Polynomials `ax^(3)+3x^(2)-3 " and" 2x^(3)-5x+a` when divided by `(x-4)` leave the same remainder. Find the value of `a`.

Options

A. `4`
B. `3`
C. `2`
D. `1`

Answer

To find the value of \( a \) such that the polynomials \( ax^3 + 3x^2 - 3 \) and \( 2x^3 - 5x + a \) leave the same remainder when divided by \( (x - 4) \), we can follow these steps:

### Step 1: Evaluate the first polynomial at \( x = 4 \)

Let \( P(x) = ax^3 + 3x^2 - 3 \). We need to find \( P(4) \):

\[ P(4) = a(4^3) + 3(4^2) - 3 \]
Calculating \( 4^3 \) and \( 4^2 \):

\[ 4^3 = 64 \quad \text{and} \quad 4^2 = 16 \]

Substituting these values in:

\[ P(4) = a(64) + 3(16) - 3 = 64a + 48 - 3 = 64a + 45 \]

### Step 2: Evaluate the second polynomial at \( x = 4 \)

Let \( Q(x) = 2x^3 - 5x + a \). We need to find \( Q(4) \):

\[ Q(4) = 2(4^3) - 5(4) + a \]
Using \( 4^3 = 64 \):

\[ Q(4) = 2(64) - 5(4) + a = 128 - 20 + a = 108 + a \]

### Step 3: Set the remainders equal to each other

Since both polynomials leave the same remainder when divided by \( (x - 4) \):

\[ P(4) = Q(4) \]

This gives us the equation:

\[ 64a + 45 = 108 + a \]

### Step 4: Solve for \( a \)

Rearranging the equation:

\[ 64a - a = 108 - 45 \]
\[ 63a = 63 \]
\[ a = 1 \]

Thus, the value of \( a \) is \( 1 \).

### Final Answer:
The value of \( a \) is \( 1 \).

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