2680.11 – Cubic to Quadratic to Linear

$$ax^3 + bx^2 + cx + d = P(x)$$

We work backwards. So had you.

$cx + d = 0$ when $x = 3$, so $3c + d = 0$, and $d = -3c$. $P(x)$ is now $ax^3 + bx^2 + cx - 3c$.

$bx^2 + cx - 3c = 0$ when $x = 2$, so $4b + 2c - 3c = 0 \rightarrow 4b - c = 0 \rightarrow c = 4b$ and $d = -3(4b) = -12b$. $P(x)$ is now $ax^3 + bx^2 + 4bx - 12b$.

$1$ is a root, so $a + b + c + d = 0 \rightarrow a = -b - c - d = -b - 4b + 12b = 7b$.

Now, let $b = 1$ (or anything else we want): $a = 7, b = 1, c = 4, d = -12$. So, at last, $P(x) = 7x^3 + x^2 + 4x -12$.

You can look at the equation and observe that 1 is a root. Put your finger on the cubic term and observe that 2 is a root of the remaining quadratic. Put your finger on the first two terms and observe that 3 is the root of the resulting linear equation. Kind of neat.

Bonus points for anyone who can identify the character in a children's book who said "So had you."