Question

Find all the zeros of $(x^4+x^3-23x^2-3x+60)$, if it is given that two of its zeros are $\sqrt{3}and-\sqrt{3}.$

Answer 1

$\sqrt{3}and-\sqrt{3}$ are the zeros of polynomial $(x^4+x^3-23x^2-3x+60)$

Therefore, $(x–\sqrt{3})(x+\sqrt{3})=x^2–3$ will divide the given polynomial completely

Dividing $(x^4+x^3-23x^2-3x+60)$ by $x^2–3$

Quotient $q\left(x\right)=x^2+x–20$
$=x^2+5x–4x–20$

$=x(x+5)-4(x+5)$

$=(x+5)(x–4)$

Other zeros of the given polynomial are the zeros of q(x)

Therefore, $q\left(x\right)=0\Rightarrow (x+5)(x–4)=0$

Or $x=-5or4$

Hence the zeros of given polynomial are $\sqrt{3},-\sqrt{3},-5,4$