It is given that
[tex]g(x)=x^3+3x^2-18x-40[/tex]
The one factor of g(x) is (x+5).
By using the synthetic method, we get
[tex]g(x)=(x+5)(x^2-2x-8)[/tex]
[tex]g(x)=(x+5)(x^2-4x+2x-8)[/tex]
[tex]g(x)=(x+5)(x(x-4)+2(x-4))[/tex]
[tex]g(x)=(x+5)(x-4)(x+2)[/tex]
Hecne zeros of g(x) is -5,4, and -2.
2)
It is given that
[tex]g(x)=x^3+x^2-17x+15[/tex]
The one zero of the given g(x) is x=1.
By using the synthetic method, we get
[tex]g(x)=(x-1)(x^2+2x-15)[/tex]
[tex]g(x)=(x-1)(x^2+3x-5x-15)[/tex]
[tex]g(x)=(x-1)(x(x+3)-5(x+3))[/tex]
[tex]g(x)=(x-1)(x+3)(x-5)[/tex]
To find zeros of g(x) by equating g(x) to zero.
[tex]g(x)=(x-1)(x+3)(x-5)=0[/tex]
[tex](x-1)=0;(x+3)=0;(x-5)=0[/tex]
[tex]x=1;x=-3;\text{ x=5}[/tex]
Hence the zeros of the given function g(x) are 1,-3, and 5.
