Solving for a and b in Polynomial Factors

Determining the values of a and b when x-1 and x-2 are factors of the polynomial x3-ax2bx-8 involves the application of the Factor Theorem. This theorem states that if x-r is a factor of a polynomial P(x), then P(r) 0. Let's walk through the solution step-by-step.

Applying the Factor Theorem

First, we denote the polynomial as:

P(x) x3-ax2bx-8

Given that x-1 is a factor, we substitute x1 and set P(1) 0:

P(1) 13-a(1)2b(1)-8 0

This simplifies to:

1 - a - b - 8 0

Which further simplifies to:

-a - b - 7 0 or b - a 7 (1)

Using the Second Factor

Similarly, since x-2 is also a factor, we substitute x2 and set P(2) 0:

P(2) 23-a(2)2b(2)-8 0

This simplifies to:

8 - 4a - 2b - 8 0

Which further simplifies to:

-4a - 2b 0 or 2b 4a or b 2a (2)

Formulating and Solving Equations

Now, we have two equations:

b - a 7 (1)

b 2a (2)

We can substitute equation (2) into equation (1) to solve for a:

b - a 7

2a - a 7

a 7

Now substituting a 7 back into equation (2) to solve for b:

b 2(7) 14

Thus, the values of a and b are:

boxed{7} boxed{14}

Verification

Let's verify the values:

b - a 14 - 7 7 (Correct)

b 2a 2(7) 14 (Correct)

Addendum

The second approach, as an alternative, uses polynomial division:

x3-ax2bx-8 (x-1)(x-2)(x2-3x-4)

From the factorization, we can directly determine:

b 14

And,

-a -7 or a 7

Conclusion

The solution set is:

boxed{a7, b14}