If `(x-1)^3` is a factor of `x^4+ax^3+bx^2+cx-1=0` then the other factor is

Options

A. x - 3
B. x + 1
C. x + 2
D. x - 1

Answer

To find the other factor of the polynomial \( x^4 + ax^3 + bx^2 + cx - 1 \) given that \( (x-1)^3 \) is a factor, we can follow these steps:

### Step 1: Write down the polynomial and the factor
We have the polynomial:
\[ P(x) = x^4 + ax^3 + bx^2 + cx - 1 \]
and we know that \( (x-1)^3 \) is a factor of \( P(x) \).

### Step 2: Express \( (x-1)^3 \)
Using the binomial expansion, we can express \( (x-1)^3 \) as:
\[ (x-1)^3 = x^3 - 3x^2 + 3x - 1 \]

### Step 3: Set up the equation
Since \( (x-1)^3 \) is a factor of \( P(x) \), we can write:
\[ P(x) = (x-1)^3 \cdot Q(x) \]
where \( Q(x) \) is the other factor we need to find. Since \( P(x) \) is a polynomial of degree 4 and \( (x-1)^3 \) is of degree 3, \( Q(x) \) must be of degree 1. We can express it as:
\[ Q(x) = x + p \]
where \( p \) is a constant.

### Step 4: Multiply the factors
Now we multiply \( (x-1)^3 \) and \( (x+p) \):
\[ P(x) = (x^3 - 3x^2 + 3x - 1)(x + p) \]

### Step 5: Expand the product
Expanding this product:
\[ P(x) = x^3(x + p) - 3x^2(x + p) + 3x(x + p) - 1(x + p) \]
\[ = x^4 + px^3 - 3x^3 - 3px^2 + 3x^2 + 3px - p \]
\[ = x^4 + (p - 3)x^3 + (3 - 3p)x^2 + (3p - 1)x - p \]

### Step 6: Compare coefficients
Now, we compare the coefficients of \( P(x) \) with \( x^4 + ax^3 + bx^2 + cx - 1 \):
- Coefficient of \( x^3 \): \( p - 3 = a \)
- Coefficient of \( x^2 \): \( 3 - 3p = b \)
- Coefficient of \( x \): \( 3p - 1 = c \)
- Constant term: \( -p = -1 \)

### Step 7: Solve for \( p \)
From the constant term, we have:
\[ -p = -1 \implies p = 1 \]

### Step 8: Substitute \( p \) back to find coefficients
Substituting \( p = 1 \) into the equations:
- \( a = 1 - 3 = -2 \)
- \( b = 3 - 3(1) = 0 \)
- \( c = 3(1) - 1 = 2 \)

### Conclusion
Thus, the other factor \( Q(x) \) is:
\[ Q(x) = x + 1 \]

### Final Answer
The other factor is \( x + 1 \).

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