Question

pls answer this question i will mark brainliest

Answer 1

Given f(x) = x^3 + 3x^2 - 3px + q.

Given that x + 1 and x + 2 are the factors of f(x).

By factor theorem, we get

= > x + 1 = 0

= > x = -1.

Plug x = -1 in f(x), we get

f(-1) = (-1)^3 + 3(-1)^2 - 3p(-1) + q = 0

       = -1 + 3 + 3p + q = 0

       = 2 + 3p + q = 0

       = 3p + q = -2  ----- (1)



By factor theorem, we get

= > x + 2 = 0

= > x = - 2.

Plug x = -2 in f(x), we get

f(-2) = (-2)^3 + 3(-2)^2 - 3p(-2) + q = 0

        = -8 + 12 + 6p + q = 0

        = 6p + q = -4  ----- (2)



On solving (1) * 2 & (2), we get

6p + 2q = -4

6p + q = -4

--------------------
        -q = 0

          q = 0.



Substitute q = 0 in (1), we get

= > 3p + q = -2

= > 3p + 0 = -2

= > 3p = -2

= > p = -2/3.



Therefore the value of p = -2/3 and q = 0.



Hope this helps!