Question

\frac{x^3+3x^2-x+2}{x-1}

Answer 1

[AI-generated answer — Bot GPT 3.5]

To divide the polynomial \(\frac{x^3+3x^2-x+2}{x-1}\), we can use polynomial long division.

First, we divide the first term of the numerator, \(x^3\), by the first term of the denominator, \(x\), and write the result in the quotient. The result is \(x^2\).

So far, we have \(\frac{x^3+3x^2-x+2}{x-1} = x^2 + \_\_\_\_\_\_\_\_\).

Next, we multiply the denominator by the quotient term we just found, \(x^2\), and subtract it from the numerator. \((x-1)(x^2) = x^3-x^2\), and when we subtract this from \(x^3+3x^2-x\), we get \(4x^2 - x\).

So, we now have \(\frac{x^3 + 3x^2 - x + 2}{x-1} = x^2 + \frac{4x^2 - x}{x-1} + 2\).

We repeat the process with the new simplified fraction \(\frac{4x^2 - x}{x-1}\).
Dividing the first term of the numerator, \(4x^2\), by the first term of the denominator, \(x\), gives \(4x\).

We now have \(\frac{x^3 + 3x^2 - x + 2}{x-1} = x^2 + 4x + \_\_\_\_\_\_\_\_\).

We multiply the denominator by the quotient term we just found, \(4x\), and subtract it from the numerator. \((x-1)(4x) = 4x^2-4x\), and when we subtract this from \(4x^2 - x\), we get \(3x\).

So far, we have \(\frac{x^3 + 3x^2 - x + 2}{x-1} = x^2 + 4x + \frac{3x}{x-1} + 2\).

Finally, we divide the first term of the remaining numerator, \(3x\), by the first term of the denominator, \(x\), and write the result in the quotient. The result is \(3\).

Thus, the final result is \(\frac{x^3+3x^2-x+2}{x-1} = x^2 + 4x + 3 + \frac{3}{x-1}\).