Answer:
Option B 3,-3,-6 is correct.
Step-by-step explanation:
We need to find real zeroes of [tex]x^3+6x^2-9x-54[/tex]
Solving
[tex]x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)[/tex]
Taking x^2 common from first 2 terms and -9 from last two terms we get
[tex]=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)\\[/tex]
Taking (x+6) common
[tex](x+6)(x^2-9)\\[/tex]
x^2-9 can be solved using formula a^2-b^2 = (a+b)(a-b)
[tex]=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)[/tex]
Putting it equal to zero,
[tex](x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\, x=3[/tex]
So, Option B 3,-3,-6 is correct.
