How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

Oct 16, 2016

$f \left(x\right) = 3 {x}^{3} - 5 {x}^{2} - 47 x - 15$

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of $- 3 , - \frac{1}{3} , 5$, the factors are

$\left(x + 3\right) \left(x + \frac{1}{3}\right) \left(x - 5\right)$

Let's take a look at the factor $\left(x + \frac{\textcolor{b l u e}{1}}{\textcolor{red}{3}}\right)$. Using the factor in this form will not result in integer coefficients because $\frac{1}{3}$ is not an integer.

Move the $\textcolor{red}{3}$ in front of the x and leave the $\textcolor{b l u e}{1}$ in place: $\left(\textcolor{red}{3} x + \textcolor{b l u e}{1}\right)$.

When set equal to zero and solved, both
$\left(x + \frac{1}{3}\right) = 0$ and $\left(3 x + 1\right) = 0$ result in $x = - \frac{1}{3}$.

$f \left(x\right) = \left(x + 3\right) \left(3 x + 1\right) \left(x - 5\right)$

Multiply the first two factors.

$f \left(x\right) = \left(3 {x}^{2} + 10 x + 3\right) \left(x - 5\right)$

Multiply/distribute again.

$f \left(x\right) = 3 {x}^{3} + 10 {x}^{2} + 3 x - 15 {x}^{2} - 50 x - 15$

Combine like terms.

$f \left(x\right) = 3 {x}^{3} - 5 {x}^{2} - 47 x - 15$