Question

Find a quadratic polynomial whose zeroes are $-1\text{and}3$ . Verify the relation between the coefficient and zeroes of the polynomial.

Answer 1

Step 1: Form a quadratic polynomial:

Here, the zeroes are $\alpha =1\text{and}\beta =-3$.

$\begin{array}{rcl}\text{sum of zeroes}& =& \alpha +\beta \\ & =& 1-3\\ & =& -2\\ & =& -\frac{b}{a}\end{array}$

$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& 1\times \left(-3\right)\\ & =& -3\\ & =& \frac{c}{a}\end{array}$

Therefore, we can conclude that $a=1,b=2,c=-3$.

The standard form of polynomial is, $p\left(x\right)=a{x}^2+bx+c$.

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

Step 2. Verification:

$\begin{array}{rcl}\text{sum of zeroes}& =& \alpha +\beta \\ & =& 1-3\\ & =& -2\end{array}$

$\begin{array}{rcl}-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}& =& \frac{-b}{a}\\ & =& \frac{-2}{1}\\ & =& -2\end{array}$

$\begin{array}{rcl}\text{Product of zeroes}& =& \alpha \beta \\ & =& 1\times \left(-3\right)\\ & =& -3\end{array}$

$\begin{array}{rcl}\frac{\text{constant term}}{\text{coefficient of}{x}^2}& =& \frac{c}{a}\\ & =& \frac{-3}{1}\\ & =& -3\end{array}$

Hence, $p\left(x\right)=x^2+2x-3$ is the required quadratic polynomial and the relationship between the zeroes and the coefficients is verified.