Find the zeros of p(x)= (x-3)(x+3)(2x+1) HELP PLEAEE

3, -3, -1/2

Step-by-step explanation:

There is no need to distribute, just set each parenthesis to equal to zero.

p(x)= (x-3)(x+3)(2x+1)

         X-3=0                     x+3=0                      2X+1=0

           +3  +3                     -3  -3                           -1  -1

             X=3                        X=-3                          2x=-1

                                                                                x=-1/2

So your answers are 3, -3, -1/2

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

no solution

step-by-step explanation:

there is not enough information in the problem to solve.