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Remember that    i * i  =  i^2 = -1

a)   (-4i) ( 6i) = -4 * 6 * i * i    =  -24 * -1   =  24

b (2 + i) ( 2 - i)  = 

(2 + i) * 2    -  (2 + i)*i   =

4+ 2i - 2i - i^2  =

4 - i^2

4 - (-1) =

5

c)  ( 4 - 3i) ( 5 + i) =

(4 - 3i) * 5  + (4 - 3i) * i  =

20 - 15i + 4i - 3i^2 =

20 - 11i - 3(-1) =

23 - 11i

I don't.  I just learned the very easy way of doing it.  It's pretty nice.

We have

y = (4x) ( x + 2)^(-1)        the derivative of this gives the slope of the graph at any point x

So....the derivative is

y ' (x)  =  4(x + 2)^(-1) - (4x)(x +  2)^(-2)

So...the slope at  x = 2 is given by

y ' (2)   =   2(2 + 2) ^(-1) -  (4*2)(2 + 2)^(-2)  =   1 - (1/2) = 1/2

So....using the slope and the point (2,2).....the equation of the tangent line is

y = (1/2) (x - 2) + 2

y  = (1/2)x - 1 + 2

y = (1/2)x + 1

Here's a graph : https://www.desmos.com/calculator/a65are5fgn

P.S......if you need to take the derivative using the difference quotient....let me know

Here is another way:

∑ [ 2^(-n) * 3^(2 - n), n, 1,∞] = 9/5

Remember that the second function....because of the "-" out front... just "flips" (reflects) the first graph over the x axis

But....the zeroes for the second polynomial are just the same as those of the first one....

Sorry so so sorry got carried away well if it helps let it rip then!!!!!!!!!

            x^2  - x    + 1

x + 1 [  x^3 + 0x^2 + 0x + 1 ]

            x^3  + 1x^2

           _____________________

                     -1x^2 + 0x

                     -1x^2 - 1x

                    ____________  

                                1x   + 1

                                1x   +  1

                                _________

                                          0

             x^4   - x^3  +  x^2  - x   +  1

x + 1  [  x^5  + 0x^4  + 0x^3 + 0x^2 + 0x  + 1 ]

             x^5  + 1x^4

             _______________

                       -1x^4  + 0x^3

                      -1x^4 -  1x^3

                      __________________

                                 1x ^3   + 0x^2 

                                  1x^3 +  1x^2

                                   _______________

                                           -1x^2  + 0x

                                           -1x^2 - 1x

                                             _____________

                                                      1x   + 1

                                                      1x  +  1

                                                       _______

                                                                0

Note GM that the pattern seems to be one of alternating signs on decreasing powers

(x^n + 1) / ( x + 1)  =   x^(n - 1)  - x^(n - 2) + x^(n - 3) - x^(n - 4) + ..... - x .+  1

So.....we can intuit that 

(x^7 + 1) / ( x + 1) = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1

Knowing this.....can you find     (x^9 + 1) / ( x + 1)   ?????

And finding that.....the factorization will be  ...   (your answer ) (x + 1)

1)    ∑[ (3n -2) / 3^n, n, 1,∞] =5/4

2) 

See the answer here by Max Wong:

https://web2.0calc.com/questions/the-sum-of-a-certain-infinite-geometric-series-is#r4

We have

1 + 1/2 + 1/6 + 1/12 + 1/36 + 1/72 +  ......

Notice that we can split this up like so

1 + 1/6 + 1/36 + ......+

1/2 + 1/12 +  1/72 +  ......+

The sum of the first series is            1 / (1 - 1/6)  =   1 / (5/6) =   6/5

The sum of the second series is  (1/2) / ( 1 - 1/6)  =  (1/2) /(5./6) = 6/10 =  3/5

So....the sum of the series is   6/5 + 3/5  =   9/5

Pretty easy, GM

P(x) = (x - 1 )^4

Note the logic of this

P(x) = (x -1) (x - 1) (x -1) (x - 1)

The same zero, x = 1, is "repeated" 4 times  [ a " multiplicity" of 4 ]

 The graph shifts 5 units left and 3 units up.