Question

determine the quotient and remainder.\\((3x^3 - 8x^2 - 24x - 7) \div (3x + 1)\\)

Explanation:

Step1: Divide leading terms

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

Step2: Multiply divisor by $x^2$

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

Step3: Subtract from dividend

$(3x^3 -8x^2 -24x -7) - (3x^3 + x^2) = -9x^2 -24x -7$

Step4: Divide new leading terms

$\frac{-9x^2}{3x} = -3x$

Step5: Multiply divisor by $-3x$

$-3x(3x+1) = -9x^2 -3x$

Step6: Subtract from new dividend

$(-9x^2 -24x -7) - (-9x^2 -3x) = -21x -7$

Step7: Divide new leading terms

$\frac{-21x}{3x} = -7$

Step8: Multiply divisor by $-7$

$-7(3x+1) = -21x -7$

Step9: Subtract to find remainder

$(-21x -7) - (-21x -7) = 0$

Answer:

Quotient: $x^2 - 3x - 7$
Remainder: $0$