Find all the zeros of f(x). f(x) = 2x3 + 7x2 - 28x + 12
Arrange your answers from smallest to largest. If
there is a double root, list it twice.

Plz help!

The zeroes are -6, 1/2 and 2.

Step-by-step explanation:

f(x) = 2x3 + 7x2 - 28x + 12 = 0

From the first and last coefficient 2 and 12, one guess for a zero is x = 2.

So substituting x = 2:

f(2) = 16+ 28 - 56 + 12 = 0

So x = 2 is a zero and x - 2 is a factor of f(x)

Performing long division:

x - 2)2x3 + 7x2 - 28x + 12( 2x2 + 11x - 6 < Quotient

       2x3  - 4x2

                 11x2 - 28x

                 11x2 - 22x

                          - 6x + 12

                           -6x + 12

Now we solve

2x2 + 11x - 6 = 0

(2x - 1)(x + 6) = 0

2x - 1 = 0 or x + 6 = 0, so:

x = 1/2,  x = -6.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6