Cubic to Quadratic to Linear

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, with all coefficients as integers.


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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 → 4bc = 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 = –bcd = –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.

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