MEMEG0D

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 #3
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Part (b):

 

From the previous part, we know that the cubic polynomial with roots a, b, and c is: p(t) = t^3 - 2t^2 - 41t + 108

 

To find the possible values of x+y, we need to find the possible pairs of roots that add up to (a+b).

 

We can use Vieta's formulas to find the sum of the roots and the product of the roots:

 

Sum of roots: a + b + c = 2

 

Product of roots: abc = -108

 

From the product of roots, we can see that one of the roots must be negative.

 

Let's consider the possible scenarios:

 

One root is -1:

 

If one root is -1, the other two roots must multiply to 108.

 

The possible pairs are (9, 12) and (-9, -12).

 

One root is -2:

 

If one root is -2, the other two roots must multiply to 54.

 

The possible pairs are (6, 9) and (-6, -9).

 

One root is -3:

 

If one root is -3, the other two roots must multiply to 36.

 

The possible pairs are (4, 9) and (-4, -9).

 

One root is -4:

 

If one root is -4, the other two roots must multiply to 27.

 

The possible pairs are (3, 9) and (-3, -9).

 

One root is -6:

 

If one root is -6, the other two roots must multiply to 18.

 

The possible pairs are (3, 6) and (-3, -6).

 

One root is -9:

 

If one root is -9, the other two roots must multiply to 12.

 

The possible pairs are (3, 4) and (-3, -4).

 

From these pairs, we can calculate the possible values of a+b:

 

9 + 12 = 21

 

-9 - 12 = -21

 

6 + 9 = 15

 

-6 - 9 = -15

 

4 + 9 = 13

 

-4 - 9 = -13

 

3 + 9 = 12

 

-3 - 9 = -12

 

3 + 6 = 9

 

-3 - 6 = -9

 

3 + 4 = 7

 

-3 - 4 = -7

 

Therefore, the distinct possible values of x+y (which is equivalent to (a+b)/5) are: 21/5, -21/5, 3, -3, 13/5, -13/5, 12/5, -12/5, 9/5, -9/5, 7/5, -7/5

6 hours ago