Question

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

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:
$(x^3 + 3x^2 + x + 3) - (x^3 - x^2) = 4x^2 + x + 3$

Step2: Divide new leading terms

$\frac{4x^2}{x} = 4x$
Multiply divisor by $4x$: $4x(x-1) = 4x^2 - 4x$
Subtract from current polynomial:
$(4x^2 + x + 3) - (4x^2 - 4x) = 5x + 3$

Step3: Divide new leading terms

$\frac{5x}{x} = 5$
Multiply divisor by $5$: $5(x-1) = 5x - 5$
Subtract from current polynomial:
$(5x + 3) - (5x - 5) = 8$

Step4: Verify the result

Multiply quotient by divisor, add remainder:
$(x^2 + 4x + 5)(x-1) + 8 = x^3 - x^2 + 4x^2 - 4x + 5x - 5 + 8 = x^3 + 3x^2 + x + 3$, which matches the original dividend.

Answer:

The quotient is $x^2 + 4x + 5$ with a remainder of $8$.