One root of a certain third-degree equation is 1. When the cubic term of the equation is crossed off, the resulting quadratic equation has a root of 2. When the squared term is also crossed off, the resulting linear equation has a root of 3. Reconstruct the original third-degree equation, expressing it in the form
ax3 + bx2 + cx + d = P(x)
We work backwards.
cx + d = 0 when x = 3, so 3c + d = 0, and d = -3c.
P(x) is now ax3 + bx2 + cx – 3c.
bx2 + cx – 3c = 0 when x = 2, so 4b + 2c – 3c = 0 → 4b – c = 0 → c = 4b and
d = -3(4b) = -12b.
P(x) is now ax3 + bx2 + 4bx – 12b.
1 is a root: a + b + c + d = 0 → 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) = 7x3 + x2 + 4x -12.