polynomial Q (1 Viewer)

[bobo123]

x^3+3x^2+2=0 has roots a, b, c

find the equation with roots : a+1/a, b+1/b, c+1/c

 

[spice girl]

The method of working backwards:

Assume the cubic with roots a+1/a, b+1/b, c+1/c has the equation Ax^3 + Bx^2 + Cx + D = 0

Now let x = y + 1/y (note that there are two values of y: if x=a+1/a, y=a or y=1/a both work)

Hence the expression:
A(y + 1/y)^3 + B(y + 1/y)^2 + C(y + 1/y) + D = 0 has 6 roots. These are a, b, c, 1/a, 1/b, 1/c.

Note that none of the roots equal zero (constant term of original cubic =/= 0)

So A(y + 1/y)^3 + B(y + 1/y)^2 + C(y + 1/y) + D = 0
becomes
Ay^3 + By^2 + (C+3A)y + (D+2B) + (C+3A)y^-1 + By^-2 + Ay^-3
= 0
which becomes:
Ay^6 + By^5 + (C+3A)y^4 + (D+2B)y^3 + (C+3A)y^2 + By + A
= 0

Now the polynomial with the 3 roots 1/a, 1/b, 1/c has coefficients reversed to that of the original, i.e. 2y^3 + 3^y + 1 = 0

hence the polynomial with the 6 roots a, b, c, 1/a, 1/b, 1/c is the polynomial:
(y^3+3y^2+2)(2y^3 + 3^y + 1) = 0
which becomes
2y^6 + 6y^5 + 3y^4 + 14y^3 + 3y^2 + 6y + 2 = 0

equating coefficients:
A = 2
B = 6
C + 3A = 3 => C = -3
D + 2B = 14 => D = 2

hence the polynomial with roots a+1/a, b+1/b, c+1/c is the polynomial 2x^3 + 6x^2 - 3x + 2 = 0

 

[spice girl]

if i were to get this question in an exam, i'd seriously skip it

 

[bobo123]

thankyou very much

and i thought it was just me with a mental block

 

[ezzy85]

its similar to the question on this post:

http://www.boredofstudies.org/community/showthread.php?threadid=5700

 

[spice girl]

Not quite the same

the other question is for roots (1-a)/a, (1-b)/b, (1-c)/c
= 1/a - 1, 1/b - 1, 1/c - 1

this basically boils down to doing it two steps
1) finding the polynomial roots 1/a, 1/b, 1/c
2) from that, shift the polynomial to the right by one unit. i.e. find the polynomial with all roots one less than that in polynomial (1)

a damn lot easier

 

[OLDMAN]

Interesting.

"Conventional" method is still useful :

y=x+1/x ...... (1)
x^2=xy-1 ..... (2)
sub into x^3+3x^2+2=0 gives
x^2y-x+3xy-1=0 ... (3)
sub (2) in (3)
xy^2-y-x+3xy-1=0, giving
x=(1+y)/(y^2+3y-1)
sub this x into (1),
y=(1+y)/(y^2+3y-1)+(y^2+3y-1)/(1+y)
Multiplying by (y^2+3y-1)((1+y) to get rid of denominators
y^4+4y^3+2y^2-y=y^4+6Y^3+8y^2-4y+2
giving 2y^3+6y^2-3y+2=0

Following "old fashioned" method is messy, but a serious 4u student should master it :

helpful hints and notation-

a=a+b+c
ab=ab+ac+bc
(a)^2=a^2+b^2+c^2+2ab .......3X3=9 terms
(a/b) = a/b+b/a+a/c+c/a+b/c+c/b ......6 terms(6 ordered pairs)
(a^2b)= a^2b+a^2c+b^2a+etc. ......6 terms
ab*a = 3abc+(a^2b) ......3X3=9terms "of degree 3"

Solution :
a=a+b+c=-3
ab=0
abc=-2

for sum of roots,
(a+1/a)=a+(1/a)=-3+(ab)/abc =-3
for sum of product-pairs,
[(a+1/a)(b+1/b)]=(ab+b/a+a/b+1/ab)=
ab+(1/ab)+(a/b) ...note that the last sum has 6 terms, ordered pairs a/b.
=0+(a)/abc+((a^2c))/abc
=0+3/2+[(ab)(a)-3abc]/abc=0+3/2-3=-3/2
for product of roots,
(a+1/a)(b+1/b)(c+1/c)=......this will have 2X2X2=8 terms
=abc+1/abc+(bc/a)+(a/bc) ..... 8 terms.
now (bc/a)=(b^2c^2/abc)=[(ab)^2-2(a^2bc)]/abc
=0-2abc(a)/abc=6
and (a/bc)=(a^2)/abc=[(a)^2-2(ab)]/abc=-9/2
hence product of roots is,
=-2-1/2+6-9/2=-1
Thus required equation is
x^3+3x^2-(3/2)X+1=0
or
2x^3+6x^2-3x+2=0

 

[OLDMAN]

The shortcut for the "sigma" notation didn't seem to work when I posted the above solution. Anyway, means the "sigma" sum symbol.