Question

Find the zeroes of the polynomial $x^2-3$ and verify them

Answer 1

Given that: $p\left(x\right)=x^2-3$ By factorisation, $x^2-3=x^2-(\sqrt{3}{)}^2=(x+\sqrt{3}\left)\right(x-\sqrt{3})$

So, the value of $(x^2-3)$ is zero when $x=\sqrt{3}$ or $x=-\sqrt{3}$

$\therefore$ the zeroes of $(x^2-3)$ are $\sqrt{3}$ and $(-\sqrt{3})$

verification: zeroes of $(x^2-3)$ are $\sqrt{3}$ and $-\sqrt{3}$

$p\left(x\right)=x^2-3\therefore p\left(\sqrt{3}\right)=(\sqrt{3}{)}^2-3=3-3=0$ and $p(-\sqrt{3})=(-\sqrt{3}{)}^2-3=3-3=0$