Question

One root of f(x)=x^3-9x^2+26x-24 is x = 2. What are all the roots of the function?

Answer 1

Answer:

Solutions are

$x_{1}=4\\x_{2}=2\\ x_{3}=3$

Step-by-step explanation:

The given expression is

$f(x)=x^{3}-9x^{2} +26x-24$

First, we need the divisors of the independent term, which is 24.

24 divisors: 1, 2, 3, 4, 6, 8, 12, 24.

Now, we replace each divisor in the function, and those which gives zero as result, those are gonna be the roots of the equation.

For $x=1$

$f(1)=(1)^{3}-9(1)^{2} +26(1)-24=1-9+26-24=-6$

This means $x=1$ is not a solution.

For $x=2$

$f(2)=(2)^{3}-9(2)^{2} +26(2)-24=8-36+52-24=0$

So, $x=2$ is the first solution.

For $x=3$

$f(3)=(3)^{3}-9(3)^{2} +26(3)-24=27-81+78-24=0$

It's solution.

For $x=4$

$f(4)=(4)^{3}-9(4)^{2} +26(4)-24=64-144+104-24=0$

Therefore, all roots are

$x_{1}=4\\x_{2}=2\\ x_{3}=3$

Answer 2

The roots of the function is 26