CPhill

Correct,RP!!!...good job  !!!

[Just a note...remember to write the remainder as   -1 /[9n + 4]....what you have is  -1/9n + 4...which translates to :

(-1/9) n   +  4 .... ]

Otherwise....very nice!!!

Different roads....same answer....good job  !!!

[You divided the polynomial by 8 right away.....and realized that the  "7"  in the denominator of the "remainder fraction" represented the true remainder....if you like that method better, stick with it...I just learned it a different way....at least you understand these, now....that's the important thing  !!! ] 

That's OK, tertre.....synthetic division is a little bit tricky in this kind of problem when the linear divisor  is in the form ax +/- b    !!!

Set 8x  - 5  = 0  ⇒    x =  5/8....this is what we need to divide by

Note that the polynomial is really   80x^3  - 50x^2  + 0x   + 7

5/8  [  80      -50       0        7  ]

                     50       0        0

         _____________________

         80         0        0         7

The remainder is correct

The apparent remaining polynomial is  80x^2

Note....RP...that all we really need to do to find the correct remaining polynomial is just to divide  the apparent remaining polynomial  by   the "a"  coefiicient  of the  divisor, ax - b.....in this case.....ax - b  = 8x  - 5...so we can divide

80x^2  by   8  = 10x^2  and this is the correct residual polynomial

So...the answer   10x^2 R [ 7 /(8x - 5) ] 

Does that help????

b)  sin2x  = cos (pi/2  - x)     

Note   cos(pi/2 - x)   =  sin x

So we have

sin2x  = sinx

2sinxcosx  =  sinx

2sinx cosx  - sinx  = 0

sinx ( 2cosx - 1)  = 0

So...either   sin x  = 0    and this happens  at    0,  pi  and 2pi

Or

2cosx - 1  = 0

cosx  =  1/2   and this happens at     pi/3   and 5pi/3

Here's a graph :  https://www.desmos.com/calculator/y1o4mppidr

a) 4sin x  =  cosx        square both sides

16sin^2x  =  cos^2x

16sin^2x  = 1 -  sin^2x

17sin^2x  = 1

sin^2x   = 1/17               take both roots

sinx  = ±√ (1/17)

To find both values in degrees...take the sine inverse

arcsin (√[1/17])  ≈  14.04°

arcsin (-√ [ 1/17] )  ≈ -14.04°.......however.....we must   subtract this from -180° to get the correct answer

-180 - - 14.04  ≈  -165.96°

Here's a graph that shows the intersection of these curves : https://www.desmos.com/calculator/c0dyooszqg

Excellent....!!!!

Are you beginning to see this  ???

Again...set   2r + 1  = 0   ⇒   r =  -1/2  ....this is what we need to divide by

 [ note.....we need the  "placeholder" 0   to denote that there is no r^2 term...!!!]

-1/2  [  2      1       0       20      12    ]    

                   -1      0        0       -10

        ________________________

          2      0       0        20       2

The remainder,2,  is correct....

The apparent remaining polynomial is  2r^3  + 20

If we performed polynomial division....the first term would be r^3...as follows

               r^3

2r + 1  [ 2r^4

             -(2r^4)

            _________

So...this indicates that we need to divide each term of the apparent  residual polynomal by 2, the get the correct answer of

r^3  + 10  R  [ 2 / (2r + 1) ]

OK.....we'll carry it through....

Let her sister's age  = x  and let Kim's age  = 4x

So  (a)  is

x + 4x  = 60    simplify

5x  = 60    divide both sides by 5

x  = 12

(b)

Her sister is 12 years old and Kim  is  4(12)  =  48 years old