Question

The cubic polynomial $f\left(x\right)=x^3-8x^2+19x-12$ has distinct zeros.

• A. $2$
• B. $3$
• C. $1$

Answer 1

The correct option is B $3$

Given: $f\left(x\right)=x^3-8x^2+19x-12$

We look for $x$ such that $f\left(x\right)=0$
Putting $x=1$, we get $f\left(1\right)=1^3-8(1{)}^2+19(1)-12=0$
$\therefore x=1$ is a root of $f\left(x\right)$ or $(x-1)$ is a factor of $f\left(x\right)$
$\therefore f\left(x\right)=(x+1)(x^2-7x+12)$
$f\left(x\right)=(x+1)(x^2-(3+4)x+(3\left)\right(4\left)\right)$
$f\left(x\right)=(x+1)(x-3)(x-4)$

Hence the roots of $f\left(x\right)\text{are}x=1,3,4$
So we have have $3$ distinct roots.