Q5(i):

Factorise : (i) $x^3 – 2x^2 – x + 2$

Initial Setup & Given Polynomial

We are tasked with finding the complete linear factorization of the given cubic polynomial. Let the polynomial be defined as:

$P(x) = x^3 - 2x^2 - x + 2$

Because the degree of the polynomial is $3$ (a cubic polynomial), the Fundamental Theorem of Algebra dictates that it will have exactly $3$ roots in the complex plane, which corresponds to up to $3$ linear factors of the form $(x - r_i)$, where $r_i$ are the roots.

Step 1: Method of Grouping Terms

For polynomials with four terms, the most efficient initial strategy is factorisation by grouping. We partition the polynomial into two distinct binomial groups to identify common algebraic structures.

Group the first two terms and the last two terms:

$P(x) = (x^3 - 2x^2) - (x - 2)$

[Note: A negative sign is factored out from the third and fourth terms, which reverses the sign of the constant term inside the parenthesis, ensuring mathematical equivalence to the original expression.]

Step 2: Extracting the Greatest Common Factor (GCF)

Next, we extract the Greatest Common Factor from each individual binomial group.

• From the first group $(x^3 - 2x^2)$, the GCF is $x^2$. Factoring this out yields: $x^2(x - 2)$.
• From the second group $(x - 2)$, the GCF is simply $1$. Factoring this out yields: $1(x - 2)$.

Substituting these back into the polynomial expression:

$P(x) = x^2(x - 2) - 1(x - 2)$

Step 3: Factoring Out the Common Binomial

Observe that the binomial $(x - 2)$ is now a common factor to both terms in the expression. [Per the Distributive Property of Multiplication over Addition, $ab - cb = (a - c)b$]. We factor out $(x - 2)$:

$P(x) = (x - 2)(x^2 - 1)$

Step 4: Applying Algebraic Identities

The expression is now a product of a linear binomial and a quadratic binomial. The quadratic factor $(x^2 - 1)$ is a classic "Difference of Two Squares".

[Per the standard algebraic identity: $a^2 - b^2 = (a - b)(a + b)$]

By setting $a = x$ and $b = 1$, we can expand the quadratic term:

$x^2 - 1 = (x - 1)(x + 1)$

Substituting this complete factorization back into our equation for $P(x)$:

$P(x) = (x - 2)(x - 1)(x + 1)$

Alternative Verification: The Factor Theorem

To ensure absolute rigor, we can verify this result using the Rational Root Theorem and the Factor Theorem.

The Rational Root Theorem states that any rational root $\frac{p}{q}$ must have $p$ as a factor of the constant term ($2$) and $q$ as a factor of the leading coefficient ($1$). The possible rational roots are $\pm 1, \pm 2$.

Testing $x = 1$:

$P(1) = (1)^3 - 2(1)^2 - (1) + 2 = 1 - 2 - 1 + 2 = 0$

[Per the Factor Theorem, since $P(1) = 0$, $(x - 1)$ is a guaranteed factor of $P(x)$].

Performing polynomial long division or synthetic division of $(x^3 - 2x^2 - x + 2)$ by $(x - 1)$ yields the quotient $(x^2 - x - 2)$. Factoring this resulting quadratic equation yields $(x - 2)(x + 1)$, perfectly corroborating our grouping method.

Visual Representation: Factorization Tree

The following diagram illustrates the hierarchical breakdown of the polynomial into its constituent linear factors.

Final Solution: The complete linear factorization of the polynomial $x^3 - 2x^2 - x + 2$ is $(x - 2)(x - 1)(x + 1)$.