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1. Suppose g(x) is a polynomial of degree five for which g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 5, g(5) = 6, and g(6) = -113. Find g(0).

 

2. 6 is the remainder when g(x) = x^9 - k x^5 + 4 is divided by x - 1. Find k.

 

3. Find a monic quadratic polynomial f(x) such that the remainder when f(x) is divided by x-1 is 2 and the remainder when f(x) is divided by x-3 is 4.

 

4. Find an ordered pair of constants (a,b) such that the polynomial f(x)=x^3+ax^2+(b+2)x+1 is divisible by x^2-1.

 

5. Suppose the polynomial f(x) is of degree 3 and satisfies f(3)=2, f(4)=4, f(5)=-3, and f(6)=8. Determine the value of f(0).

 

Challenge: 

Suppose f(x) is a polynomial of degree 4 or greater such that f(1)=2, f(2)=3, and f(3)=5.

Find the remainder when f(x) is divided by (x-1)(x-2)(x-3).

 Dec 15, 2018
 #1
avatar+100519 
+1

I only have time for a couple.....

 

2. 6 is the remainder when g(x) = x^9 - k x^5 + 4 is divided by x - 1. Find k

 

Using synthetic division and accounting for missing powers, we have....

 

 

1     [    1     0     0     0   - k       0        0       0      0      4  ]

                   1     1     1    1      1- k     1-k   1-k   1-k   1-k

           ________________________________________

             1     1    1     1   1-k   1-k      1-k    1-k  1-k   5-k

 

5 - k   is the remainder

 

So

 

5 - k  =  6    

 

So

 

k = -1

 

 

cool cool cool

 Dec 15, 2018
 #2
avatar+100519 
+1

3. Find a monic quadratic polynomial f(x) such that the remainder when f(x) is divided by x-1 is 2 and the remainder when f(x) is divided by x-3 is 4.

 

Let the polynomial be    x^2 + bx + c

 

Use synthetic divsion

 

 

1     [    1      b               c    ]                      3  [  1         b          c    ]

                    1             b + 1                                        3       3b + 9              

         __________________                           ___________________

           1     b + 1    [1 + b + c]                          1       b+ 3   [ 9 + 3b + c]

 

 

We have this system

 

1 + b + c  =   2           ⇒  b + c   = 1

9  + 3b + c  =  4     ⇒   3b + c  =  -5

 

Subtract the 1st equation from the second and we get that

 

2b  =  -6

b = -3

 

So

b + c  = 1

-3 + c = 1

c = 4

 

 

So....the polynomial is

 

 

x^2 - 3x + 4

 

 

 

cool cool cool

 Dec 15, 2018
 #3
avatar+100519 
+1

4. Find an ordered pair of constants (a,b) such that the polynomial f(x)=x^3+ax^2+(b+2)x+1 is divisible by x^2-1.

 

 

If the polynomial is divisible by  x^2 - 1

Then it is divisible by    (x -1)  (x + 1)

This means that     x = 1     and   x  = - 1    are roots

 

Therefore

 

(1)^3 + a(1)^2 + (b+2)(1) + 1 = 0    →   1 + a + b+2 + 1 = 0 →  a + b = -4     (1)

And

(-1)^3 + a(-1)^2 + (b + 2)(-1) + 1 = 0 → -1 + a - (b + 2)  + 1 = 0  → a - b = 2   (2)

 

Add (1)  and (2)  and we get that

 

2a = -2

a = -1

 

And

a - b = 2

-1 - b = 2

b = -3

 

The polynomial is

 

x^3 - x^2 - x + 1

 

 

 

 

cool cool cool

 Dec 15, 2018
 #4
avatar+100519 
+1

5. Suppose the polynomial f(x) is of degree 3 and satisfies f(3)=2, f(4)=4, f(5)=-3, and f(6)=8. Determine the value of f(0).

 

I don't have time to go through this one fully....but...we have this system

 

27a + 9b + 3c + d =  2

64a + 16b + 4c + d = 4

125a + 25b + 5c + d = -3

216a + 36b + 6c + d = 8

 

a = 9/2

b = -117/ 2

c = 245

d = - 328

 

The polynomial is

 

(9/2)x^3  - (117/2)x^2  + 245 - 328

 

So

 

f(0)  =  d    =   -328

 

NOTE    :    #1   is solved similarly.....6 equations....6 unknowns.....a computer algebra system may be the fastest way to go......that's a lot of work to solve such a large system by hand....!!!

See if you can do it based on this one

 

 

 

cool cool cool

 Dec 15, 2018
edited by CPhill  Dec 15, 2018
 #5
avatar+647 
0

Wow, smart! I never knew you could do it that way... helpful!

CoolStuffYT  Dec 15, 2018

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