <-- these are factors you don't have to solve for x (x+6)(x-7) are factors x=-6,7 are zeros(x-intercept ,solutions )

28. Nnesha

yes that's right

29. johnweldon1993

$\large x^2 -x - 42=0$ We have -7 and 6 as zeros...meaning $\large (x+7)(x-6)=0$ right?

30. YanaSidlinskiy

Yeah.

31. johnweldon1993

Make sure if you mean "yeah" or not...because we have 2 different answers :P

32. johnweldon1993

And that was a test :P You have it correct @YanaSidlinskiy

33. YanaSidlinskiy

Ok, well for my next question.... 7.) Solve by factoring. $b^2=32-4b$ My answer: $-8,4$ OMG @johnweldon1993 I'M CALLING YOU ON THAT ONE!!!!! After all the struggle...... YOU JUST HAD TOO!!!!

34. johnweldon1993

I was trying to test you because you've been doing so good!! You are correct The REAL zeros are -6 and 7 So indeed $\large (x+6)(x-7)$ is correct

35. johnweldon1993

36. YanaSidlinskiy

LOL. I was about to go off and say that -6+7 is not -1 lol.. Omg that was a heartbeat. Say sorry hhaa;) Jkjk.. Anyways..

37. YanaSidlinskiy

I've posted... #7..... Is it right?

38. YanaSidlinskiy

39. Nnesha

$$\huge\color{green}{\checkmark}$$

40. johnweldon1993

Oh sorry...looked away...indeed!

41. YanaSidlinskiy

Why would you look away when it's me... SUNSHINE hahha;;) Jkjk.... Anyways.. 9.) $x^2-4x+4=0$ My answer: The equation has no real solutions..

42. YanaSidlinskiy

^ I was supposed to be solving that by the quadratic formula..

43. johnweldon1993

Hmm...are you sure? You didnt get ANY zeros?

44. YanaSidlinskiy

Is this a test John?

45. johnweldon1993

Lol no, a legitimate question...lets do the formula out $\large \frac{4 \pm \sqrt{16 - 4(1)(4)}}{2}$

46. YanaSidlinskiy

$\frac{ 4\pm \sqrt{16-16} }{ 2 }$ I tried ignoring the "-"..... Lol. Even if this is right, it'd factor out 16-16=0..

47. YanaSidlinskiy

Ohh... great... Here comes the PRO.

48. johnweldon1993

No that is correct So if we have $\large \frac{4 \pm \sqrt{0}}{2}$ Whats the square root of 0?

49. YanaSidlinskiy

0..

50. johnweldon1993

Right, so it is 4 plus or minus 0 (which is just 4 in either case) divided by 2 $\large \frac{4}{2} = ?$

51. YanaSidlinskiy

2.

52. johnweldon1993

And that is your zero :)

53. YanaSidlinskiy

So, wait... x=0?

54. YanaSidlinskiy

I mEAN 2=X? Hhaa.

55. YanaSidlinskiy

Or more like x=2 lol...

56. johnweldon1993

There you go :) You can always check all these too...plug in 2 into your equation for 'x' and see if it comes out right

57. YanaSidlinskiy

10.) If S=-16t^2+64t represents the height of a rocket, in feet, t seconds after it was fired, when will the rocket hit the ground? I got 4 seconds...

58. YanaSidlinskiy

Lol, okie.

59. johnweldon1993

Ehh instead of going into a whole thing...yes correct lol

60. YanaSidlinskiy

Lol, okie. Ridiculous one for the last haha;) So I'm supposed to be solving this equation using the quadratic formula... I do not know like what to do.. I mean I do but. Idk lol.. $y^2=3y+18$

61. YanaSidlinskiy

I can do this, right? $y^2-3y-18 = 0$

62. johnweldon1993

Its the same thing...just using 'y' instead of 'x' But same process And yes...that is exactly what you will do!

63. YanaSidlinskiy

Ok, you can have a break while I solve this... And tell me if I'm right..

64. johnweldon1993

"Makes a sandwich" :P

65. YanaSidlinskiy

LOL.

66. YanaSidlinskiy

6?

67. johnweldon1993

68. YanaSidlinskiy

(6,-6)

69. YanaSidlinskiy

*-3

70. johnweldon1993

Not quite... $\large \frac{3\pm \sqrt{9 - 4(1)(-18)}}{2}$ $\large \frac{3\pm \sqrt{9 + 72}}{2}$ $\large \frac{3 \pm \sqrt{81}}{2}$ Ahh there you go...you got it The answers are 6 and -3

71. YanaSidlinskiy

So, (6,-3)?

72. johnweldon1993

Just dont write it like that...it is not a coordinate Just y = 6 and y = -3 are solutions to this equation

73. YanaSidlinskiy

|dw:1438903906614:dw|

74. johnweldon1993

Lol <3

75. YanaSidlinskiy

Okie, thanks again. I gotta go cuz college closes in 30 minutes and that's where I'm at haha. So I gotta go home rn..

76. johnweldon1993

Of course :) have fun!