Can `x^2-1` be the quotient on division of `x^6 + 2x^3 + x -1` by a polynomial in x of degree 5?

Answer

To determine whether \( x^2 - 1 \) can be the quotient when dividing \( x^6 + 2x^3 + x - 1 \) by a polynomial of degree 5, we can follow these steps:

### Step 1: Identify the degrees of the polynomials
The polynomial we are dividing is:
\[ f(x) = x^6 + 2x^3 + x - 1 \]
The degree of \( f(x) \) is 6.

The polynomial we are dividing by, denoted as \( g(x) \), is of degree 5.

### Step 2: Determine the degree of the quotient
We are given that the quotient \( q(x) \) is:
\[ q(x) = x^2 - 1 \]
The degree of \( q(x) \) is 2.

### Step 3: Use the relationship between degrees
When dividing polynomials, the relationship between the degrees can be expressed as:
\[ \text{Degree of } f(x) = \text{Degree of } g(x) + \text{Degree of } q(x) + \text{Degree of remainder} \]
Here, the degree of the remainder is typically less than the degree of the divisor \( g(x) \).

### Step 4: Set up the equation
Substituting the degrees into the equation:
\[ 6 = 5 + 2 + r \]
where \( r \) is the degree of the remainder.

### Step 5: Solve for the degree of the remainder
Rearranging gives:
\[ r = 6 - 5 - 2 = -1 \]
Since the degree of a polynomial cannot be negative, this indicates that our assumption about the quotient being \( x^2 - 1 \) is incorrect.

### Conclusion
Thus, \( x^2 - 1 \) cannot be the quotient when dividing \( x^6 + 2x^3 + x - 1 \) by a polynomial of degree 5.

### Final Answer
No, \( x^2 - 1 \) cannot be the quotient.

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