# How do you find a polynomial function that has zeros 0, -2, -3?

Jan 16, 2017

$f \left(x\right) = {x}^{3} + 5 {x}^{2} + 6 x$

#### Explanation:

Since we are given the zeroes of the polynomial function, we can write the solution in terms of factors.

In general, given 3 zeroes of a polynomial function, a, b, and c, we can write the function as the multiplication of the factors $\left(x - a\right) , \left(x - b\right) , \mathmr{and} \left(x - c\right)$

Simply:

$f \left(x\right) = \left(x - a\right) \left(x - b\right) \left(x - c\right)$

In this case, we can show that each of a, b, and c are zeroes of the function:

$f \left(a\right) = \left(a - a\right) \left(a - b\right) \left(a - c\right) = \left(0\right) \left(a - b\right) \left(a - c\right) = 0$

$f \left(b\right) = \left(b - a\right) \left(b - b\right) \left(b - c\right) = \left(b - a\right) \left(0\right) \left(b - c\right) = 0$

$f \left(a\right) = \left(c - a\right) \left(c - b\right) \left(c - c\right) = \left(c - a\right) \left(c - b\right) \left(0\right) = 0$

Since the value of the function at x=a, b and c is equal to 0, then the function $f \left(x\right) = \left(x - a\right) \left(x - b\right) \left(x - c\right)$ has zeroes at a, b, and c.

With the generalized form, we can substitute for the given zeroes, $x = 0 , - 2 , \mathmr{and} - 3$, where $a = 0 , b = - 2 , \mathmr{and} c = - 3$.

$f \left(x\right) = \left(x - 0\right) \left(x - \left(- 2\right)\right) \left(x - \left(- 3\right)\right)$

Simplifying gives:

$f \left(x\right) = x \left(x + 2\right) \left(x + 3\right)$

From here, we can put it in standard polynomial form by foiling the right side:

$f \left(x\right) = x \left({x}^{2} + 5 x + 6\right)$

And distributing the x yields a final answer of:

$f \left(x\right) = {x}^{3} + 5 {x}^{2} + 6 x$

To double check the answer, just plug in the given zeroes, and ensure the value of the function at those points is equal to 0.

$f \left(0\right) = {\left(0\right)}^{3} + 5 {\left(0\right)}^{2} + 6 \left(0\right) = 0$

$f \left(- 2\right) = {\left(- 2\right)}^{3} + 5 {\left(- 2\right)}^{2} + 6 \left(- 2\right) = - 8 + 20 - 12 = 0$

$f \left(- 3\right) = {\left(- 3\right)}^{3} + 5 {\left(- 3\right)}^{2} + 6 \left(- 3\right) = - 27 + 45 - 18 = 0$

Thus, the function has zeroes as given by x=0, -2, and -3.