## Mimi_x3 4 years ago Can you please help me with this maths question 1. The cubic polynomial P(x)= x^3 +rs^2 +sx+t , where r,s and t are real numbers has three real zeroes,1,a, and -a (a) find the value of s + t

1. myininaya

is the polynomial written correctly?

2. Mimi_x3

woops my bad sorry P(x) = x^3 +rx^2 +sx +t

3. helpme1234

Colorful pi can you help me?

4. myininaya

let me try to help him and then i will look at your question

5. Mimi_x3

him ? are yu referring to me ?

6. myininaya

is this algebra?

7. Mimi_x3

noo , its a polynomial question

8. myininaya

so what class is it then?

9. Mimi_x3

class ? what do yu mean ? Its from the topic polynomials

10. myininaya

$P(x)=(x-1)(x-a)(x+a)=x^3+rs^2+sx+t$

11. Mimi_x3

12. myininaya

so I guess you are doing it for fun? $P(x)=(x-1)(x^2-a^2)=x^3+rs^2+sx+t$ no i just thought u would like to see the steps

13. Mimi_x3

no , im not doing it for fun , its my maths assignment LOL

14. myininaya

$P(x)=x(x^2-a^2)-1(x^2-a^2)=x^3+rsx^2+sx+t$

15. myininaya

so the class is called polynomials lol?

16. Mimi_x3

what does yu mean by class ?

17. myininaya

$P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t$ people go to class to learn and get assignments lol

18. Mimi_x3

no the class is called maths LOL

19. myininaya

$P(x)=x^3-x^2-a^2x+a^2=x^3+rsx^2+sx+t$

20. Mimi_x3

2 unit maths to be specific

21. myininaya

ok we are almost done with this problem can you guess what to do next?

22. myininaya

we want both sides to be the same

23. Mimi_x3

collecting like terms ?

24. myininaya

ok the the coefficient of x^2 on left hand side is -1 the coefficient of x^2 on the other sides is rs so what can you say about rs and -1?

25. myininaya

they must be equal

26. myininaya

so we need that for the other terms as well

27. myininaya

we have -1=rs -a^2=s a^2=t

28. polpak

I seriously think I'm being trolled here.. but just to ease my tired brain.. $\sqrt{25} \ne -5$

29. myininaya

you are right polpak

30. Mimi_x3

wait , dnt yu factorise it further to (x^2-a^2)(x-1)

31. myininaya

that =5

32. myininaya

factor?

33. polpak

Someone was trying to tell me that it equals both 5 and -5

34. myininaya

ok the objective is to find s+t

35. polpak

But I can only really prove it by going to the definition. Which they don't accept.

36. Mimi_x3

37. polpak

Sorry for bothering you mimi, but myin rarely watches group chat (with good reason)

38. myininaya

what do you think s+t is based on what i gave you above? hint we don't need that first equation rs=-1

39. Mimi_x3

yu talking to me or polpak ? LOL

40. myininaya

you

41. Mimi_x3

lols , tbh im lost , i dont get it xDD

42. myininaya

do you understand when I wrote P as P(x)=(x-1)(x-a)(x+a)?

43. Mimi_x3

yeah , i dont understand the part -1 = rs -a^2 = s a^2 = t where did yu get it from ?

44. myininaya

both sides had to be equal we were looking at $P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t$

45. myininaya

look at the coefficients of each term on both sides we want each term to be the same as it is on the other side

46. myininaya

so if we look at the x^3's they are good they both have coefficient 1 what is the coefficient of x^2 on left hand side ? what is the coefficient of x^2 on right hand side?

47. Mimi_x3

ohh i get it rs = -1

48. myininaya

yes now what is the coefficient of x on the left hand side? what is the coefficient of x on the right hand side?

49. Mimi_x3

x on the left hand side: 1 i think x on the right hand side 1 i think

50. myininaya

no what is the number in front of x on boths?

51. myininaya

both sides*

52. Mimi_x3

there's no other number

53. myininaya

-a^2? s?

54. Mimi_x3

lols , its a letter not a number

55. myininaya

they are numbers

56. Mimi_x3

they are ? arent they are called pronumerals ?

57. Mimi_x3

are they*

58. myininaya

never heard of this term

59. Mimi_x3

ohh ok

60. myininaya

when you are doing math and you don't see a word and you see a letter it usually represents a number unless it is the word a or i

61. myininaya

becauce a and i can be words

62. Mimi_x3

ohh ok , i never knew that they are numbers xD

63. myininaya

but usually the book will make it clear to you what they are

64. myininaya

yep

65. Mimi_x3

ok , so x= a^3

66. myininaya

a^3 is nowhere to be found we aren;t solving equation for x

67. Mimi_x3

ok , so x= a^3

68. Mimi_x3

ohh ok

69. myininaya

ok so look again at the polynomial what is the coefficent of x on both sides?

70. Mimi_x3

ok , so x= a^3

71. Mimi_x3

ok , so x= a^3

72. Mimi_x3

x= s

73. myininaya

ok no $P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t$ ^ ^ | | -a^2 = s

74. Mimi_x3

ok , so x= a^3

75. Mimi_x3

ohh lols , sorry im so slow of understanding it

76. myininaya

ok do you see the terms that don't an x in em? there terms are called constants and they must be equal of that we have the same polynomial on both sides

77. myininaya

god my english is getting bad

78. Mimi_x3

ok , so x= a^3

79. Mimi_x3

lols , probably getting annoyed of me not getting it xD

80. myininaya

no its just late here i been up for along time today

81. myininaya

a long*

82. polpak

Ditto!

83. myininaya

so what are the constants on both sides?

84. myininaya

the terms that don't have a x in them

85. Mimi_x3

ok , so x= a^3

86. Mimi_x3

t

87. Mimi_x3

ok , so x= a^3

88. myininaya

t and what is it on the other side?

89. myininaya

remember you are looking for non-x terms

90. Mimi_x3

ok , so x= a^3

91. Mimi_x3

a

92. myininaya

a or a^2?

93. Mimi_x3

a

94. myininaya

$P(x)=x^3-a^2x-x^2+a^2=x^3+rsx^2+sx+t$ ^ ^ | | a^2 = t

95. Mimi_x3

96. myininaya

so what is s again? and what is t again?

97. Mimi_x3

s = 1 t = a^2

98. myininaya

s is not 1

99. Mimi_x3

a^3

100. myininaya

not a^3 scroll up we have it above

101. Mimi_x3

-a^2

102. Mimi_x3

lolss , im so~ slow

103. myininaya

YES! :)

104. Mimi_x3

lols , finally i get it

105. myininaya

ok now s+t is -a^2+a^2 right?

106. Mimi_x3

yeah

107. myininaya

but what is -a^2+a^2?

108. Mimi_x3

so substitute it in s and t

109. Mimi_x3

idk

110. myininaya

-5+5 is

111. myininaya

-6+6 is -24+24 is -999+999 is

112. Mimi_x3

where did 5 come from ?

113. myininaya

114. Mimi_x3

0 ?

115. myininaya

yes! :)

116. myininaya

and now you are done

117. Mimi_x3

really ? omg ! yay i got it right xD

118. Mimi_x3

thank you very mcuh for helping (:

119. myininaya

np