Center of symmetry of a cubic

Problem

A cubic polynomial $f(x)$ with real coefficients is divided by $x+6$. The quotient is written $q(x)$, and the remainder is 3. It is given that $q(x)$ attains a maximum of 8 at $x=-6$. The question asks for the center of symmetry of the graph $y=f(x)$.

The full paper is the CEEC 114 GSAT Mathematics A paper.

Where it goes wrong

The CEEC item statistics give a correct rate of $48\%$ for this item, $88\%$ in the high-scoring group and only $10\%$ in the low-scoring group. The step that sticks is treating the high point $(-6,8)$ of $q(x)$ as the center of $f(x)$, and forgetting that 8 belongs to the quotient. The value of the original function at $x=-6$ is fixed by the remainder 3.

Idea

By polynomial division,

$q(x)$ is a quadratic that opens downward, with vertex $(-6,8)$, so

Substitute back:

Set $t=x+6$. Then $f(-6+t)-3=a{t}^3+8t$. Replacing $t$ by $-t$ only changes the sign of the right-hand side, so

The average of the two $x$-coordinates is $-6$, and the average of the two $y$-coordinates is 3, so the center of symmetry is $(-6,3)$. The argument uses only polynomial division, the vertex form of a quadratic, and point symmetry. It does not use derivatives.

What the figure shows

• Pick a center. Compare (-6, 8) with (-6, 3), and separate the high point of the quotient from the center of symmetry of $f$.
• Drag h. On the left, the two points of the quotient stay at the same height, and on the right $P$ and $Q$ move with it.
• The midpoint $M$ of the segment $PQ$ on the right stays at $(-6,3)$.