Question

Write the polynomial of whose zeroes are $-2\text{and}\frac{1}{3}$.

Answer 1

Step 1: Find the sum and product of zeroes:

If $\alpha \text{and}\beta$ are the zeroes of any quadratic polynomial, then quadreatic polynomial= $x^2-\left(\alpha +\beta \right)x+\alpha \beta =0$.

$\alpha +\beta =\frac{-b}{a}$,

$\alpha \beta =\frac{c}{a}$

Here, $\alpha =-2$, $\beta =\frac{1}{3}$.

$\begin{array}{rcl}\text{Sum of zeroes}& =& \alpha +\beta \\ & =& -2+\frac{1}{3}\\ & =& \frac{-5}{3}\end{array}$
$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& \left(-2\right)\left(\frac{1}{3}\right)\\ & =& \frac{-2}{3}\end{array}$

Step 2: Find the polynomial whose zeroes are $-2\text{and}\frac{1}{3}$:
Putting the value of the sum of zeroes and the product of zeroes in the standard formula: $x^2-\left(\alpha +\beta \right)x+\alpha \beta =0$.

We get,

$\begin{array}{rcl}& & \begin{array}{cc}\Rightarrow & x^2-\left(-\frac{5}{3}\right)x-\frac{2}{3}=0\\ \Rightarrow & 3x^2+5x-2=0\end{array}\end{array}$

Hence, the polynomial is $p\left(x\right)=3x^2+5x-2$.