Question 6, 2.4.15 points Points: 0 of 1 Divide using long division. State the quotient, $\mathrm{q}(\mathrm{x})$, and the remainder, $\mathrm{r}(\mathrm{x})$. \[ \frac{6 x^{4}+16 x^{3}+x^{2}}{2 x^{2}+3} \] \[ \frac{6 x^{4}+16 x^{3}+x^{2}}{2 x^{2}+3}=\square+\frac{\square}{2 x^{2}+3} \] (Simplify your answers. Do not factor. Use integers or fractions for any numbers in the expressions.)
Solution
Step 1 ：\(6x^4\) divided by \(2x^2\) is \(3x^2\). This is the first term of the quotient.
Step 2 ：Multiply the divisor \((2x^2 + 3)\) by the first term of the quotient \(3x^2\) to get \(6x^4 + 9x^2\). Subtract this from the dividend to get \(16x^3 - 8x^2\).
Step 3 ：\(16x^3\) divided by \(2x^2\) is \(8x\). This is the next term of the quotient.
Step 4 ：Multiply the divisor \((2x^2 + 3)\) by the next term of the quotient \(8x\) to get \(16x^3 + 24x\). Subtract this from the new dividend to get \(-8x^2 - 24x\).
Step 5 ：\(-8x^2\) divided by \(2x^2\) is \(-4\). This is the next term of the quotient.
Step 6 ：Multiply the divisor \((2x^2 + 3)\) by the next term of the quotient \(-4\) to get \(-8x^2 - 12\). Subtract this from the new dividend to get \(-24x + 12\).
Step 7 ：Since the degree of the new dividend is less than the degree of the divisor, we stop here. The quotient is \(3x^2 + 8x - 4\) and the remainder is \(-24x + 12\).
Step 8 ：\(\boxed{\frac{6 x^{4}+16 x^{3}+x^{2}}{2 x^{2}+3} = 3x^2 + 8x - 4 + \frac{-24x + 12}{2 x^{2}+3}}\)
