Question

The number of distinct zeros of the cubic polynomial $f\left(x\right)=x^3-9x^2+26x-24$ is

• A. 03
• B. 3.00
• C. 3.0
• D. 3

Answer 1

Answer: (A), (B), (C), (D)

$f\left(x\right)=x^3-9x^2+26x-24$

Let's put $x=1$ in $f\left(x\right)$, we get
$f\left(1\right)=1^3-9(1{)}^2+26(1)-24=-6\ne 0$

Let's put $x=2$ in $f\left(x\right)$, we get
$f\left(2\right)=2^3-9(2{)}^2+26(2)-24=60-60=0$

Hence,$x=2$ is a root of $f\left(x\right)$ or $(x-2)$ is its factor.
Using division algorithm $f\left(x\right)=(x-2)(x^2-7x+12)$
$f\left(x\right)=(x-2)(x^2-3x-4x+(-3\left)\right(-4\left)\right)$
$\therefore f\left(x\right)=(x-2)(x-3)(x-4)$

Distinct zeros of f(x) are $2,3,4$