Question

find all the zeros, given that f(-3) = 0
f(x) = 2x³+5x²-6x-9
3,3,3
-1,-3,3/2
-1,-3,-3
-3,2/3,-1

Explanation:

Step1: Use factor theorem

Since $f(-3)=0$, $(x+3)$ is a factor of $f(x)=2x^3+5x^2-6x-9$.

Step2: Polynomial division

Divide $2x^3+5x^2-6x-9$ by $(x+3)$:
$$\frac{2x^3+5x^2-6x-9}{x+3}=2x^2-x-3$$

Step3: Factor quadratic

Factor $2x^2-x-3$:
$$2x^2-x-3=(2x-3)(x+1)$$

Step4: Find all zeros

Set each factor to 0:
$x+3=0 \implies x=-3$
$2x-3=0 \implies x=\frac{3}{2}$
$x+1=0 \implies x=-1$

Answer:

-1, -3, 3/2