Question

The roots of $x^3-12x^2+39x-28=0$ are

• A. $1,-4,7$
• B. $1$,$4,-7$
• C. $1$,$4$,$7$
• D. $1,-4,-7$

Answer 1

Answer: (C)

$1$,$4$,$7$

$x^3-12x^2+39x-28=0$

$x=1$ is clearly a root.

$x^3-x^2-11x^2+11x+28x-28=0x^2(x-1)-11x(x-1)+28(x-1)=0(x-1)(x^2-11x+28)=0(x-1)(x-7)(x-4)=0$

$\therefore 1,7,4$ are the roots.