Question

The rational zeroes of the cubic function $f\left(x\right)=x^3-2x^2-5x+6=0$ are

• A. $2,-1,3$
• B. $2,-1,-3$
• C. $-2,1,3$
• D. $-2,1,-3$

Answer 1

The correct option is C $-2,1,3$

$f\left(x\right)=x^3-2x^2-5x+6$
By using Rational Root Theorem, Possible zeros/roots will be of the form $\frac{p}{q}$, where $p$ is a factor of constant term and $q$ is a factor of leading coefficient.
Here, constant term $=6$ and its factors are $\pm 1,\pm 2,\pm 3,\pm 6$.
Leading coefficient$=1$ and its factors are $\pm 1$
$\therefore$ possible rational zeros of $f\left(x\right)$ are $\pm 1,\pm 2,\pm 3,\pm 6$.
on putting $x=1$,
$f\left(1\right)=(1{)}^3-2(1{)}^2-5\left(1\right)+6=0$
$\therefore$ $(x-1)$ is a factor of $f\left(x\right)$
$\Rightarrow x^3-2x^2-5x+6=(x-1)(x^2-x-6)$
$=(x-1)(x-3)(x+2)$
$\therefore$ Roots of $f\left(x\right)$ are $-2,1,3$