Given the function $$ C(x)=4 x^{4}+20 x^{3}+24 x^{2} $$ its $$ C $$-intercept is its $$ x $$-intercepts are

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To find the $$ C $$-intercept and the $$ x $$-intercepts of the function $$ C(x) = 4x^4 + 20x^3 + 24x^2 $$, we can follow these steps:

### 1. Finding the $$ C $$-intercept:
The $$ C $$-intercept occurs when $$ x = 0 $$. We can find it by evaluating $$ C(0) $$:

$$
C(0) = 4(0)^4 + 20(0)^3 + 24(0)^2 = 0
$$

Thus, the $$ C $$-intercept is $$ (0, 0) $$.

### 2. Finding the $$ x $$-intercepts:
The $$ x $$-intercepts occur when $$ C(x) = 0 $$. We need to solve the equation:

$$
4x^4 + 20x^3 + 24x^2 = 0
$$

We can factor out the common term $$ 4x^2 $$:

$$
4x^2(x^2 + 5x + 6) = 0
$$

Setting each factor to zero gives us:

1. $$ 4x^2 = 0 $$ which leads to $$ x^2 = 0 $$ or $$ x = 0 $$.
2. $$ x^2 + 5x + 6 = 0 $$

Now, we can solve the quadratic equation $$ x^2 + 5x + 6 = 0 $$ using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 1 $$, $$ b = 5 $$, and $$ c = 6 $$:

$$
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2}
$$

Calculating the two possible values:

1. $$ x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 $$
2. $$ x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 $$

### Summary:
- The $$ C $$-intercept is $$ (0, 0) $$.
- The $$ x $$-intercepts are $$ x = 0, -2, -3 $$.

Thus, the $$ x $$-intercepts are $$ (0, 0) $$, $$ (-2, 0) $$, and $$ (-3, 0) $$.
The $$ C $$-intercept is $$ (0, 0) $$. The $$ x $$-intercepts are $$ (0, 0) $$, $$ (-2, 0) $$, and $$ (-3, 0) $$.

Given the function
its -intercept is
its -intercepts are

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