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

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Answer 1 · 2015-11-24T05:25:56

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

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

$f (x) = (x - 1) (x - 2) (x - 3)$ $f(x) = (x-1)(x-2)(x-3)$

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

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

$f (x) = 3 (x - 1) (x - 2) (x - 3)$ $f(x) = 3(x-1)(x-2)(x-3)$

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

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