Question

If $5,-7,2$ are the roots of $l{x}^3+m{x}^2+nx+p=0$, then the roots of $l{x}^3-m{x}^2+nx-p=0$ are

• A. $-5,-7,-2$
• B. $5,7,2$
• C. $7,-5,-2$
• D. $-7,-5,2$

Answer 1

The correct option is C $7,-5,-2$

As $5,-7,2$ are roots of $l{x}^3+m{x}^2+nx+p=0$
If we replace $x\to -x$, we get
$l{(-x)}^3+m{(-x)}^2+n(-x)+p=0\Rightarrow -l{x}^3+m{x}^2-nx+p=0\Rightarrow l{x}^3-m{x}^2+nx-p=0$
Hence, roots of $l{x}^3-m{x}^2+nx-p=0$ are $-5,7,-2$