Question

How do you write a polynomial function of least degree with integral coefficients that has the given zeros $-3,-\frac{1}{3},5$

Answer 1

Use the given roots to find the cubic polynomial:

Given roots are $-3,-\frac{1}{3},5$,

Since there are three roots of a polynomial function that means the degree will be $3$ or it will be a cubic polynomial,

Therefore, the polynomial can be given as,

$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$

Here, $a=-3,b=-\frac{1}{3},c=5$

$$\begin{gathered} a+b+c=-3+\left(\frac{-1}{3}\right)+5 \\ =\frac{5}{3} \\ abc=-3\times \left(\frac{-1}{3}\right)\times 5 \\ =5 \\ ab+bc+ca=-3\times \left(\frac{-1}{3}\right)+\left(\frac{-1}{3}\right)\times 5-3\times 5 \\ =-\frac{47}{3} \end{gathered}$$

Thus, $x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=x^3-\frac{5}{3}{x}^2-\frac{47}{3}x-\frac{15}{3}$

Hence, the polynomial can be given as $\mathbf{3}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{47}\mathit{x}\mathbf{-}\mathbf{15}$.