Answer:
f(x) = (x -1)(x + 2)(x -3) is the factored form.
Step-by-step explanation:
The given function is f(x) = x³ -2x² - 5x + 6
We have to write it in the completely factored form.
For this we will use the rational roots theorem first.
In the equation x³ - 2x² - 5x + 6 = 0
P = ± multiples of constant term 6 = ± 1, ±2, ±3, ±6
Q = ± multiples of the coefficient of highest degree term = ±1
By theorem factors will be P/Q
Possible rational roots will be 1, ±2, ±3, ±6
Therefore 1 is a confirm rational root. Now we will find the depressed polynomial from synthetic division to find the other rational roots.
1 | 1 -2 -5 6
1 -1 -6
-------------------------
1 -1 -6 0
So the depressed polynomial will be (x² - x - 6).
Now can easily factorize this polynomial to get the rational roots.
x² - x - 6 = x² - 3x + 2x - 6
= x(x - 3) + 2(x - 3) = (x+2)(x-3)
Therefore whole factorized form of the polynomial function will be
f(x) = (x - 1)(x + 2)(x - 3)
The completely factored form of f(x) is [tex]\rm (x-1) (x-3)(x+2)[/tex].
Given;
Equation; [tex]\rm f(x)=x^3-2x^2-5x+6[/tex]
A polynomial is expressed as a product of the simplest possible form and it can be written in small parts so that the polynomial is going for factorization.
To determine the factored form of the equation following all the steps given below.
The factored form of the equation is;
[tex]\rm x^3-2x^2-5x+6=0[/tex]
Consider (x-1) as a factor of f(x).
Let us check (x-1) is a factor of f(x).
x-1 = 0
x = 1
Then,
[tex]\rm x^3-2x^2-5x+6=0\\\\(1)^3-2(1)^2-5(1)+6=0\\\\1-2-5+6=0\\\\-6+6=0\\\\0=0[/tex]
f(1) =0
Then,
The factor of [tex]\rm x^3-2x^2-5x+6=0[/tex]
[tex]\rm x^3-2x^2-5x+6=0\\\\(x-1) (x^2 - x - 6)=0\\\\(x-1) (x^2-3x+2x-6)=0\\\\(x-1) (x(x-3)+2(x-3))=0\\\\(x-1) (x-3) (x+2)=0[/tex]
Hence, the completely factored form of f(x) is [tex]\rm (x-1) (x-3)(x+2)[/tex].
To know more about the Factor form click the link given below.
https://brainly.com/question/24380382
