CPhill

P.S.....there may be other solutions for  "a"....I don't  know.....perhaps someone else knows another way to  obtain all of them,  if there are more...!!!!

(C)   To draw this, hoiuu....note that we have the  points   (0,0 ) ,  ( 2 , 120 )  and ( 3, 180)

Plot these points.....draw a line through them and stop  the left side  of the line at (0,0)

[ You can also plot (4, 240)  and (5, 300)  if you want to....your graph will "automatically" go through  them  !!!! ] 

All the other answers are good  !!!!

 (a+1)x^2 - (a^2 + 1)x + 2a^3 - 6 = 0

By inspection...let  a  =  0   and we have

x^2  - x   - 6  =  0     factoring, we get

(x - 3) ( x + 2)  = 0

And the two  roots  are  x  = 3  and x  = -2

Jessica.....

Anna can type at "only"  54  words  per  miniute  [  the slope of the equation, y  = 54x ]

"H"  ⇒   "minus "

If 1 is a root...let's perform  some synthetic division

1  [  1     a + 17    38   - a      - 56  ]

                     1      a + 18        56

   _________________________

      1     a + 18        56            0

The remaining polynomial is    

x^2  + (a + 18)x  +  56

Factors of 56  are  1   2   4    7   8   14   28   56

Since a  is positive...only the pairs  2, 28    and  1, 56   will give us what we want

So...the possibiliites for the factorization of this are

(x +28) ( x + 2) =  x^2  + 30x  + 56       so  a  =  12       or

( x + 56) ( x + 1)  = x^2  + 57x  +  56    so   a  = 39

Checking this

x^3 + 29x^2 + 26x  - 56  = 0  produces the three roots of 1, -28  and  -2

And

x^3  + 56x^2 -1x - 56  = 0   produces the three roots   1, -1, - 56

I'm assuming  that  Contractor 1 fees are modeled by

y = 200 + 50x

And we know that Contractor 2  fees are modeled by

y = 150  + 55x

(a)  Contractor 2 charges  $5 more per hour than Contractor  1

(c) The flat fee for Contractor 1  is $50 more than that for  Contractor  2

No prob, skye  !!!!

Don't see any  formulas. hoiuu  (???)

Yep...I can see it  now  !!!

And it is correct........