## anonymous one year ago If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder

1. UsukiDoll

\[x^4+4x^3+2x^2+x+4\] by \[x^2+3x\] have you made any attempts to this problem?

2. anonymous

x4+4x3+2x2+x+4byx2+3x =x4+4x3+2x2+x+4bx2y+3x Combine Like Terms: =x4+4x3+2x2+x+4bx2y+3x =(4bx2y)+(x4)+(4x3)+(2x2)+(x+3x) =4bx2y+x4+4x3+2x2+4x Answer: =4bx2y+x4+4x3+2x2+4x right?

3. UsukiDoll

whoa man.. direct answers are against openstudy's code of conduct

4. UsukiDoll

and this is long division... I'm not doing that until the official poster responds

5. anonymous

6. UsukiDoll

so you had tried the long division method to this problem?

7. anonymous

thats not a direct answer i said it step by step and just to let you know i did this on the other post and they said that they under stood it completely now

8. anonymous

yes sir i have. would you like to see the answer choices

9. UsukiDoll

yeah I like to see the choices

10. anonymous

A) x2 + 3x B) 4x + 4 C) x2 + x – 2 D) 4 E) 0 That is all of the answer choices

11. UsukiDoll

ok... so this problem is asking to divide... so we need long division. |dw:1435149580706:dw|

12. UsukiDoll

wow.. this is long.. so first of all we need to get rid of the x^4

13. anonymous

yea it sucks i'm in Algebra 2.

14. UsukiDoll

I've learned this in Algebra 1 O_O anyway , are you aware of the exponential rules?

15. anonymous

Don't you combine them? If its not that then I don't know about it

16. UsukiDoll

one of the exponential rules is \[n^{a+b}=n^{a}n^{b}\] let n = x and a = 2 \[x^{2+b}\] what does b need to equal in order to get \[x^4 \]?

17. anonymous

2 if i'm not mistaking

18. UsukiDoll

yes because \[x^{2+2} =x^2x^2\]

19. UsukiDoll

so we need an x^2

20. UsukiDoll

|dw:1435149984739:dw|

21. anonymous

Okay. Inside of the sqrt there is a 2x^2

22. UsukiDoll

we have to go slowly.. getting rid of the x^4 and x^3 is a great idea at this point

23. anonymous

Alright so would we divide x^2+3x to x^4

24. UsukiDoll

we need to multiply (x^2+3x) by x^2

25. UsukiDoll

|dw:1435150150946:dw|

26. anonymous

Okay so its going to be x^4+3x^3

27. UsukiDoll

yeah...one problem we need to switch the signs.

28. UsukiDoll

29. UsukiDoll

let's just distribute the - -(x^4+3x^3) = -x^4-3x^3

30. UsukiDoll

|dw:1435150318360:dw|

31. UsukiDoll

oh the plus sign is invisible for long division. I'll just put it in () |dw:1435150362975:dw|

32. UsukiDoll

so what's x^4-x^4 and what's 4x^3-3x^3 ?

33. anonymous

Okay so if we are trying to change the sign wouldn't it be x^8 and x^6

34. UsukiDoll

whoa no...

35. UsukiDoll

you only use the exponent rule to figure out what's missing. Like what I did earlier. with the x^4 = x^{2+b}

36. anonymous

ok. sorry im not much of a math person. I love history

37. anonymous

Oh Okay.

38. UsukiDoll

|dw:1435150555870:dw| let's try this again what's x^4-x^4 and 4x^3-3x^3 ?

39. anonymous

x and x. right.

40. anonymous

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder? Here is the full question

41. UsukiDoll

...... let's make it easier I'm going to factor out the x^3 x^3(4-3)

42. UsukiDoll

well we need to do long division...this is the right method

43. UsukiDoll

or synthetic division which is faster but I forgot

44. UsukiDoll

x^4(1-1) x^3(4-3) try solving those two ^

45. anonymous

You forgot how to do synthetic division

46. UsukiDoll

I've practiced more with long division...it stuck in my head.

47. anonymous

Oh okay, and i got x^4-x^4 and 4x^3-3x^3

48. UsukiDoll

wow @_@!

49. UsukiDoll

try again... x^4-x^4 is the same as x^4(1-1) what's 1-1 ?

50. anonymous

sorry i suck at math.

51. anonymous

1-1=0

52. UsukiDoll

yes and what's x^4(0) also known as what's x^4 multiplied by 0

53. anonymous

0

54. anonymous

cause anything that is multiplyed by 0 is 0

55. UsukiDoll

yes! so the x^4's are gone! SImilarly what's x^3(4-3) ? what;s 4-3?!

56. UsukiDoll

what's 4-3?

57. anonymous

1

58. UsukiDoll

yes and x^3 times 1 is?

59. anonymous

x^3

60. UsukiDoll

yay!

61. anonymous

lol im getting it slowly but surely

62. UsukiDoll

|dw:1435151077510:dw| so after we solve that we bring the rest of the terms down

63. UsukiDoll

now our new goal is to get rid of the x^3

64. UsukiDoll

remember the exponent rule?

65. anonymous

na+b=nanb let n = x and a = 2 x2+b what does b need to equal in order to get x4 that one

66. UsukiDoll

\[x^{2+b} =x^2x^b\]

67. UsukiDoll

yup now what value of b is needed to produce x^3?

68. anonymous

1

69. UsukiDoll

yes.

70. anonymous

ok so it would be x^3-x^3

71. anonymous

which would get rid of each other

72. UsukiDoll

|dw:1435151215690:dw|

73. UsukiDoll

now this time x(x^2+3x) and switch signs after distributing

74. anonymous

ok so it is 4x^3

75. UsukiDoll

|dw:1435151316916:dw|

76. UsukiDoll

?!?!?!! |dw:1435151357797:dw|

77. anonymous

78. UsukiDoll

something is off.. are you sure the x^2 part of your original question is correct?

79. anonymous

yes it is ill post the whole question in here

80. anonymous

If we divide the polynomial x4 + 4x3 + 2x2 + x + 4 by x2 + 3x, what will be the remainder?

81. UsukiDoll

let's try continuing and see what happens..

82. anonymous

Alright

83. UsukiDoll

|dw:1435151602784:dw|

84. anonymous

so the x^3 would cancel and it would be -x^2

85. UsukiDoll

I see the question is asking for the remainder XD. we're not done xD we're doing it right heheeh so what's x^3-x^3 and 2x^2-3x^2 ... yes x^3 is gone and we will have -x^2

86. anonymous

so i did that part right

87. UsukiDoll

|dw:1435151714443:dw|

88. UsukiDoll

the last part is a digit, but it has to be negative

89. UsukiDoll

because when switching the sign it will be positive.. and I have a negative -x^2 right at me

90. anonymous

what do yo mean by that

91. UsukiDoll

we've reached the end of the problem.. after all the exponents are used up, it's down to digits. like 1 2 3 4 ...

92. UsukiDoll

there can only be 1 that will produce x^2

93. anonymous

ok so the only one that can be right is 4 then. cause i have 0 and 4 as a answer

94. UsukiDoll

0 is no remainder.. I doubt that will happen

95. anonymous

A) x2 + 3x B) 4x + 4 C) x2 + x – 2 D) 4 E) 0 Those are the answer

96. UsukiDoll

|dw:1435151939146:dw|

97. UsukiDoll

it's not going to be 0 . trust me.. I did the final step in my head.

98. UsukiDoll

quick question what's -1(x^2) ?

99. anonymous

-x^2 right

100. UsukiDoll

yeah.. noticed that I needed the -1 to produce the -x^2-3x but I need to switch signs because we already have a -x^2 present!

101. anonymous

ok so it becomes positive

102. UsukiDoll

|dw:1435152046028:dw|

103. UsukiDoll

yup you just need to compute x+3x and 4+0 that's your remainder

104. anonymous

SO it would be 4

105. UsukiDoll

oh I saw what happened I confused regular division with polynomial division. In polynomial division we can have bigger or smaller than the original numbers... can't have that in regular division xD (that was for the x^2 remark) you're missing x+3x!!!!!!!!

106. UsukiDoll

x(1+3) what's 1+3 ?

107. anonymous

4

108. UsukiDoll

yeah and 4(x) is ?

109. anonymous

4x

110. UsukiDoll

yeah with 4x and 4 combined together we have 4+4x

111. UsukiDoll

|dw:1435152315128:dw|

112. anonymous

Ok so thats the remainder

113. UsukiDoll

anything left over on the bottom is the remainder.. what aloud did was way off.

114. UsukiDoll

so that's done... yay x)