If 2 is a root of the equation `x^4+2x^3-3x^2+kx-4=0` , then k=

Options

A. -12
B. -10
C. -8
D. 4

Explanation

To find the value of \( k \) such that \( 2 \) is a root of the equation \( x^4 + 2x^3 - 3x^2 + kx - 4 = 0 \), we can follow these steps:

### Step 1: Substitute \( x = 2 \) into the equation
Since \( 2 \) is a root, we can substitute \( x = 2 \) into the equation:

\[ (2)^4 + 2(2)^3 - 3(2)^2 + k(2) - 4 = 0 \]

### Step 2: Calculate each term
Now, we calculate each term:

- \( (2)^4 = 16 \)
- \( 2(2)^3 = 2 \times 8 = 16 \)
- \( -3(2)^2 = -3 \times 4 = -12 \)
- \( k(2) = 2k \)
- The constant term is \( -4 \)

### Step 3: Combine the terms
Now, we combine these results into the equation:

\[ 16 + 16 - 12 + 2k - 4 = 0 \]

### Step 4: Simplify the equation
Combine the constants:

\[ (16 + 16 - 12 - 4) + 2k = 0 \]
\[ 16 + 2k = 0 \]

### Step 5: Solve for \( k \)
Now, isolate \( k \):

\[ 2k = -16 \]
\[ k = \frac{-16}{2} = -8 \]

Thus, the value of \( k \) is \( -8 \).

### Final Answer
\[ \boxed{-8} \]

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