Question

Show that (x+1),(x+2) and (x-2) are zeroes of p(x) = x3 + x2 - 4x - 4​

Answer 1

Answer:

this is your answer

Step-by-step explanation:

I think

Answer 2

Answer:

All three are the zeroes of the equation.

Step-by-step explanation:

(x+1) = 0

x= -1

p(x) = x^3 + x^2 - 4x - 4​

p(-1) = (-1)^3 + (-1)^2 - 4(-1) - 4

       = (-1) + (1) + 4 - 4

       = 0

So x+1 is a zero.

now,

x+2=0

x = -2

p(x) = x^3 + x^2 - 4x - 4​

p(-2) = (-2)^3 + (-2)^2 - 4(-2) - 4

        = -8 + 4 + 8 - 4

        = 0

So x+2 is a zero.

Now,

x-2=0

x = 2

p(x) = x^3 + x^2 - 4x - 4​

p(x) = (2)^3 + (2)^2 -4(2) - 4

      = 8 + 4 - 8 - 4

      = 0

So x-2 is also a zero.

Please do mark me the brainliest as i had typed and done this myself:)