Answer :
To factorize the polynomial [tex]\( x^3 - 2x^2 + x - 2 \)[/tex], we need to follow a systematic approach. Here’s a step-by-step detailed solution:
### Step 1: Check for Rational Roots
We'll use the Rational Root Theorem, which states that any rational root of the polynomial [tex]\( p(x) \)[/tex] with integer coefficients must be of the form [tex]\( \pm \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (here, -2), and [tex]\( q \)[/tex] is a factor of the leading coefficient (here, 1).
Factors of [tex]\(-2\)[/tex]: [tex]\( \pm 1, \pm 2 \)[/tex]
Factors of [tex]\(1\)[/tex]: [tex]\( \pm 1 \)[/tex]
Possible rational roots: [tex]\(\pm 1, \pm 2 \)[/tex]
### Step 2: Test the Possible Rational Roots
We test these possible roots by substituting them into the polynomial:
1. Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ 1^3 - 2 \cdot 1^2 + 1 - 2 = 1 - 2 + 1 - 2 = -2 \quad (\text{not a root}) \][/tex]
2. Substituting [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)^3 - 2 \cdot (-1)^2 + (-1) - 2 = -1 - 2 - 1 - 2 = -6 \quad (\text{not a root}) \][/tex]
3. Substituting [tex]\( x = 2 \)[/tex]:
[tex]\[ 2^3 - 2 \cdot 2^2 + 2 - 2 = 8 - 8 + 2 - 2 = 0 \quad (\text{a root}) \][/tex]
Therefore, [tex]\( x - 2 \)[/tex] is a factor of the polynomial.
### Step 3: Factor Out [tex]\( x - 2 \)[/tex]
Since [tex]\( x - 2 \)[/tex] is a factor, we perform polynomial division to find the quotient:
[tex]\[ \frac{x^3 - 2x^2 + x - 2}{x - 2} \][/tex]
Performing the division:
1. [tex]\( x^3 \div x = x^2 \)[/tex]
2. [tex]\( (x^2)(x - 2) = x^3 - 2x^2 \)[/tex]
3. Subtract: [tex]\( x^3 - 2x^2 + x - 2 - (x^3 - 2x^2) = x - 2 \)[/tex]
Next term:
1. [tex]\( x \div x = 1 \)[/tex]
2. [tex]\( (1)(x - 2) = x - 2 \)[/tex]
3. Subtract: [tex]\( x - 2 - (x - 2) = 0 \)[/tex]
Thus, the polynomial division confirms:
[tex]\[ x^3 - 2x^2 + x - 2 = (x - 2)(x^2 + 1) \][/tex]
### Step 4: Factor the Quadratic
Now, we need to factor the quadratic [tex]\( x^2 + 1 \)[/tex]. We notice that [tex]\( x^2 + 1 \)[/tex] cannot be factored further with real numbers; it is irreducible over the reals. However, it factors over the complex numbers:
[tex]\[ x^2 + 1 = (x + i)(x - i) \][/tex]
### Step 5: Complete Factorization
Combining all the factors, the polynomial can be written as:
[tex]\[ x^3 - 2x^2 + x - 2 = (x - 2)(x + i)(x - i) \][/tex]
### Step 6: Verify with Answer Choices
Examining the given options:
A. [tex]\( (x - 2)(x + i)(x - 1) \)[/tex] - Incorrect
B. [tex]\( (x - 2)(x - 1)(x - 1) \)[/tex] - Incorrect
C. [tex]\( (x + 2)(x + i)(x - 1) \)[/tex] - Incorrect
D. [tex]\( (x + 2)(x - 1)(x - 1) \)[/tex] - Incorrect
None of the provided choices exactly match our computed factorization [tex]\( (x - 2)(x + i)(x - i) \)[/tex]. Therefore, the correct factorization given our steps is different from the choices provided.
### Step 1: Check for Rational Roots
We'll use the Rational Root Theorem, which states that any rational root of the polynomial [tex]\( p(x) \)[/tex] with integer coefficients must be of the form [tex]\( \pm \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term (here, -2), and [tex]\( q \)[/tex] is a factor of the leading coefficient (here, 1).
Factors of [tex]\(-2\)[/tex]: [tex]\( \pm 1, \pm 2 \)[/tex]
Factors of [tex]\(1\)[/tex]: [tex]\( \pm 1 \)[/tex]
Possible rational roots: [tex]\(\pm 1, \pm 2 \)[/tex]
### Step 2: Test the Possible Rational Roots
We test these possible roots by substituting them into the polynomial:
1. Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ 1^3 - 2 \cdot 1^2 + 1 - 2 = 1 - 2 + 1 - 2 = -2 \quad (\text{not a root}) \][/tex]
2. Substituting [tex]\( x = -1 \)[/tex]:
[tex]\[ (-1)^3 - 2 \cdot (-1)^2 + (-1) - 2 = -1 - 2 - 1 - 2 = -6 \quad (\text{not a root}) \][/tex]
3. Substituting [tex]\( x = 2 \)[/tex]:
[tex]\[ 2^3 - 2 \cdot 2^2 + 2 - 2 = 8 - 8 + 2 - 2 = 0 \quad (\text{a root}) \][/tex]
Therefore, [tex]\( x - 2 \)[/tex] is a factor of the polynomial.
### Step 3: Factor Out [tex]\( x - 2 \)[/tex]
Since [tex]\( x - 2 \)[/tex] is a factor, we perform polynomial division to find the quotient:
[tex]\[ \frac{x^3 - 2x^2 + x - 2}{x - 2} \][/tex]
Performing the division:
1. [tex]\( x^3 \div x = x^2 \)[/tex]
2. [tex]\( (x^2)(x - 2) = x^3 - 2x^2 \)[/tex]
3. Subtract: [tex]\( x^3 - 2x^2 + x - 2 - (x^3 - 2x^2) = x - 2 \)[/tex]
Next term:
1. [tex]\( x \div x = 1 \)[/tex]
2. [tex]\( (1)(x - 2) = x - 2 \)[/tex]
3. Subtract: [tex]\( x - 2 - (x - 2) = 0 \)[/tex]
Thus, the polynomial division confirms:
[tex]\[ x^3 - 2x^2 + x - 2 = (x - 2)(x^2 + 1) \][/tex]
### Step 4: Factor the Quadratic
Now, we need to factor the quadratic [tex]\( x^2 + 1 \)[/tex]. We notice that [tex]\( x^2 + 1 \)[/tex] cannot be factored further with real numbers; it is irreducible over the reals. However, it factors over the complex numbers:
[tex]\[ x^2 + 1 = (x + i)(x - i) \][/tex]
### Step 5: Complete Factorization
Combining all the factors, the polynomial can be written as:
[tex]\[ x^3 - 2x^2 + x - 2 = (x - 2)(x + i)(x - i) \][/tex]
### Step 6: Verify with Answer Choices
Examining the given options:
A. [tex]\( (x - 2)(x + i)(x - 1) \)[/tex] - Incorrect
B. [tex]\( (x - 2)(x - 1)(x - 1) \)[/tex] - Incorrect
C. [tex]\( (x + 2)(x + i)(x - 1) \)[/tex] - Incorrect
D. [tex]\( (x + 2)(x - 1)(x - 1) \)[/tex] - Incorrect
None of the provided choices exactly match our computed factorization [tex]\( (x - 2)(x + i)(x - i) \)[/tex]. Therefore, the correct factorization given our steps is different from the choices provided.