# What is the third–degree polynomial function such that f(0) = –18 and whose zeros are 1, 2, and 3?

Nov 24, 2015

$f \left(x\right) = 3 {x}^{3} - 18 {x}^{2} + 33 x - 18$

#### Explanation:

To set the zeros of the function, we can first write the function as

$f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

This gives us $f \left(1\right) = f \left(2\right) = f \left(3\right) = 0$, however $f \left(0\right) = - 1 \cdot - 2 \cdot - 3 = - 6$

To fix this, we can multiply by a constant.
$- 18 = 3 \cdot - 6$ so we multiply by $3$ to obtain

$f \left(x\right) = 3 \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

which has the desired properties. Multiplying that out gives us the cubic function

$f \left(x\right) = 3 {x}^{3} - 18 {x}^{2} + 33 x - 18$