Question

Find the quadratic polynomial whose zeroes are $3$ and $-5$.

Answer 1

Step 1: Define the quadratic polynomial in terms of its zeros

A quadratic polynomial can be expressed using its zeros as follows,

$x^2-\left(\alpha +\beta \right)x+\left(\alpha \beta \right)$ $...\left(\mathrm{i}\right)$

Where, $\alpha ,\beta$ are the zeroes (roots) of the quadratic polynomial.

It is given that the zeroes of the required quadratic polynomial are $3$ and $-5$, i.e., $\alpha =3$ and $\beta =-5$

So, the sum of zeroes $=\left(\alpha +\beta \right)=3+\left(-5\right)=3-5=-2$

And, the product of zeroes $=\left(\alpha \beta \right)=\left(3\right)\left(-5\right)=-15$

Step 2: Form the quadratic polynomial

Substitute the values of the sum and product of zeroes in equation $\left(\mathrm{i}\right)$ to get the required quadratic polynomial,

$x^2-\left(\alpha +\beta \right)x+\left(\alpha \beta \right)$

$=x^2-\left(-2\right)x+\left(-15\right)$

$=x^2+2x-15$ $\left[\because -\left(-a\right)=+a,+\left(-a\right)=-a\right]$

Hence, the required quadratic polynomial is $x^2+2x-15$.