Use the remainder theorem to determine if the given number $$ c $$ is a zero of the polynomial. $$ f(x)=4 x^{4}+8 x^{3}-18 x^{2}+6 x+72 $$ (a) $$ c=-3 $$ (b) $$ c=2 $$

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**Step 1. Recall the Remainder Theorem**

If $$ f(x) $$ is divided by $$ x - c $$, then the remainder is $$ f(c) $$. Therefore, if $$ f(c)=0 $$, then $$ c $$ is a zero of the polynomial.

**Step 2. Evaluate $$ f(-3) $$ for $$ c=-3 $$**

Given  
$$
f(x)=4x^4+8x^3-18x^2+6x+72,
$$
we substitute $$ x=-3 $$:

1. $$ 4(-3)^4 = 4 \times 81 = 324 $$  
2. $$ 8(-3)^3 = 8 \times (-27) = -216 $$  
3. $$ -18(-3)^2 = -18 \times 9 = -162 $$  
4. $$ 6(-3) = -18 $$  
5. Constant term: $$ 72 $$

Now, sum the values:
$$
f(-3) = 324 - 216 - 162 - 18 + 72
$$
Combine step by step:
$$
324 - 216 = 108,
$$
$$
108 - 162 = -54,
$$
$$
-54 - 18 = -72,
$$
$$
-72 + 72 = 0.
$$
Since $$ f(-3)=0 $$, $$ c=-3 $$ is a zero of the polynomial.

**Step 3. Evaluate $$ f(2) $$ for $$ c=2 $$**

Substitute $$ x=2 $$:
$$
f(2)=4(2)^4+8(2)^3-18(2)^2+6(2)+72.
$$

Calculate each term:
1. $$ 4(2)^4 = 4 \times 16 = 64 $$  
2. $$ 8(2)^3 = 8 \times 8 = 64 $$  
3. $$ -18(2)^2 = -18 \times 4 = -72 $$  
4. $$ 6(2) = 12 $$  
5. Constant term: $$ 72 $$

Now, sum them:
$$
f(2)=64+64-72+12+72.
$$
Add:
$$
64+64=128,
$$
$$
128-72=56,
$$
$$
56+12=68,
$$
$$
68+72=140.
$$
Since $$ f(2)=140 \neq 0 $$, $$ c=2 $$ is not a zero of the polynomial.

**Final Answer**

- For $$ c=-3 $$, $$ f(-3)=0 $$, so $$ c=-3 $$ is a zero.  
- For $$ c=2 $$, $$ f(2)=140 \neq 0 $$, so $$ c=2 $$ is not a zero.
$$ c=-3 $$ is a zero of the polynomial, and $$ c=2 $$ is not a zero.

Use the remainder theorem to determine if the given number \( c \) is a zero of the polynomial. \( f(x)=4 x^{4}+8 x^{3}-18 x^{2}+6 x+72 \) (a) \( c=-3 \) (b) \( c=2 \)

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