The values of a and b so that the polynomial `x^(3)-ax^(2)-13x+b` has `(x-1)` and `(x+3)` as factors respectively are

Options

A. `3,15`
B. `5,13`
C. `15,3`
D. `5,10`

Answer

To find the values of \( a \) and \( b \) such that the polynomial \( P(x) = x^3 - ax^2 - 13x + b \) has \( (x-1) \) and \( (x+3) \) as factors, we will use the fact that if \( (x-1) \) and \( (x+3) \) are factors, then \( P(1) = 0 \) and \( P(-3) = 0 \).

### Step 1: Set up the equations

1. Substitute \( x = 1 \) into the polynomial:
\[ P(1) = 1^3 - a(1^2) - 13(1) + b = 0 \]
This simplifies to:
\[ 1 - a - 13 + b = 0 \]
Rearranging gives us:
\[ -a + b - 12 = 0 \quad \text{(Equation 1)} \]

2. Substitute \( x = -3 \) into the polynomial:
\[ P(-3) = (-3)^3 - a(-3)^2 - 13(-3) + b = 0 \]
This simplifies to:
\[ -27 - 9a + 39 + b = 0 \]
Rearranging gives us:
\[ -9a + b + 12 = 0 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations

Now we have the following system of equations:
1. \( -a + b = 12 \)
2. \( -9a + b = -12 \)

We can solve these equations by elimination or substitution. Let's subtract Equation 1 from Equation 2:

\[ (-9a + b) - (-a + b) = -12 - 12 \]
This simplifies to:
\[ -9a + b + a - b = -24 \]
\[ -8a = -24 \]
Dividing both sides by -8 gives:
\[ a = 3 \]

### Step 3: Find \( b \)

Now that we have \( a \), we can substitute \( a = 3 \) back into Equation 1 to find \( b \):
\[ -a + b = 12 \]
Substituting \( a = 3 \):
\[ -3 + b = 12 \]
Adding 3 to both sides gives:
\[ b = 15 \]

### Final Answer

Thus, the values are:
\[ a = 3 \quad \text{and} \quad b = 15 \]