If 2 is one of the zeros of [tex]x^{3}-3x^{2} -4x-12[/tex] then find the remaining two zeroes. About the author Adeline

Step-by-step explanation: [tex]x = 2 \\ x – 2 = 0 \\ {x}^{3} – 3 {x}^{2} – 4x + 12 = 0 \\ {x}^{3} – 2 {x}^{2} – {x}^{2} + 2x – 6x + 12 = 0 \\ {x}^{2} (x – 2) – x(x – 2) – 6(x – 2) = 0 \\ (x – 2)( {x}^{2} – x – 6) = 0 \\ {x}^{2} – 3x + 2x – 6 = 0 \\ x(x – 3) + 2(x – 3) = 0 \\ (x + 2)(x – 3) = 0 \\ x = 3 , – 2[/tex]other two roots are 3,-2 Reply

Step-by-step explanation:[tex]x = 2 \\ x – 2 = 0 \\ {x}^{3} – 3 {x}^{2} – 4x + 12 = 0 \\ {x}^{3} – 2 {x}^{2} – {x}^{2} + 2x – 6x + 12 = 0 \\ {x}^{2} (x – 2) – x(x – 2) – 6(x – 2) = 0 \\ (x – 2)( {x}^{2} – x – 6) = 0 \\ {x}^{2} – 3x + 2x – 6 = 0 \\ x(x – 3) + 2(x – 3) = 0 \\ (x + 2)(x – 3) = 0 \\ x = 3 , – 2[/tex]other two roots are 3,-2