2011/064) Solve 6x^5-41x^4+97x^3-97x^2 + 41x – 6 = 0

This as the coefficients are afternates signs and symmetric so they can be combined as below

6(x^5-1) – 41(x^4- x) + 97(x^3-1)
= 6(x-1)(x^4 + x^3 + x^2 + x + 1) – 41x(x-1)(x^2+x + 1) + 97x^2(x-1)
= (x-1)( 6x^4 + 6x^3 + 6x ^2 + 6x + 1 – 41x^3- 41 x^2 – 41 + 97 x^2 )
= (x-1)(6x^4 – 35x^3 + 62x^3 – 35 + 6)
So x= 1 is a solution and other factor is
(6x^4 – 35x^3 + 62x^2 – 35 + 6)
As the coefficients as symmetric we have combining similarly

6(x^4+1) – 35x(x^2+1) + 62 x^2 ...1
Taking x^2 out we get symmetric

X^2(6(x^2+ 1/x^2) - 35(x+1/x) + 62)

Put x+ 1/x = t o get
X^2(6(t^2-2) - 35 t + 62) = 0 and as x is not zero we get
(6(t^2-2) - 35 t + 62 = 0
Or 6t^2 - 35 t + 50 = 0
Ot (2t-5) (3t – 10) = 0
t= 5/2 or 10/3
now x + 1/x = 5/2 gives x ^2 + 1- 5/2 x= 0 or (4x^2 + 4x- 5) = 0 or (2x-1)(x-2) or x = ½ or 2
x + 1/x = 10/3 gives x ^2 + 1- 10/3 x= 0 or (3x^2 + 3x- 10) = 0 or (3x-1)(x-3) or x = 1/3 or 3

so solution = ½ , 1/3 , 1 , 2 , 3

you can also proceed from 1 putting x^2+1 = tx

2011/063) factor x^5 + x^4 + 1

This is a qunitic polynomial for which there is no direct method
(
one of the methods commonly used is given below at the end but a short cut can be used

(since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

as it is of the form x^(3n+2) + x^(3m+1) + 1 so
x = w and x = w^2 are zeros ( w is cube root of 1)
(x-w)(x-w^2) or (x^2 + x + 1) is a factor and by division

x^5 + x^4 + 1 = (x^2+x+1)(x³ - x + 1)


The common method
It's easy to check that x^5 + x^4 + 1 has no linear factors (since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

So, if x^5 + x^4 + 1 factors, then it must be the product of a quadratic factor and a cubic factor.
Hence, we assume that x^5 + x^4 + 1 = (x² + Ax + B)(x³ + Cx² + Dx + E).

Expanding yields
x^5 + x^4 + 1 = x^5 + (A + C)x^4 + (B + AC + D)x³ + (AD + BC + E)x² + (BD + AE)x + BE

Equate like coefficients:
A + C = 1
B + AC + D = 0
AD + BC + E = 0
BD + AE = 0
BE = 1.

Note that C = 1 - A and E = 1/B.
We can say more; since these variables should be integers, we know that B = E = ±1.

Thus, we obtain
±1 + A(1 - A) + D = 0
AD ± (1 - A) ± 1 = 0
±D ± A = 0

Let's look at the +1 case (it turns out that we don't need the -1 case...)
1 + A(1 - A) + D = 0
AD + (1 - A) + 1 = 0
D + A = 0

Since D = -A, we have
1 + A(1 - A) - A = 0 ==> A^2 - 1 = 0
-A^2 + (1 - A) + 1 = 0 ==> A^2 + A - 2 = 0.
These two equations are both true when A = 1.

Hence assuming that B = E = 1, we have A = 1, D = -1 ==> C = 0.
------------------
So, we have the factorization x^5 + x^4 + 1 = (x² + x + 1)(x³ - x + 1).)

2011/062) The sum of two natural numbers when added to their LCM, gives a total of 143.? How many such pairs of numbers exis

let the numbers be mx and my with x and y co prime

so LCM = mxy

so mx + my + mxy = 143

or m(x+y+xy) = 143

add m on both sides

m(x+y+xy+1) = 143+m

m(x+1)(y+1) = 143 +m or (x+1)(y+1) = 143/m + 1

so m is factor or 143 can be 1,11,13, 143

m = 1=> (x+1)(y+1) = 144
144 = 2 * 72( x=1 y = 71)
= 3 * 48 (2,47)
= 4 * 36(3,35)
= 6 * 24( 5,23)
= 8 * 18 (7,17)
= 9 * 16(8,15)
= 12 * 12 (11,11)
all except (11,11) are coprimes so solution

m = 11 => (x+1)(y+1) = 14 , x + 1 = 2, x = 1 and y+1 = 7 y = 6 so 11 and 66
m = 13 => (x+1)(y+1) = 12 (x=1, y = 5), (x=2, y = 3) so 13, 65 or 26,39
m = 143 => (x+1)(y+1) = 2 so no solution

so 9 solutions (1,71),(5,23), (8,15), (7,17),(3,35), (2,47) (11,66), (13,65), (26,39)

2011/061) if x = 0.666... (base 8) and y = 0.555... (base 6), what is the value of y - x in base 7?

we need to compute x and y say in base 10

x = .6666 base 8
= (6/8)( 1 + (1/8)^2 + (1/8)^2 + ....
= (6/8) /(1-(1/8)) = (6/7)

y = .55555 (base 6) = 5/6( 1+ 1/6 + (1/6)^2 ....) = (5/6)/(1-1/6) = 1
hence
y - x = 1- 6/7 = 1/7 and hence .1 in base 7

2011/060)If a^2=5a-3, b^2=5b-3, then equation having a/b and b/a as its roots is___?

a is not equal to b
A. 3x^2+19x+3
B. 3x^2-19x+3
C. 3x^2-19x-3
D. x^2-16x+1
Ans:
since a^2=5a-3, b^2=5b-3
a,b are soln's for the eqn. x^2=5x-3
ie x^2-5x+3=0
and they are not same so they are 2 roots of x^2-5x+3=0

hence sum of roots =a+b= -(-5/1) =5
and product of roots=ab=+( 3/1)=3

now if a/b and b/a are roots then

sum of the roots is
a/b + b/a = a^2 + b^2 /ab =(a+b)^2 -2ab /ab =5^2 -2*3 /3 =25-6 /3=19/3

and product is a/b *b/a =1

so eqn wth roots a/b&*b/a is
x^2 -(19/3)x +1 =0
=>3x^2-19x+3=0

hence the ans is B