Answer :
The real zeroes of this function [tex] x^{3} +2 x^{2} -9x-18[/tex] are: -2,-3 and +3
By hit and trial method, we find that x = -2 will the root of this polynomial.
We do the factorisation of the given polynomial:
[tex]x^{3} +2 x^{2} -9x-18 \\ \\ = x^{2} (x+2)-9 (x+2) \\ \\
we put the expression to zero, to find the zeroes or roots of the polynomial,
= ([tex](x+2) = 0 \\ and \\ (x^{2} -9)=0 \\ \\ x= -2 \\ and \\ x = +3,-3[/tex] \\ \\ [/tex]
By hit and trial method, we find that x = -2 will the root of this polynomial.
We do the factorisation of the given polynomial:
[tex]x^{3} +2 x^{2} -9x-18 \\ \\ = x^{2} (x+2)-9 (x+2) \\ \\
we put the expression to zero, to find the zeroes or roots of the polynomial,
= ([tex](x+2) = 0 \\ and \\ (x^{2} -9)=0 \\ \\ x= -2 \\ and \\ x = +3,-3[/tex] \\ \\ [/tex]