Question

What are the real zeroes of x3 + 6 x2 – 9x - 54?

Answer 1

Answer:

Option B 3,-3,-6 is correct.

Step-by-step explanation:

We need to find real zeroes of $x^3+6x^2-9x-54$

Solving

$x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)$

Taking x^2 common from first 2 terms and -9 from last two terms we get

$=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)$

Taking (x+6) common

$(x+6)(x^2-9)$

x^2-9 can be solved using formula a^2-b^2 = (a+b)(a-b)

$=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)$

Putting it equal to zero,

$(x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\, x=3$

So, Option B 3,-3,-6 is correct.