Divide. (6x^3-7x^2-9x-6)div (3x+1) Your answer should give the quotient and the remainder. Quotient: III Remainder: square

To divide the polynomial $(6x^{3}-7x^{2}-9x-6)$ by $(3x+1)$, we can use polynomial long division.<br /><br />Step 1: Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.<br />$\frac{6x^{3}}{3x} = 2x^{2}$<br /><br />Step 2: Multiply the divisor by the first term of the quotient and subtract the result from the dividend to get the new dividend.<br />$(3x+1) \cdot 2x^{2} = 6x^{3} + 2x^{2}$<br />$(6x^{3}-7x^{2}-9x-6) - (6x^{3} + 2x^{2}) = -9x^{2}-9x-6$<br /><br />Step 3: Repeat steps 1 and 2 with the new dividend.<br />$\frac{-9x^{2}}{3x} = -3x$<br />$(3x+1) \cdot -3x = -9x^{2} - 3x$<br />$(-9x^{2}-9x-6) - (-9x^{2} - 3x) = -6x-6$<br /><br />Step 4: Repeat steps 1 and 2 with the new dividend.<br />$\frac{-6x}{3x} = -2$<br />$(3x+1) \cdot -2 = -6x - 2$<br />$(-6x-6) - (-6x - 2) = -4$<br /><br />Therefore, the quotient is $2x^{2} - 3x - 2$ and the remainder is $-4$.<br /><br />Quotient: $2x^{2} - 3x - 2$<br />Remainder: $-4$