If all the zeros of polynomial function `f(x)=2x^5+5x^4+10x^3+10x^2+10x+10` lies in `(a,a+1)` where `a in I ` then find `absa`

Answer

To solve the problem, we need to analyze the polynomial function \( f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10 \) and determine the integer \( a \) such that all zeros of the polynomial lie in the interval \( (a, a+1) \).

### Step-by-Step Solution:

1. **Identify the Polynomial**:
We have the polynomial \( f(x) = 2x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 10 \).

2. **Check the Degree and Behavior**:
The polynomial is of degree 5, which means it can have up to 5 real roots. Since the leading coefficient (2) is positive, the polynomial will approach \( +\infty \) as \( x \to +\infty \) and \( -\infty \) as \( x \to -\infty \).

3. **Determine the Nature of the Function**:
We can see that the polynomial is strictly increasing. This is because the derivative \( f'(x) = 10x^4 + 20x^3 + 30x^2 + 10 \) is always positive for all \( x \) (since all terms are non-negative).

4. **Evaluate the Function at Specific Points**:
We will evaluate \( f(x) \) at \( x = -2 \) and \( x = -1 \):
- Calculate \( f(-2) \):
\[ f(-2) = 2(-2)^5 + 5(-2)^4 + 10(-2)^3 + 10(-2)^2 + 10(-2) + 10 \]
\[ = 2(-32) + 5(16) + 10(-8) + 10(4) + 10(-2) + 10 \]
\[ = -64 + 80 - 80 + 40 - 20 + 10 = -34 \]
- Calculate \( f(-1) \):
\[ f(-1) = 2(-1)^5 + 5(-1)^4 + 10(-1)^3 + 10(-1)^2 + 10(-1) + 10 \]
\[ = 2(-1) + 5(1) + 10(-1) + 10(1) + 10(-1) + 10 \]
\[ = -2 + 5 - 10 + 10 - 10 + 10 = 3 \]

5. **Identify the Interval for Roots**:
Since \( f(-2) < 0 \) and \( f(-1) > 0 \), and knowing that \( f(x) \) is strictly increasing, we can conclude that there is exactly one root in the interval \( (-2, -1) \).

6. **Determine the Integer \( a \)**:
The root lies in the interval \( (a, a+1) \). Since the root is between \( -2 \) and \( -1 \), we can deduce that \( a = -2 \).

7. **Find \( |a| \)**:
The problem asks for \( |a| \). Thus, we calculate:
\[ |a| = |-2| = 2 \]

### Final Answer:
The value of \( |a| \) is \( 2 \).