Answer
$1,-2,-3i,3i$
Work Step by Step
Consider $P(x)=x^4+x^3+7x^2+9x-18=0$
In order to get the zeroes of the polynomial $P(x)$ we will have to solve as follows:
$x^4+x^3+7x^2+9x-18=(x-1)(x^3+2x^2+9x+18)=0$
This can be re-written as:$(x-1)[x(x^2+9)+2(x^2+9)]=0$
This gives: $(x-1)(x^2+9)(x+2)=0$
Here, $x-1=0 \implies x=-1\\x+2=0 \implies x=-2\\x^2+9=0 \implies x^2=-9 \\ or, x=\pm i 3$
Thus the zeroes of $P(x)$ are: $1,-2,-3i,3i$
