Divide x^3 - 7x + 6 by -2x + 1. Use synthetic division, handling the fact that the divisor does not have leading coefficient 1.

Step-by-step solution

1. 1

   Recognise why plain synthetic division does not apply yet. The method is built for divisors of the shape $x-c$, with leading coefficient exactly $1$. Here the divisor is $-2x+1$, so first pull the leading coefficient out:

   $-2x+1=-2\left(x-\tfrac{1}{2}\right)$

   That rewriting is the whole trick: divide by the monic part $x-\tfrac12$ with the table, then account for the factor $-2$ afterwards.

2. 2

   Divide by $x-\tfrac12$ with $c=\tfrac12$. Pad the missing $x^2$ term with a zero, giving coefficients $1,\;0,\;-7,\;6$:

   $\begin{array}{r|rrrr}\tfrac12 & 1 & 0 & -7 & 6\\[2pt] & & \tfrac12 & \tfrac14 & -\tfrac{27}{8}\\ \hline & 1 & \tfrac12 & -\tfrac{27}{4} & \tfrac{21}{8}\end{array}$

   So $x^3-7x+6=\left(x-\tfrac12\right)\left(x^2+\tfrac12 x-\tfrac{27}{4}\right)+\tfrac{21}{8}$, and the remainder agrees with $f\!\left(\tfrac12\right)=\tfrac18-\tfrac72+6=\tfrac{21}{8}$.

3. 3

   Divide the quotient — and only the quotient — by $-2$. Since $x^3-7x+6=-2\left(x-\tfrac12\right)\cdot Q(x)+R$, the quotient with respect to $-2x+1$ is the previous quotient divided by $-2$:

   $Q(x)=\frac{x^2+\tfrac12 x-\tfrac{27}{4}}{-2}=-\frac{1}{2}x^2-\frac{1}{4}x+\frac{27}{8}$

   The remainder is untouched. This asymmetry is the part people get wrong: scaling the divisor scales the quotient reciprocally but leaves the remainder alone, because the remainder is a constant tied to the point $x=\tfrac12$.

4. 4

   State the division and note the remainder is legitimate. Putting the pieces together,

   $x^3-7x+6=(-2x+1)\left(-\frac{1}{2}x^2-\frac{1}{4}x+\frac{27}{8}\right)+\frac{21}{8}$

   The remainder $\tfrac{21}{8}$ has degree $0$, which is below the degree of the divisor, so the division is complete.

5. 5

   Verify by expanding. Multiply out the right-hand side term by term:

   $(-2x)\left(-\tfrac12 x^2-\tfrac14 x+\tfrac{27}{8}\right)=x^3+\tfrac12 x^2-\tfrac{27}{4}x$

   $(1)\left(-\tfrac12 x^2-\tfrac14 x+\tfrac{27}{8}\right)=-\tfrac12 x^2-\tfrac14 x+\tfrac{27}{8}$

   Adding, the $x^2$ terms cancel and $-\tfrac{27}{4}x-\tfrac14 x=-7x$, leaving $x^3-7x+\tfrac{27}{8}$; plus the remainder $\tfrac{21}{8}$ gives constant $\tfrac{48}{8}=6$. So the identity reproduces $x^3-7x+6\;\checkmark$.