Davide wants to find the zeros of the polynomial p(x)=x3−2x2−9x+18. His work is shown below.

Question

Step 1: p(x)=(x+3)(x−3)(x−2)
Step 2: x+3=0
x−3=0
x−2=0
Step 3: x=2, x=3, x=−3

Explain why Davide’s work shows that 2, 3, and −3 are zeros of p(x) by matching each step with the appropriate justification.
Step 1
Step 2
Step 3

The choices to fill each step are
A)Davide gets rid of the x in each of the equations from Step 2 by setting x=0 . The zeros of a polynomial occur when x=0 .
B)Davide sets each linear factor equal to 0 because the x -value for the zero of a polynomial is 0 .
C)Davide solves for x in each of the equations from Step 2. The zeros of a polynomial are the x -values for which p(x)=0 .
D)Davide factors the polynomial p(x) into linear factors using factoring by grouping and the difference of squares.
E)Davide factors the polynomial p(x) into linear factors using the difference of cubes method.
F)Davide sets each linear factor equal to 0 because of the Zero Product Property.

I said
Step 1: Davide gets rid of the x in each of the equations from Step 2 by setting x=0. The zeros of a polynomial occur when x=0.
Step 2:Davide sets each linear factor equal to 0 because of the Zero Product Property.
Step 3:Davide factors the polynomial p(x) into linear factors using factoring by grouping and the difference of squares.

Please help me and give me feedback. Thank you

Answer 1

step1 is false. The zeroes of p(x) occur when p=0

I like D,F,C

Answer 2

I can verify that the answer is indeed:
Step 1:D
Step 2:F
Step 3:C