Question

divide using long division. state the quotient, q(x), and the remainder, r(x).

\\((x^3 + 2x^2 - 5x + 2) \div (x - 1)\\)

\\((x^3 + 2x^2 - 5x + 2) \div (x - 1) = \square + \frac{\square}{x - 1}\\)

(simplify your answers. do not factor.)

Explanation:

Set up the division

We want to divide the polynomial \(x^3 + 2x^2 - 5x + 2\) by the linear binomial \(x - 1\). Using Polynomial Functions concepts, we identify the dividend as \(x^3 + 2x^2 - 5x + 2\) and the divisor as \(x - 1\).

Divide the first term

Divide the leading term of the dividend by the leading term of the divisor:

$$ \frac{x^3}{x} = x^2 $$

Multiply \(x^2\) by the divisor \(x - 1\):

$$ x^2(x - 1) = x^3 - x^2 $$

Subtract this from the dividend:

$$ (x^3 + 2x^2) - (x^3 - x^2) = 3x^2 $$

Bring down the next term, \(-5x\), to get \(3x^2 - 5x\).

Divide the second term

Divide the leading term of the new expression by the leading term of the divisor:

$$ \frac{3x^2}{x} = 3x $$

Multiply \(3x\) by the divisor \(x - 1\):

$$ 3x(x - 1) = 3x^2 - 3x $$

Subtract this from the current expression:

$$ (3x^2 - 5x) - (3x^2 - 3x) = -2x $$

Bring down the final term, \(+2\), to get \(-2x + 2\).

Divide the third term

Divide the leading term of the current expression by the leading term of the divisor:

$$ \frac{-2x}{x} = -2 $$

Multiply \(-2\) by the divisor \(x - 1\):

$$ -2(x - 1) = -2x + 2 $$

Subtract this from the current expression:

$$ (-2x + 2) - (-2x + 2) = 0 $$

The remainder is \(0\).

State the final quotient and remainder

The quotient is \(q(x) = x^2 + 3x - 2\) and the remainder is \(r(x) = 0\).
We write the final expression in the requested form:

$$ (x^3 + 2x^2 - 5x + 2) \div (x - 1) = x^2 + 3x - 2 + \frac{0}{x - 1} $$

Answer:

\((x^3 + 2x^2 - 5x + 2) \div (x - 1) =\) <blank>\(x^2 + 3x - 2\)</blank> \(+ \frac{\text{}0\text{}}{x - 1}\)