+116 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).
+128408 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
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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
+128408 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
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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
+128408 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
+128408 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
