CPhill

Simplify as

x^4 + 2x^3 + 16x^2 + 20x  - 31

(pqr  + pqs + prs + qrs)  =  -20

pqrs = -31

1/p  + 1/q + 1/r + 1/s =  

[ pqr + pqs + prs + qrs ]  /  pqrs  =    -20  / -31   =  20 / 31

2mn  - 18m + 7n - 3mn + 15m  + 8n   + _____  simplify

-mn - 3m   + 15n +  ____

Take the product of the absolute value of the coefficients  on m, n  =   3 * 15 =  45

This is the added constant

The factrorization  is  

(-m  + 15) (n + 3)  =  -mn -3m + 15n  +  45

r + s =  2m  

(r + s)^2 = 4m^2 =  r^2  + s^2  + 2rs  →   r^2 + s^2  =  4m^2 - 2rs

And

rs  = (m^2 - 6m + 11)

So

(r - s)^2  =  (r^2 + s^2) - 2rs  =   (4m^2 - 2rs) - 2rs  =  4m^2 - 4rs =  4m^2 - 4(m^2 -6m + 11)  =

24m - 44

24m  - 44   has no min value

https://web2.0calc.com/questions/geometry_63597

Boris's speed  =    Distance walked /  Time  = [75 - 30 ] / [ 5 - 0 ]  =  45 / 5 =  9 km / hr 

\(\frac{1}{r - 2s - 2t} + \frac{1}{s - 2r - 2t} + \frac{1}{t - 2r - 2s} \)

r + s + t =  -2                   

2r + 2s + 2t  =  -4

2r =  -4 - 2s - 2t

2r + 4 = -2 -2t

And

2s + 4 = -2r -2t

2t + 4 =  -2r - 2s

And

rs + rt + st  = -5

rst  = -15

Simplifying, we have

          1               +        1         +       1

_________               ______        ______  =

     3r + 4                    3s +4          3t  + 4

(3s + 4) + (3t + 4)  +  ( 3s + 4)

________________________  =

(3r + 4) (3s + 4) (3t + 4)

            3(r + s + t)  + 12

____________________________________   =

27rst  + 36 (rs + rt + st)  + 48 ( r + s + t)  + 64

     3 (-2)  +  12

________________________    =

27(-15) + 36(-5) + 48(-2) + 64

               -   6

             _______

                 617

\(\frac{rs}{t^2} + \frac{rt}{s^2} + \frac{st}{r^2} \)

Simplify as

rs (rs)^2  + rt (rt)^2  + st (st)^2

________________________       =

            (rst)^2

(rs)^3  + (rt)^3   +  (st)^3

_____________________

           (rst)^2

By Viete

r + s + t =  4

(rs + rt + st)  = -7

(rst)  =  -12

And

(rs + rt + st)^3  =  (-7)^3 = -343 

( rs + rt + st)^3 =

[ (rs)^3 + (rt)^3  + (st)^3 ]  +   3rst [ r^2s + r^2t] + 3rst [ s^2r + s^2t] + 3rst [ t^2r + t^2s] + 6(rst)^2  = 

[ (rs)^3 + (rt)^3 + (st)^3 ]  +   3(-12) [ r ( rs + rt)] + 3(-12) [ s (rs + st)] + 3(-12) [ t(rt + st] + 6(-12)^2  =

[ (rs)^3 + (rt)^3 + (st)^3 ]  -  36 [ r(rs + rt) + s(rs + st)  + t(rt + st) ] + 864

Note: [ rs + rt + st] = -7

So 

rs + rt = -7 - st

rs + st  = -7 - rt

rt + st =  -7 - rs

So

 [(rs)^3 + (rt)^3 + (st)^3 ] -  36 ( r [ -7 - st]  + s [-7 - rt] + t [ -7 -rs] )  + 864  =   -343

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36 ( -7 [ r + s + t ] - 3 (rst) ]  =   -1207

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36 [ -7(4) - 3(-12) ]  =  -1207

[ (rs)^3 + (rt)^3 + (st)^3 ] - 36 [ -28 + 36 ]  =  -1207

[ (rs)^3 + (rt)^3 + (st)^3 ]  - 36(8)   =-1207

[ (rs)^3 + (rt)^3 + (st)^3] =  - 288 = -1207

[ (rs)^3 + (rt)^3 + (st)^3]  = -919

So

rs       +     rt      +     st

__           ___          ___  =

r^2           s^2          r^2

[ rs (rs)^2  + rt(rt)^2 + st(st)^2]

________________________   =

       (rst)^2

[ (rs)^3 + (rt)^3 + (st)^3]

___________________  =

       (rst)^2

-919 

____

 144

Simplify as

x^4 + 2x^3 + 16x^2 + 20x - 31

By Viete :

p + q + r + s  =  -2

 (pr + pq + ps + qr + qs + rs) = 16

So

( p + q + r + s)^2  =     (-2)^2  =  4  =

(p^2 + q^2 + r^2 + s^2)  +   2(pr + pq + ps + qr + qs + rs)  

And

2 (pr + pq + ps + qr + qs + rs) =  2(16) = 32

So

( p + q + r + s)^2  =     

(p^2 + q^2 + r^2 + s^2)  +   2(pr + pq + ps + qr + qs + rs)  =

(p^2 + q^2 + r^2 + s^2) +  32  =   4

(p^2 + q^2 + r^2 + s^2) =   -28

See the following

Let D=(0,0)  C = (4,0)  B  (4,3)  A   = (0,3)   and M = (2,3)

AM = 2

AD = 3

And  triangle AMD  is right with hypotenuse MD

By Pythagoras  MD  =  sqrt [ AM^2 + AD^2 ] =  sqrt [ 2^2 + 3^2] = sqrt [13]  = MX = MC

Since MDX = 77° and MD = MX , then angle MXD also = 77°

And in triangle MDX, angle DMX =  180 - 77 - 77 =  26°

Construct a circle at M with a radius = MD  = MX  = MC

Angle  DMX   is a  central angle in the circle  intercepting the minor arc DX

But angle XCD is an inscribed angle in the  same  circle and intercepts the same  minor arc

So, angle XCD =  (1/2)measure  of  central angle DMX   = (1/2) (26°)   = 13°

3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.

Simplify  as

x^4 + 2x^3 + 16x^2  + 20x - 31

Note that

p^2 qrs + pq^2 rs + pqr^2 s + pqrs^2  can be written as

pqrs  ( p + q + r + s)

The sum of the roots  =  -2/1   =  -2

The product of the roots   -31/1   =   -31

So

-31 (-2)  =  62