What are the real zeroes of x3 + 6 x2 – 9x - 54? A. 1,2, 27B. 3, -3, -6 c. -6,3, -6D. 2,-1, 18E. 3,3, -6
Accepted Solution
A:
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.