Question

Find all the zeros of (x⁴ + x³ − 23x² − 3x + 60), if it is given that two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$.

Answer 1

$\begin{array}{l}\text{Let}f\left(x\right)=x^4+x^3-23x^2-3x+60\\ \text{Since}\sqrt{3}\text{and}-\sqrt{3}\text{are the zeroes of}f\left(x\right),\text{it follows that each one of}(x-\sqrt{3})\text{and}(x+\sqrt{3})\text{is a factor of}f\left(x\right).\\ \text{Consequently,}(x-\sqrt{3})(x+\sqrt{3})=(x^2-3)\text{is a factor of}f\left(x\right).\\ \text{On dividing}f\left(x\right)\text{by}(x^2-3)\text{, we get:}\\ \end{array}$


$\begin{array}{l}\therefore f\left(x\right)=0\\ =>(x^2+x-20)(x^2-3)=0\\ \begin{array}{l}=>(x^2+5x-4x-20)(x^2-3)\\ =>\left[x\right(x+5)-4(x+5\left)\right](x^2-3)\\ \begin{array}{l}\begin{array}{l}=>(x-4)(x+5)(x-\sqrt{3})(x+\sqrt{3})=0\\ =>x=4\text{or}x=-5\mathrm{or}x=\sqrt{3}\text{or}x=-\sqrt{3}\end{array}\\ \text{Hence, all the zeroes are}\sqrt{3}\text{,}-\sqrt{3},4\mathrm{and}-5.\end{array}\end{array}\\ \end{array}$