Polynomials (1 Viewer)

[bras91]

o please o please o please ....

If u think ur so good with polynomials please try these.

Try those arent circled for practice.

Try those that are circled if u think u are so talented coz i couldnt do over 3 days 2 nights.

But if u think u got an answer please post it here XD

Thank you sooooooooooooooooomuch

 

[Trebla]

I don't think I'm that talented in Polynomials lol....but I'll take a stab at Q13.
Q13
If multiple roots then:
P(x) = x⁵ - px² + q = 0
P'(x) = 5x⁴ - 2px = 0
=> x⁴ = 2px/5
=> x⁵ = 2px²/5
Sub into P(x):
2px²/5 - px² + q = 0
3px²/5 = q
x² = 5q/3p
x = | √(5q/3p) |
Sub into P'(x)
5(5q/3p)² - 2p| √(5q/3p) | = 0
125q²/9p² = 2p| √(5q/3p) |
Square both sides
15625q⁴/81p⁴ = 4p²(5q/3p)
15625q³/81p⁴ = 20p/3
15625q³ = 540p⁵
.: 3125q³ = 108⁵

 

[cutemouse]

lol, what's the likelihood of being asked those questions in an exam? :|

 

[bras91]

Very likely.

 

[wolfhunter2]

Stuck on Q.12, so close that i could see the finish line, but im stuck. :mad1:
Any help.

Here's what i have so far:

Ax^4 + x^3(-6A+B) + x^2(11A-3B) + x(-6A+2B+C+D) + E - C

therefore:
A =1
B =3
C = -D
E - C = -1

Is it even possible??!!

Thanks for the questions..

 

[bras91]

Stuck on Q.12, so close that i could see the finish line, but im stuck. :mad1:
Any help.

Here's what i have so far:
A =1
B =3
C = -D
E - C = -1

Thanks for the questions..

No problem for the question.
My friend and i cant do those questions too.
dammit couldnt help you

 

[SH33P]

polynomials = sad
thats the answer

 

[gurmies]

x^4 - 3x^3 + 2x^2 - 1 = Ax^4 - 6Ax^3 + 11Ax^2 - 6Ax + Bx^3 - 3Bx^2 + 2Bx + Cx - Dx + E - C

Equating the leading coefficients:

A = 1

x^4 - 3x^3 + 2x^2 - 1 = x^4 - 6x^3 + 11x^2 - 6x + Bx^3 - 3Bx^2 + 2Bx + Cx - Dx + E - C

Equating the cubic coefficients:

B - 6 = -3

B = 3

x^4 - 3x^3 + 2x^2 - 1 = x^4 - 3x^3 + 2x^2 + 12x + Cx - Dx + E - C

12 + C - D = 0

C-D = -12

If you sub in x = 0 and x = 1 you end up with:

D + E = -1 [1]

E - C = -1 [2]

Subtracting the two expressions

[1] - [2] produces D + C = 0

D = - C

So, 2C = -12

C = -6

D = 6

E = -7

Q12.
This is a similar method to finding partial fractions...
x⁴ - 3x³ + 2x² - 1 = Ax(x - 1)(x - 2)(x - 3) + Bx(x - 1)(x - 2) + Cx(x - 1) + Dx + E
Obviously A = 1 as it is the leading co-efficient
Let x = 0
.: E = - 1
Let x = 1
D + E = 1 - 3 + 2 - 1
D + E = - 1
D - 1 = - 1
.: D = 0
Let x = 2
C(2)(1) + 2D + E = 16 - 24 + 8 - 1
2C - 1 = - 1
.: C = 0
Let x = 3
B(3)(2)(1) + C(3)(2) + D(3) + E = 81 - 81 + 18 - 1
6B + 6C + 3D + E = 17
6B - 1 = 17
.: B = 3
Hence we have:
A = 1, B = 3, C = D = 0, E = - 1

Trebla, there is no Cx in the original question - just C(x-1)

 

[Trebla]

Yeh, my bad...I thought it followed a nice pattern lol.
Btw you can determine the coefficients of each power of x without having to expand it completely by using inspection...(a skill you develop more in Binomial Theorem)

 

[cutemouse]

Too bad that we're doing the binomial thereom last (ie. AFTER the trials). Would it help with other topics if I learn it now?

 

[cyl123]

Question 10
From x^3 +3x +9 = 0 where roots are a,b,c
First find the equation where roots are a^3, b^3, c^3:
So sub x^(1/3) for x in the original equation gives:

x + 3x^(1/3) + 9 =0
x+9 = -3x^(1/3)
Cubing both sides gives:
x^3 + 27x^2 +243x + 729 = -27x
x^3 + 27x^2 +270x + 729 = 0 <---- equation where roots are a^3, b^3, c^3

So thus a^6 + b^6 + c^6 = (a^3+ b^3+ c^3)^2 - 2(a^3b^3+ a^3c^3+c^3b^3)
= (-27)^2 - 2(270)
= 189

So sub a^3, b^3, c^3 into x^3 + 27x^2 +270x + 729 = 0 gives
a^9 + 27a^6 +270a^3 + 729 = 0
b^9 + 27b^6 +270b^3 + 729 = 0
c^9 + 27c^6 +270c^3 + 729 = 0

Adding them up gives
a^9 + b^9 +c^9 +27(a^6+b^6+c^6) +270(a^3+b^3+c^3) +3*729 = 0
thus a^9 + b^9 +c^9 +27(189) +270(-27) +2187= 0
Thus a^9 +b^9 +c^9 = 0

sorry about the crappy formating... there is prob an easier way lol.

 

[bored of sc]

Too bad that we're doing the binomial thereom last (ie. AFTER the trials). Would it help with other topics if I learn it now?

Complex Numbers has binomial expansions when you find expressions for cos5@, tan4@ etc.

 

[cyl123]

Question 9

Let a,b,c,d be root of x^4 - 8x^3+21x^2-20x+5=0 where a+b=c+d

So using sum of roots a+b+c+d = 8, so a+b = c+d = 4
From product of roots abcd = 5 .....(1)
From sum of roots 2 at a time:
ab +cd +ac+ad+bc+bd = 21
ab+cd + a(c+d) +b(c+d) =21
ab+cd+(a+b)(c+d) = 21
Since a+b = c+d = 4
then ab+cd = 21 - 16 =5
From (1) we have ab = 5/cd
so 5/cd +cd =5
(cd)^2 - 5cd +5 =0
Using quad eq gives cd = (5 + sqrt(5))/2 or (5 - sqrt(5))/2

Take cd = (5 + sqrt(5))/2 so clearly ab = (5 - sqrt(5))/2
Then using c+d =4 ----> c=4-d
gives d(4-d) = (5 + sqrt(5))/2
4d-d^2=(5 + sqrt(5))/2
Completing square gives
d^2-4d+4 = (-5 -sqrt(5))/2 +4
So (d-2)^2 = (3 - sqrt(5))/2
So d = sqrt[(3 - sqrt(5))/2] + 2 or -sqrt[(3 - sqrt(5))/2] + 2

Take d= sqrt[(3 - sqrt(5))/2] + 2 so clearly c= -sqrt[(3 - sqrt(5))/2] + 2
Using a similar technique to solve a and b gives
a= sqrt[(3 + sqrt(5))/2] + 2 and b = -sqrt[(3 + sqrt(5))/2] + 2

Please correct me if i made a mistake somewhere...I hope these are the right answers

 

[Michaelmoo]

Question 9

Let a,b,c,d be root of x^4 - 8x^3+21x^2-20x+5=0 where a+b=c+d

So using sum of roots a+b+c+d = 8, so a+b = c+d = 4
From product of roots abcd = 5 .....(1)
From sum of roots 2 at a time:
ab +cd +ac+ad+bc+bd = 21
ab+cd + a(c+d) +b(c+d) =21
ab+cd+(a+b)(c+d) = 21
Since a+b = c+d = 4
then ab+cd = 21 - 16 =5
From (1) we have ab = 5/cd
so 5/cd +cd =5
(cd)^2 - 5cd +5 =0
Using quad eq gives cd = (5 + sqrt(5))/2 or (5 - sqrt(5))/2

Take cd = (5 + sqrt(5))/2 so clearly ab = (5 - sqrt(5))/2
Then using c+d =4 ----> c=4-d
gives d(4-d) = (5 + sqrt(5))/2
4d-d^2=(5 + sqrt(5))/2
Completing square gives
d^2-4d+4 = (-5 -sqrt(5))/2 +4
So (d-2)^2 = (3 - sqrt(5))/2
So d = sqrt[(3 - sqrt(5))/2] + 2 or -sqrt[(3 - sqrt(5))/2] + 2

Take d= sqrt[(3 - sqrt(5))/2] + 2 so clearly c= -sqrt[(3 - sqrt(5))/2] + 2
Using a similar technique to solve a and b gives
a= sqrt[(3 + sqrt(5))/2] + 2 and b = -sqrt[(3 + sqrt(5))/2] + 2

Please correct me if i made a mistake somewhere...I hope these are the right answers

Yep. I did that question two and that was the fastest method I could come up with.

 

[bras91]

sorry does sqrt stand for square root?

 

[cyl123]

sqrt does stand for square root

use ab+cd =5
so if cd = (5+sqrt(5))/2 then ab = 5 - 5/2 - sqrt(5)/2 = (5-sqrt(5))/2

Note if you use abcd = 5 then you should get the same result for ab.
if ab = (5-sqrt(5)) then abcd = (5+sqrt(5))(5-sqrt(5)) /2 = (25-5)/2 = 10 which is clearly not the right result
On the otherhand if if ab = (5-sqrt(5))/2 then abcd = (5+sqrt(5))(5-sqrt(5)) /4 = (25-5)/4 = 5

(Also i meant clearly cos i have solved the quadratic equation using ab+cd =5 and abcd =5. So the quadratic has sum of roots = 5 and product of roots =5. hence the 2 roots from solving the quadratic should give ab and cd so if i make cd one of them, then ab must be the other)

 

[bras91]

o i c o i c thank you thank you