Question

determine the quotient and remainder.\\((3x^{3} - 5x^{2} - 23x - 7) div (3x + 1)\\)\
quotient\
remainder

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

$(3x^3-5x^2-23x-7)-(3x^3+x^2) = -6x^2-23x-7$

Step4: Divide new leading terms

$\frac{-6x^2}{3x} = -2x$

Step5: Multiply divisor by $-2x$

$-2x(3x+1) = -6x^2-2x$

Step6: Subtract from current polynomial

$(-6x^2-23x-7)-(-6x^2-2x) = -21x-7$

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