Answer :
You have the following expression given in the problem:
f(x) = x³ – 2x² – x + 2
Therefore, to find the roots, you must apply the proccedure shown below:
1. You have:
0 = x3 – 2x2 – x + 2
2. Then, when you factor, you obtain:
(x-2)(x-1)(x+1)=0
3. Therefore, you have that the roots are the following:
x1=-1
x2=1
x3=2
f(x) = x³ – 2x² – x + 2
Therefore, to find the roots, you must apply the proccedure shown below:
1. You have:
0 = x3 – 2x2 – x + 2
2. Then, when you factor, you obtain:
(x-2)(x-1)(x+1)=0
3. Therefore, you have that the roots are the following:
x1=-1
x2=1
x3=2
The zeros of the given polynomial function are 1, -1 and 2
Given the polynomial function [tex]f(x) = x^3 - 2x^2 - x + 2,[/tex]
First, we need to assume a value of x to check if it will give us zero after substituting:
Assuming x = 1
[tex]f(1) = 1^3 - 2(1)^2 - 1 + 2\\f(1) = 1 - 2 - 1 + 2\\f(1) = 0[/tex]
Since f(1) =0, hence x - 1 is a factor
Dividing x - 1 by the given polynomial function, we will have:
[tex]\frac{x^3-2x^2-x+2}{x-1} \\= \frac{x^2(x-2)-1(x-2)}{x-1} \\=\frac{(x^2-1)(x-2)}{x-1}\\=\frac{(x+1)(x-1)(x-2)}{x-1}\\=(x+1)(x-2)[/tex]
Equating the result to zero
x + 1 = 0 and x - 2 = 0
x = -1 and 2
Hence the zeros of the given polynomial function are 1, -1 and 2
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