Question

divide using long division. check your answer.
$(x^3 + 3x^2 + x + 7) \div (x - 1)$

the quotient is \\(\square\\) with a remainder of \\(\square\\).
(simplify your answers. do not factor.)

Explanation:

Step1: Divide leading terms

$\frac{x^3}{x} = x^2$
Multiply divisor by $x^2$: $x^2(x-1) = x^3 - x^2$
Subtract from dividend:
$$(x^3 + 3x^2 + x + 7) - (x^3 - x^2) = 4x^2 + x + 7$$

Step2: Divide new leading terms

$\frac{4x^2}{x} = 4x$
Multiply divisor by $4x$: $4x(x-1) = 4x^2 - 4x$
Subtract from current polynomial:
$$(4x^2 + x + 7) - (4x^2 - 4x) = 5x + 7$$

Step3: Divide new leading terms

$\frac{5x}{x} = 5$
Multiply divisor by $5$: $5(x-1) = 5x - 5$
Subtract from current polynomial:
$$(5x + 7) - (5x - 5) = 12$$

Step4: Verify the result

Use the formula: $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$
$$(x-1)(x^2 + 4x + 5) + 12 = x^3 + 4x^2 + 5x - x^2 - 4x - 5 + 12 = x^3 + 3x^2 + x + 7$$
This matches the original dividend.

Answer:

The quotient is $x^2 + 4x + 5$ with a remainder of $12$.