If `a_0, a_1, a_2, a_3` are all the positive, then `4a_0x^3+3a_1x^2+2a_2x+a_3=0` has least one root in `(-1,0)` if

Options

A. `a_(0) + a_(2) = a_(1) + a_(3) and 4_(a_(0)) + 2a_(2) gt 3a_(1) + a_(3)`
B. `4a_(0) + 2a_(2) lt 3a_(1) + a _(3)`
C. `4a_(0) + 2a_(2) = 3a_(1) + a_(3) and a_(0) + a_(2) lt a_(1) + a_(3)`
D. none of these

Explanation

To determine the conditions under which the polynomial \(4a_0x^3 + 3a_1x^2 + 2a_2x + a_3 = 0\) has at least one root in the interval \((-1, 0)\), we can use the Intermediate Value Theorem. This theorem states that if a continuous function changes sign over an interval, then there exists at least one root in that interval.

### Step-by-Step Solution:

1. **Define the Polynomial**:
Let \( p(x) = 4a_0x^3 + 3a_1x^2 + 2a_2x + a_3 \).

2. **Evaluate the Polynomial at the Endpoints**:
- Calculate \( p(-1) \):
\[ p(-1) = 4a_0(-1)^3 + 3a_1(-1)^2 + 2a_2(-1) + a_3 = -4a_0 + 3a_1 - 2a_2 + a_3 \]
- Calculate \( p(0) \):
\[ p(0) = a_3 \]

3. **Determine the Sign of \( p(0) \)**:
Since \( a_3 > 0 \) (given that all \( a_i \) are positive), we have:
\[ p(0) > 0 \]

4. **Set Up the Inequality for \( p(-1) \)**:
For \( p(x) \) to have at least one root in \((-1, 0)\), we need:
\[ p(-1) < 0 \]
This gives us the inequality:
\[ -4a_0 + 3a_1 - 2a_2 + a_3 < 0 \]
Rearranging this, we get:
\[ 4a_0 + 2a_2 > 3a_1 + a_3 \]

5. **Combine with the Condition from Step 3**:
Since \( p(0) > 0 \) and \( p(-1) < 0 \), we can conclude that:
\[ 4a_0 + 2a_2 > 3a_1 + a_3 \]

6. **Final Condition**:
The condition we derived can be expressed as:
\[ a_0 + a_2 = a_1 + a_3 \]
and
\[ 4a_0 + 2a_2 > 3a_1 + a_3 \]
This indicates that the first option \( a_0 + a_2 = a_1 + a_3 \) is necessary for the polynomial to have a root in the interval \((-1, 0)\).

### Conclusion:
The polynomial \( 4a_0x^3 + 3a_1x^2 + 2a_2x + a_3 = 0 \) has at least one root in the interval \((-1, 0)\) if the condition \( a_0 + a_2 = a_1 + a_3 \) holds true along with the inequality \( 4a_0 + 2a_2 > 3a_1 + a_3 \).