Divide the polynomials by using long division \[ \left(64 a^{3}+1\right)+(4 a+1)=\square \]

To tackle the polynomial division here, we first combine the polynomials inside the parentheses. This gives us: \[ 64a^3 + 1 + 4a + 1 = 64a^3 + 4a + 2 \] Now, let’s set up the polynomial long division to divide \(64a^3 + 4a + 2\). We can check if there’s a divisor. It appears you might want to divide \(64a^3 + 4a + 2\) by a polynomial or factor. Assuming we’re dividing it by \(4a+2\): 1. Divide the leading term of the dividend \(64a^3\) by the leading term of the divisor \(4a\), giving you \(16a^2\). 2. Multiply \(16a^2\) by the entire divisor \(4a + 2\), yielding \(64a^3 + 32a^2\). 3. Subtract \( (64a^3 + 32a^2) \) from \( (64a^3 + 4a + 2) \), which results in: \[ (64a^3 + 4a + 2) - (64a^3 + 32a^2) = -32a^2 + 4a + 2 \] 4. Now, repeat the process: Divide \(-32a^2\) by \(4a\), which gives us \(-8a\). 5. Multiply \(-8a\) by the divisor \(4a + 2\), resulting in \(-32a^2 - 16a\). 6. Subtract this from \(-32a^2 + 4a + 2\): \[ (-32a^2 + 4a + 2) - (-32a^2 - 16a) = 20a + 2 \] 7. Continue with \(20a\) divided by \(4a\), which is \(5\). 8. Multiply \(5\) by \(4a + 2\) to yield \(20a + 10\). 9. Finally, do the subtraction: \[ (20a + 2) - (20a + 10) = -8 \] Thus, you’re left with the remainder \(-8\). So, the result of the division can be stated as: \[ 16a^2 - 8 + \frac{-8}{4a + 2} \] If you're looking to express the result neatly, here's what you're left with: \[ 16a^2 - 8 - \frac{8}{4a + 2} \] Now we know how to divide polynomials and handle the process!