## anonymous one year ago given the function f(x)=4x^8+3x^6+2x^4+2 what is the value of f(1)

1. UsukiDoll

f(x) =f(1) means that our x =1 so we plug in x =1 throughout the function $\LARGE f(x) =4x^8+3x^6+2x^4+2$

2. UsukiDoll

so at f(1) our function should look like this $\LARGE f(1) =4(1)^8+3(1)^6+2(1)^4+2$

3. anonymous

4. UsukiDoll

YEAH 1 to any exponent number will just be a number so $\LARGE f(1) =4+3+2+2 =11$

5. anonymous

could you help me with a few more please ?

6. UsukiDoll

ack one to any exponent number will just be 1 so then we have 4(1) which is 4 3(1) which is 3 2(1) which is 2 si 4+3+2+2 is 11. I could, but I am on a netbook, and it's a bit buggy, so you'll have to be a bit patient with me.

7. anonymous

thats fine , im a very patient person ! Given the function f(x)=x3−8x2+21x−18 what are the factors of of f(x) ?

8. UsukiDoll

$\LARGE f(x) =x^3-8x^2+21x-18$ like this?

9. anonymous

yes

10. UsukiDoll

hmmm...I wish I could use factor by grouping...

11. anonymous

i wish i was good at algebra :(

12. UsukiDoll

ok factor by grouping is out of the question. I got to use something that I haven't done in years. Using the rational root theorem.

13. anonymous

okay?

14. UsukiDoll

it's like setting that problem to 0 and breaking this function into at least 3 parts

15. UsukiDoll

$\LARGE 0 =x^3-8x^2+21x-18$

16. UsukiDoll

sigh... ok... we have to find all factors of 18 which is 1 2 3 6 9 18 and each of them has + or - signs so we got a bunch of choices to choose from 1 2 3 6 9 18 -1 -2 -3 -6 -9 -18

17. UsukiDoll

so we need to plug those values in until we get our 0... so we need to let x =2, x = 3

18. UsukiDoll

$\LARGE f(2) =(2)^3-8(2)^2+21(2)-18$ can you evaluate this function?

19. UsukiDoll

the reason why I ask this is that we need an x that can produce a 0 and then afterwards we have to use the Factor theorem which states that if p/q is that root in the polynomial then it can be divided by qx-p

20. anonymous

i think so , is in -2?

21. UsukiDoll

shouldn't be -2.

22. anonymous

0?

23. UsukiDoll

yes so now that we got our x or in that theorem p/q which is 2/1 (yeah I forgot to write the roots in fraction form), Then it's possible to divide that original function with x-2. That would give us the second polynomial we need

24. UsukiDoll

|dw:1437005051966:dw| sorry it's hard to draw on a touchpad mouse

25. UsukiDoll

so now we need to use long division . so let's start with |dw:1437005170681:dw| that portion only.

26. UsukiDoll

what we need to do is to get rid of that x^3 so I have to multiply a variable with x-2 so which variable do we need?

27. UsukiDoll

what produces x^3 $\large x(x^{?}) \rightarrow x^{1+?}$

28. UsukiDoll

similar to solveing 1 + x =3

29. anonymous

2

30. anonymous

a. (x - 3)(x + 2)(x + 3) b.. (x + 3)(x + 2)(x + 3) c. (x + 3)(x - 2)(x - 3) d. (x - 3)(x - 2)(x - 3) these are the answer choices !!

31. UsukiDoll

yes so we need to distribute (multiply) x^2 x^2(x-2) which is x^3-2x^2 however we need to switch the signs when dividing with polynomials so -x^3+2x^2

32. UsukiDoll

|dw:1437005603335:dw|

33. UsukiDoll

|dw:1437005651594:dw|

34. UsukiDoll

ok... so now I need to get rid of the -6x^2 and I have (x-2) so which choice do we need. If I can get a copycat version of -6x^2 that would be great.

35. UsukiDoll

|dw:1437005826577:dw|

36. UsukiDoll

so right now I have x -2 what do I need to distribute to have -6x^2? I need to have at least one x

37. anonymous

i dont know ?

38. UsukiDoll

[?] (x-2) well we need to produce a -6x^2 so we need at least one x but what number do we need (and it has to be negative because we already have a negative number inside the divison and we want that to go) ?x(x-2)

39. anonymous

-3?

40. UsukiDoll

no I would have -3x^2 and our goal is to get rid of the -6x^2 similar to -7+7 =0

41. anonymous

i dont know :(

42. UsukiDoll

well there is a -6 in the division. ?x(x-2) perhaps using that number -6 would work -6x(x-2) try distribute and tell me if you can get a -6x^2 ?

43. UsukiDoll

we also need a x^2 so $x \cdot x =x^{1+1}$ what's 1+1?

44. anonymous

2

45. UsukiDoll

yes so we have our x^2 x(x-2) when we distribute... but we also need a number so we can create -6x^2 as well... if we have -6 already in the division, then we need that same number -6x(x-2) so using the distributive property -6x^2+12x switching the signs due to dividing polynomials 6x^2-12x

46. UsukiDoll

|dw:1437006719993:dw|

47. UsukiDoll

at this point since all variables are used up, we just need a number

48. UsukiDoll

?(x-2) what number do I need to produce a 9x?

49. anonymous

11? or 3?

50. UsukiDoll

no.... 9 times ? = 9x

51. UsukiDoll

OH MY I just typed it XD

52. UsukiDoll

we need a 9 9(x-2) so that's 9x-18 switching the signs -9x+18

53. UsukiDoll

|dw:1437007066892:dw|

54. UsukiDoll

so our remainder is 10 but that's not important. What's important is that we have found the second polynomial which is x^2-6x+9 so from the rational root test we had x-2 (x-2)(x^2-6x+9) the second polynomial is obtained through long division can you factor (x^2-6x+9) ?

55. anonymous

(x-3)^2

56. UsukiDoll

nice! (x-2)(x-3)^2 or (x-2)(x-3)(x-3) now we can solve for x (x-2)(x-3)(x-3) =0 since one factor is repeating we just have two cases x-3=0 and x-2=0 what is our x values for those 2 equations?

57. anonymous

3,2?

58. UsukiDoll

yes 3 and 2 so that is our roots.

59. anonymous

so the answers ( x - 3)(x - 2)(x - 3)

60. anonymous

can you help me with 4 or 5 more that is much easier than that one ? 1.A polynomial is to be constructed that has 5 turning points. What is the minimum degree of the polynomial?

61. UsukiDoll

yes it is (x-3)(x-3)(x-2).... ummm I really don't know how that next questino would work. .__.

62. anonymous

okay what about 2.Given the polynomial f(x)=6x5−4x3+4x+10 What one of the following is a possible rational root?

63. UsukiDoll

we have to do the rational root test again so we find all factors of 10

64. UsukiDoll

so all factors of 10 is 1 2 5 10 but we also have to consider the negative versions as well so -1, -2, -5, -10 b

65. UsukiDoll

try -1 and see if it works $\LARGE f(-1)=6(-1)^5-4(-1)^3+4(-1)+10$

66. UsukiDoll

if our result isn't 0, we need another root...

67. anonymous

i got 4

68. UsukiDoll

yeah time to toss that one out

69. UsukiDoll

so 1 and -1 are fails. try 2 and -2

70. anonymous

it wasnt zero there either

71. UsukiDoll

5 and -5

72. anonymous

nope

73. UsukiDoll

then the last two choices are -10 , 10 either one of them works or a math error happened. I do know that 1,-1 don't work at all

74. anonymous

that didnt work either !

75. UsukiDoll

right... none of the roots worked which means there aren't any with the rational root test.

76. anonymous

okay lets just skip that one , i have two more . Given the following polynomial find the maximum possible number of turning points. f(x)=x8+x14+x6+x13

77. UsukiDoll

well the answer to that previous question should be none because the rational root test failed and we have proven that by finding all roots and testing them out. And, again I don't know what a turning point is

78. anonymous

okay then , one more?Given the function f(x)=x3+10x2+31x+30 what one the following is a factor of f(x) ?

79. UsukiDoll

the rational root test is used again and this time it's going to work all factors of 30 1 2 3 5 6 10 15 30 -1 -2 -3 -5 -6 -10 -15 -30

80. UsukiDoll

try x=-2

81. UsukiDoll

$\LARGE f(-2) = (-2)^3+10(-2)^2+31(-2)+30$

82. anonymous

i got 0!

83. UsukiDoll

woo HOO! which means we an divide that polynomial by qx-p from the Factor theorem which is p/q=-2/1 x-(-2) = x+2

84. UsukiDoll

|dw:1437009518274:dw|

85. UsukiDoll

so we need to produce an x^3 but we have already done that earlier x^2(x+2) x^3+2x^2 switching signs -x^3-2x^2

86. UsukiDoll

|dw:1437009682298:dw|

87. UsukiDoll

so now we need an 8x^2 so we need an 8x 8x(x+2) =8x^2+16x switching signs -8x^2-16x

88. UsukiDoll

|dw:1437009861228:dw|

89. UsukiDoll

well this works out nicely .. if we use the number 15 everything cancels out and there is no remainder so we have found our second polynomial (x+2)(x^2+8x+15)

90. UsukiDoll

can you factor x^2+8x+15?

91. anonymous

(x+3) (x+5)

92. UsukiDoll

mhm so now we have (x+2)(x+3)(x+5) we can solve for x to obtain the roots x+2=0 x+3=0 x+5=0

93. anonymous

the answer was just (x+2) ! THANK YOU , you're a life saver!!

94. UsukiDoll

you're welcome :)