PLEASE HELPP PLEASE find the zeros and factored form of c(x)=3x^2-3

3 ( x-1) (x+1) is the factored form

zeros are x=1, -1

Step-by-step explanation:

3x^2-3

Factor out a 3

3 ( x^2 -1)

What is inside the parentheses is the difference of squares

( a^2 -b^2) = ( a-b)(a+b)

3 ( x-1) (x+1)

To find the zeros, set this equal to zero

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

Using the zero product property

x-1 =0  x+1 =0

x=1,   x=-1

see explanation

Step-by-step explanation:

Given

c(x) = 3x² - 3 ← factor out 3 from each term

      = 3(x² - 1) ← x² - 1 is a difference of squares

      = 3(x - 1)(x + 1) ← in factored form

To find the zeros equate c(x) to zero, that is

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

Equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

x - 1 = 0 ⇒ x = 1

zeros are x = - 1, x = 1

1. 34

2. 268

3. x=14.4

step-by-step explanation: