We are given that $x^2 - x - 2$ is a factor of the polynomial $3x^3 + ax^2 - 10x + b$, and we need to find the values of $a$ and $b$.

Step-by-step solution:

1. Factor the quadratic expression: $x^2 - x - 2 = (x - 2)(x + 1)$ So, we know that $x - 2$ and $x + 1$ are factors of the cubic polynomial $3x^3 + ax^2 - 10x + b$.

2. Polynomial division: To find $a$ and $b$, we divide $3x^3 + ax^2 - 10x + b$ by $x^2 - x - 2$.

   We assume the quotient to be of the form $q(x) = dx + e$, where $d$ and $e$ are constants, since dividing a cubic polynomial by a quadratic gives a linear quotient.

   $(x^2 - x - 2)(dx + e) = 3x^3 + ax^2 - 10x + b$

3. Expand the left-hand side: Expanding $(x^2 - x - 2)(dx + e)$: $(x^2 - x - 2)(dx + e) = dx^3 + ex^2 - dx^2 - ex - 2dx - 2e$ Simplify: $dx^3 + (e - d)x^2 + (-e - 2d)x - 2e$

4. Equate coefficients: Now, compare the coefficients of $3x^3 + ax^2 - 10x + b$ with the expanded form:

   • $dx^3 = 3x^3$ → $d = 3$
   • $(e - d)x^2 = ax^2$ → $e - 3 = a$
   • $(-e - 2d)x = -10x$ → $-e - 6 = -10$
   • $-2e = b$

5. Solve for $e$ and $a$: From $-e - 6 = -10$, we get: $-e = -4 \quad \Rightarrow \quad e = 4$ Now, substitute $e = 4$ into $e - 3 = a$: $4 - 3 = a \quad \Rightarrow \quad a = 1$

6. Solve for $b$: From $-2e = b$: $-2(4) = b \quad \Rightarrow \quad b = -8$

Final Answer:

• $a = 1$
• $b = -8$

Let me know if you would like further details or have questions!

Here are 5 related questions for further practice:

1. Can you find the quotient of the division $3x^3 + x^2 - 10x - 8$ by $x^2 - x - 2$?
2. What happens if $x^2 - x - 3$ is a factor of a polynomial? Can you solve a similar problem?
3. How would you handle polynomial division if the degree of the divisor was greater than 2?
4. What method would you use to solve a polynomial equation if you can't factor it easily?
5. Can you generalize how to find unknown coefficients in a polynomial given its factors?

Tip: When solving for unknowns in polynomial division, always carefully expand and match coefficients systematically to avoid errors!