CPhill

1/x +  1/y  = 1/7   simplify

[x + y ] / xy  =  1/7

7[x + y]   = xy

7x + 7y - xy   = 0          rewrite as

xy - 7x - 7y   = 0        add 49 to both sides

xy - 7x - 7y +  49  =  49      factor

(x - 7) ( y - 7)   = 49       (1)

Note that the following  ordered pairs of integers would work :

(14, 14)

(8, 56)

(56, 8)

(-42, 6)

(6, -42)

Note that (0,0)   would  also satisfy (1), but  ......  x and y cannot be 0 in the original problem  [division by 0 ]

Here is a more "mathematical"  way of doing this......

2^27   +  3^27   =

[ (2⁹)³ + (3⁹)³ ]  =  

[ 2⁹ + 3⁹ ] [ 2¹⁸  - (2 * 3)⁹  + 3¹⁸ ]  =

[ (2³)³ + (3³)³ ]  [ 2¹⁸  - (2*3)⁹  + 3¹⁸ ]  =

[ 2³ + 3³] [ 2⁶ - (2 * 3)³ + 3⁶] [ 2¹⁸  - (2*3)⁹  + 3¹⁸ ]  =

[2 + 3] [ 2² - 6 + 3² ]   [ 2⁶ - (2 * 3)³ + 3⁶] [ 2¹⁸  - (2*3)⁹  + 3¹⁸ ]  =

[5] [ 13 - 6] [577] [377604937]  =

[ 5 ] [ 7 ]  [577] [377604937]  

So.......the sum of the two smallest distinct prime factors of  2^27 + 3^27    =    12

2^27 + 3^27   =  7625731702715  

This number is not divisible by 2

It is also not divisible by 3 because the sum of its digits is not divisible by 3

It is divisible by 5     because   it ends in  a "5" 

To test whether it is divisible by 7, double the last interger  and subtract the remaining part.....if the result is divisible by 7, then the number itself is divisible by 7....so we have.....

[2* 5 -  762573170271] / 7  = -108939024323

So.......5   and 7 are the smallest distinct prime divisors  of  7625731702715  

And their sum is 12

(a) Factor 3xy-4x^2-36y+48x.

(x  - 12)    (3y - 4x )

Yep.....I goofed this one ......40 lashes with a wet spaghetti noodle for me.........

This proves that CDD sometimes triumphs over reading comprehension........

 |x-1| + |x-5| = |x-9|         we can write this as

√[(x -1)^2] + √ [(x - 5)^2]   = √ [(x - 9)^2 ]         square both sides

(x - 1)^2 + 2√[ (x -5)^2 * (x -1)^2] + (x -5)^2   =  (x - 9)^2           simplify

x^2 - 2x + 1 + 2√[ (x -5)^2 * (x -1)^2]  + x^2 - 10x + 25   =  x^2 - 18x + 81

 2√[ (x -5)^2 * (x -1)^2]  =   -x^2 - 6x + 55

 2√[ (x -5)^2 * (x -1)^2]  =   - [x^2 + 6x - 55]   factor the right side

 2√[ (x -5)^2 * (x -1)^2] =  - (x + 11)(x - 5)        square both sides

4 (x - 5)^2 * (x - 1)^2   = (x + 11)^2 (x - 5)^2

4(x - 5)^2 * (x - 1)^2  - (x - 5)^2 (x + 11)^2   = 0         factor out (x - 5)^2

(x - 5)^2   [ 4(x - 1)^2 - (x + 11)^2 ]   =  0        simplify

(x - 5)^2 [ 4x^2 - 8x + 4  - x^2 - 22x - 121]   = 0

(x - 5)^2 [ 3x^2 - 30x - 117]  = 0

(x - 5)^2 [ x^2 - 10x - 39] * 3  = 0        divide through by 3   and factor

( x - 5)^2 [ (x - 13) ( x + 3) ]  =  0

So.........setting each factor to 0 we have that    x = 5  or x = 13   or x = -3

x = 13   is an extraneous solution produced by squaring both sides........reject it

The other two solutions x = 5   or x = -3      are good

See if this helps, Tony

                x^4 + 2x^3 + 2x^2 + 9x  + 10

x  - 2    [   x^5 + 0x^4 - 2x^ 3 + 5x^2 - 8x - 20 ]

                x^5  - 2x^4

                ----------------------------------------------

                         2x^4  - 2x^3

                         2x^4 -  4x^3

                        -----------------

                                    2x^3  + 5x^2

                                    2x^3  - 4x^2

                                    -----------------

                                               9x^2 -  8x

                                               9x^2 -18x

                                              --------------

                                                        10x  - 20

                                                        10x -  20

                                                        ------------

                                                                0

               x^3 + 2x  + 5

x + 2   [   x^4 + 2x^3 + 2x^2 + 9x  + 10 ]

               x^4 + 2x^3

              --------------------------------------

                                   2x^2 + 9x

                                  2x^2  + 4x

                                 ---------------

                                              5x  + 10  

                                              5x +  10

                                              ________

                                                      0

The remaining polynomial, x^3 + 2x + 5      doesn't have any rational roots........just one "real" root and 2 complex roots

If you have two rational roots, you shouldn't end up with any remainder......after dividing by (x  - 2)   and (x + 2), you should have either another polynomial [or 0]  as a remainder

Can you give us an example of what you have???

2. Factor 4x^2y^2-(x^2+y^2-z^2)^2 as the product of four polynomials of degree 1. 

4x^2y^2-(x^2+y^2-z^2)^2  =

[2xy + ( x^2 + y^2 - z^2) ]  [  2xy - (x^2 + y^2 - z^2)]  =

[[ x^2 + 2xy + y^2] - z^2 ] [ 2xy - x^2 -y^2 + z^2 ]  =

[ (x + y)^2 - z^2]  (-1)[ x^2 - 2xy + y^2 - z^2]  =

[ (x + y)^2 - z^2] (-1) [ ( x - y)^2 - z^2]  =

[ (x + y)^2 - z^2] [ z^2 - (x - y)^2]  =

[(x + y) + z ] [ (x + y - z] [ z + (x - y)] [ z - (x - y)]  =

[ x + y + z ] [ x + y - z ] [ x - y + z] [ -x + y + z ]

1. Factor p^2-2p+1-q^2-2qr-r^2 as the product of two polynomials of degree 1.

p^2-2p+1-q^2-2qr-r^2 a  =

(p - 1)^2 - (q  + r)^2  =

[ (p -1)  + (q + r) ] [ ( p - 1) - (q + r)]  =

[ p + q + r - 1]  [ p - q - r - 1 ]