help

3)

g(x) = 0 when x = 3, (-sqrt(33)-5)/2, (sqrt(33)-5)/2

1)

Applying the quadratic forumla gives us (x-3)(x^2+5x-2)
 

2)

(x-3)(x^2+5x-2)

1 and 2)

Using the Rational root theorem

\(\left(x-3\right)\frac{x^3+2x^2-17x+6}{x-3}\)

\(\frac{x^3+2x^2-17x+6}{x-3} = x^2+5x-2\)

\(\left(x-3\right)\left(x^2+5x-2\right)\)

3)

 \(0^3+0x^2-17\cdot \:0+6=6\)

-Vinculum