Question

We want to find the zeros of this polynomial: p(x) = (x+2) (2x^2 + 3x-9)

Answer 1

Answer:

x = - 2, x = - 3, x = $\frac{3}{2}$

Step-by-step explanation:

to find the zeros equate p(x) to zero , then

(x + 2)(2x² + 3x - 9) = 0 ← factor the quadratic

(x + 2)(x + 3)(2x - 3) = 0

equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

x + 3 = 0 ⇒ x = - 3

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = $\frac{3}{2}$