5. If $$ f(2)=f(-3)=0 $$, use the identity $$ f(x) \equiv x^{3}+2 x^{2}+a x+b $$ to find the values of $$ a $$ and $$ b $$. Hence, the remainder when $$ x^{3}+2 x^{2}+a x+b $$ is divided by $$ x-4 $$.

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Given that $$ f(2) = f(-3) = 0 $$, we can use the identity $$ f(x) \equiv x^{3} + 2x^{2} + ax + b $$ to find the values of $$ a $$ and $$ b $$.

Since $$ f(2) = 0 $$, we substitute $$ x = 2 $$ into the identity:
$$ f(2) = 2^{3} + 2(2)^{2} + a(2) + b = 0 $$
$$ 8 + 8 + 2a + b = 0 $$
$$ 16 + 2a + b = 0 $$

Since $$ f(-3) = 0 $$, we substitute $$ x = -3 $$ into the identity:
$$ f(-3) = (-3)^{3} + 2(-3)^{2} + a(-3) + b = 0 $$
$$ -27 + 18 - 3a + b = 0 $$
$$ -9 - 3a + b = 0 $$

Now we have a system of two equations:
$$ 16 + 2a + b = 0 $$
$$ -9 - 3a + b = 0 $$

We can solve this system of equations to find the values of $$ a $$ and $$ b $$.

Let's solve the system of equations.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}16+2a+b=0\-9-3a+b=0\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}b=-16-2a\-9-3a+b=0\end{array}\right.$$
- step2: Substitute the value of $$b:$$
  $$-9-3a-16-2a=0$$
- step3: Simplify:
  $$-25-5a=0$$
- step4: Move the constant to the right side:
  $$-5a=0+25$$
- step5: Remove 0:
  $$-5a=25$$
- step6: Change the signs:
  $$5a=-25$$
- step7: Divide both sides:
  $$\frac{5a}{5}=\frac{-25}{5}$$
- step8: Divide the numbers:
  $$a=-5$$
- step9: Substitute the value of $$a:$$
  $$b=-16-2\left(-5\right)$$
- step10: Simplify:
  $$b=-6$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}a=-5\b=-6\end{array}\right.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}a=-5\b=-6\end{array}\right.$$
- step13: Rewrite:
  $$\left(a,b\right) = \left(-5,-6\right)$$
The values of $$ a $$ and $$ b $$ are $$ a = -5 $$ and $$ b = -6 $$.

Now, we need to find the remainder when $$ x^{3} + 2x^{2} + ax + b $$ is divided by $$ x - 4 $$.

The remainder theorem states that the remainder of a polynomial $$ f(x) $$ when divided by $$ x - c $$ is $$ f(c) $$.

Therefore, the remainder when $$ x^{3} + 2x^{2} + ax + b $$ is divided by $$ x - 4 $$ is $$ f(4) $$.

Substitute $$ x = 4 $$ into the polynomial:
$$ f(4) = 4^{3} + 2(4)^{2} + a(4) + b $$
$$ f(4) = 64 + 32 - 20 - 6 $$
$$ f(4) = 70 $$

Therefore, the remainder when $$ x^{3} + 2x^{2} + ax + b $$ is divided by $$ x - 4 $$ is 70.
$$ a = -5 $$, $$ b = -6 $$, and the remainder is 70.

5. If \( f(2)=f(-3)=0 \), use the identity \( f(x) \equiv x^{3}+2 x^{2}+a x+b \) to find the values of \( a \) and \( b \). Hence, the remainder when \( x^{3}+2 x^{2}+a x+b \) is divided by \( x-4 \).

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