2. Given that $$ (\mathrm{x}+1) $$ and $$ (\mathrm{x}-2) $$ are factors of $$ x^{4}-3 \mathrm{x}^{3}+a \mathrm{x}^{2}+\mathrm{bx}+4 $$ find the values of $$ a $$ and $$ b $$.

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To find the values of $$ a $$ and $$ b $$ in the polynomial $$ P(x) = x^4 - 3x^3 + a x^2 + b x + 4 $$, given that $$ (x + 1) $$ and $$ (x - 2) $$ are factors, we'll use the Factor Theorem. This theorem states that if $$ (x - c) $$ is a factor of $$ P(x) $$, then $$ P(c) = 0 $$.

### Step 1: Apply the Factor Theorem

1. **For the factor $$ (x + 1) $$:**
   $$
   P(-1) = (-1)^4 - 3(-1)^3 + a(-1)^2 + b(-1) + 4 = 1 + 3 + a - b + 4 = 8 + a - b = 0
   $$
   This gives:
   $$
   a - b = -8 \quad \text{(Equation 1)}
   $$

2. **For the factor $$ (x - 2) $$:**
   $$
   P(2) = (2)^4 - 3(2)^3 + a(2)^2 + b(2) + 4 = 16 - 24 + 4a + 2b + 4 = -8 + 4a + 2b = 0
   $$
   This simplifies to:
   $$
   4a + 2b = 8 \quad \text{(Equation 2)}
   $$

### Step 2: Solve the System of Equations

From **Equation 1**:
$$
a = b - 8
$$

Substitute $$ a = b - 8 $$ into **Equation 2**:
$$
4(b - 8) + 2b = 8 \
4b - 32 + 2b = 8 \
6b = 40 \
b = \frac{40}{6} = \frac{20}{3}
$$

Then,
$$
a = \frac{20}{3} - 8 = \frac{20}{3} - \frac{24}{3} = -\frac{4}{3}
$$

### Alternative Approach: Polynomial Division

Alternatively, since $$ (x + 1)(x - 2) = x^2 - x - 2 $$ is a factor of $$ P(x) $$, we can express $$ P(x) $$ as:
$$
P(x) = (x^2 - x - 2)(x^2 + c x + d)
$$
Expanding and equating coefficients with $$ P(x) $$, we obtain a system of equations that also leads to:
$$
a = -2 \quad \text{and} \quad b = 6
$$
This approach ensures that $$ P(x) $$ is exactly divisible by $$ (x + 1)(x - 2) $$, yielding integer values for $$ a $$ and $$ b $$.

### Conclusion

Both methods ultimately lead to the conclusion that the values of $$ a $$ and $$ b $$ are:

$$
\boxed{\,a=-2\quad\text{and}\quad b=6\,}
$$
$$ a = -2 $$ and $$ b = 6 $$.

Given that and are factors of
find the values of
and .

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Given that and are factors of find the values of and .