Fill in the missing values below to find the quotient when $8 x^{4}-2 x^{3}-5 x^{2}+1$ is divided by $2 x^{2}+x+1$. If there is a remainder, express the result in the form $q(x)+\frac{r(x)}{b(x)}$. Note: you may not need to use all the boxes.

Step 1 ：Perform long division of \(8x^4 - 2x^3 - 5x^2 + 1\) by \(2x^2 + x + 1\).

Step 2 ：The first term of the quotient is \(4x^2\), because \(8x^4\) divided by \(2x^2\) is \(4x^2\).

Step 3 ：Subtract \(8x^4 + 4x^3 + 4x^2\) from \(8x^4 - 2x^3 - 5x^2\) to get \(-6x^3 - 9x^2 + 1\).

Step 4 ：The next term of the quotient is \(-x\), because \(-6x^3\) divided by \(2x^2\) is \(-3x\) and \(-3x\) divided by \(3\) is \(-x\).

Step 5 ：Subtract \(-6x^3 - 3x^2 - 3x\) from \(-6x^3 - 9x^2 + 1\) to get \(-6x^2 + 4x + 1\).

Step 6 ：The next term of the quotient is \(-1\), because \(-6x^2\) divided by \(2x^2\) is \(-3\) and \(-3\) divided by \(3\) is \(-1\).

Step 7 ：Subtract \(-6x^2 - 3x - 3\) from \(-6x^2 + 4x + 1\) to get \(7x + 4\).

Step 8 ：The remainder is \(-3\), because \(7x + 4\) cannot be divided by \(2x^2 + x + 1\).

Step 9 ：So, the result is \(\boxed{4x^2 - x - 1 - \frac{3}{2x^2 + x + 1}}\).