Question

Find the zeros and number of zeros of $p\left(x\right)=x^2+9x+18$.

Answer 1

The given polynomial is $x^2+9x+18=0$

$x^2+9x+18=0\Rightarrow x^2+3x+6x+18=0\Rightarrow x(x+3)+6(x+3)=0\Rightarrow (x+3)(x+6)=0\Rightarrow (x+3)=0,(x+6)=0\Rightarrow x=-3,x=-6$

Hence, the zeroes of the polynomial $x^2+9x+18$ are $-3,-6$ and therefore, the number of zeroes is two.