\begin{tabular}{l} Use synthetic division to divide the two polynomials. \\ \( \qquad-2 x+1)-2 x^{5}+9 x^{4}-12 x^{3}+10 x^{2}+x \) \\ Is the divisor given in \( (x-r) \) form? No \\ What do you need to multiply the divisor by to get a coefficient of 1 ? \\ How many terms are in the dividend? 6 \\ Enter the quotient and remainder. \\ \hline Quotient \\ \hline\end{tabular}

To perform synthetic division, we first need to express the division in a suitable form. The divisor \(-2x + 1\) can be rearranged to \(x - \frac{1}{2}\) after multiplying by \(-1/2\) to get the coefficient of \(1\). Now, let's set up the synthetic division using the root \(r = \frac{1}{2}\). The dividend, \(-2x^5 + 9x^4 - 12x^3 + 10x^2 + x\), can be expressed as the coefficients \([-2, 9, -12, 10, 1, 0]\) since there is no \(x^0\) term (which stands for the constant). Now, through synthetic division, we will perform the following steps: 1. Bring down the first coefficient, which is \(-2\). 2. Multiply \(-2\) by \(\frac{1}{2}\) to get \(-1\), and add it to \(9\) to get \(8\). 3. Multiply \(8\) by \(\frac{1}{2}\) to get \(4\), and add it to \(-12\) to get \(-8\). 4. Multiply \(-8\) by \(\frac{1}{2}\) to get \(-4\), and add it to \(10\) to get \(6\). 5. Multiply \(6\) by \(\frac{1}{2}\) to get \(3\), and add it to \(1\) to get \(4\). 6. Finally, multiply \(4\) by \(\frac{1}{2}\) to get \(2\), and since there is no other coefficient to add it to, the last value \(2\) is the remainder. Thus, the coefficients from synthetic division provide us with the quotient, which is: \[ -2x^4 + 8x^3 - 8x^2 + 6x + 4 \] and the remainder is \(2\). So, the output would be: **Quotient**: \(-2x^4 + 8x^3 - 8x^2 + 6x + 4\) **Remainder**: \(2\)