(9.) (i) Given that $$ 2 x^{4}+3 x^{3}+a x^{2}-2=x(x+3)(x-1) Q(x)+4 x^{2}+b x+c $$ is an identity, state the degree of the polynomial $$ Q(x) $$. (ii) Find the value of $$ a $$, of $$ b $$ and of $$ c $$. Hence, find the remainder when $$ 2 x^{4}+3 x^{3}+a x^{2}-2 $$ is divided by $$ (x+1) $$.

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### Part (i): Degree of $$ Q(x) $$

The given identity is:
$$
2x^4 + 3x^3 + a x^2 - 2 = x(x+3)(x-1) Q(x) + 4x^2 + bx + c
$$

- **Left Side**: The highest degree term is $$ 2x^4 $$, so the degree of the left side is **4**.
- **Divisor**: $$ x(x+3)(x-1) $$ is a cubic polynomial, so its degree is **3**.
- **Quotient $$ Q(x) $$**: Let’s denote the degree of $$ Q(x) $$ as $$ d $$.

For the product $$ x(x+3)(x-1) Q(x) $$ to have a degree that matches the left side (which is 4), we set:
$$
3 + d = 4 \implies d = 1
$$

**Answer to Part (i):** The degree of $$ Q(x) $$ is **1**.

### Part (ii): Finding $$ a $$, $$ b $$, and $$ c $$, and the Remainder when Divided by $$ (x+1) $$

Given:
$$
2x^4 + 3x^3 + a x^2 - 2 = x(x+3)(x-1) Q(x) + 4x^2 + bx + c
$$

To find $$ a $$, $$ b $$, and $$ c $$, we can substitute the roots of the divisor $$ x(x+3)(x-1) $$, which are $$ x = 0 $$, $$ x = -3 $$, and $$ x = 1 $$.

1. **Substitute $$ x = 0 $$:**
   $$
   2(0)^4 + 3(0)^3 + a(0)^2 - 2 = 4(0)^2 + b(0) + c \implies -2 = c
   $$
   So, $$ c = -2 $$.

2. **Substitute $$ x = -3 $$:**
   $$
   2(-3)^4 + 3(-3)^3 + a(-3)^2 - 2 = 4(-3)^2 + b(-3) + c
   $$
   $$
   2(81) + 3(-27) + 9a - 2 = 4(9) - 3b - 2
   $$
   $$
   162 - 81 + 9a - 2 = 36 - 3b - 2
   $$
   $$
   79 + 9a = 34 - 3b \implies 9a + 3b = -45 \implies 3a + b = -15 \quad \text{(Equation 1)}
   $$

3. **Substitute $$ x = 1 $$:**
   $$
   2(1)^4 + 3(1)^3 + a(1)^2 - 2 = 4(1)^2 + b(1) + c
   $$
   $$
   2 + 3 + a - 2 = 4 + b - 2
   $$
   $$
   3 + a = 2 + b \implies a - b = -1 \quad \text{(Equation 2)}
   $$

**Solving Equations 1 and 2:**
$$
\begin{cases}
3a + b = -15 \
a - b = -1
\end{cases}
$$
Adding both equations:
$$
4a = -16 \implies a = -4
$$
Substituting $$ a = -4 $$ into Equation 2:
$$
-4 - b = -1 \implies b = -3
$$

So, $$ a = -4 $$, $$ b = -3 $$, and $$ c = -2 $$.

**Finding the Remainder when Divided by $$ (x+1) $$:**

The remainder of a polynomial $$ P(x) $$ when divided by $$ (x+1) $$ is $$ P(-1) $$.

Given $$ P(x) = 2x^4 + 3x^3 -4x^2 -2 $$:
$$
P(-1) = 2(-1)^4 + 3(-1)^3 -4(-1)^2 - 2 = 2(1) + 3(-1) - 4(1) - 2 = 2 - 3 - 4 - 2 = -7
$$

**Final Answer:**

Problem i Answer:
The polynomial $$ Q(x) $$ has degree 1.

Problem ii Answer:
$$ a = -4 $$, $$ b = -3 $$, and $$ c = -2 $$. The remainder upon division by $$ (x+1) $$ is –7.
**Part (i):** The degree of $$ Q(x) $$ is **1**. **Part (ii):** - $$ a = -4 $$ - $$ b = -3 $$ - $$ c = -2 $$ - The remainder when $$ 2x^4 + 3x^3 - 4x^2 - 2 $$ is divided by $$ (x+1) $$ is **-7**.

(9.) (i) Given that \( 2 x^{4}+3 x^{3}+a x^{2}-2=x(x+3)(x-1) Q(x)+4 x^{2}+b x+c \) is an identity, state the degree of the polynomial \( Q(x) \). (ii) Find the value of \( a \), of \( b \) and of \( c \). Hence, find the remainder when \( 2 x^{4}+3 x^{3}+a x^{2}-2 \) is divided by \( (x+1) \).

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