Question

Find the quadratic polynomial, sum of whose zeros is $8$ and their product is $12$. Hence, find the zeros of the polynomial.

Answer 1

Let $\alpha$ and $\beta$ be the zeroes of the
required polynomial $f\left(x\right)$.
Then $(\alpha +\beta )=8$ and $\alpha \beta =12$
$\therefore f\left(x\right)=x^2-(\alpha +\beta )x+\alpha \beta$
$\Rightarrow f\left(x\right)=x^2-8x+12$
Hence, required polynomial
$f\left(x\right)=x^2-8x+12$
$\therefore f\left(x\right)=0$
$\Rightarrow x^2-8x+12=0$
$\Rightarrow x^2-(6x+2x)+12=0\Rightarrow x^2-6x-2x+12=0$
$\Rightarrow x(x-6)-2(x-6)=0$
$\Rightarrow (x-2(x-6)=0$
$\Rightarrow (x-2)=0or(x-6)=0$
$\Rightarrow x=2orx=6$
So, the zeroes of $f\left(x\right)$ are $2$ and $6$.