Q5(ii):

Factorise : (ii) $x^3 – 3x^2 – 9x – 5$

Initial Setup & Given Polynomial

We are tasked with factorising the cubic polynomial. Let the given polynomial be defined as:

$P(x) = x^3 - 3x^2 - 9x - 5$

Step 1: Applying the Rational Root Theorem & Factor Theorem

To find the first linear factor of the cubic polynomial, we utilize the Rational Root Theorem. The theorem states that any rational root of the polynomial must be a divisor of the constant term ($-5$) divided by a divisor of the leading coefficient ($1$).

The possible rational roots are the divisors of $-5$, which are: $\pm 1, \pm 5$.

We evaluate $P(x)$ at these test values to find a root (where $P(x) = 0$):

• Testing $x = 1$:
  $P(1) = (1)^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16$
  Since $P(1) \neq 0$, $x = 1$ is not a root.


• Testing $x = -1$:
  $P(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5$
  $P(-1) = -1 - 3(1) + 9 - 5$
  $P(-1) = -1 - 3 + 9 - 5 = 0$

Since $P(-1) = 0$, we conclude [By the Factor Theorem] that $(x + 1)$ is a factor of the polynomial $P(x)$.

Step 2: Algebraic Manipulation (Factorisation by Grouping)

Knowing that $(x + 1)$ is a factor, we can rewrite the terms of the original polynomial $P(x)$ such that $(x + 1)$ can be factored out from grouped terms. This method is mathematically more elegant than polynomial long division.

We systematically split the middle terms to force the appearance of $(x + 1)$:

$P(x) = x^3 - 3x^2 - 9x - 5$

[Split $-3x^2$ into $+x^2 - 4x^2$ to pair with $x^3$]
$P(x) = x^3 + x^2 - 4x^2 - 9x - 5$

[Split $-9x$ into $-4x - 5x$ to pair with $-4x^2$]
$P(x) = x^3 + x^2 - 4x^2 - 4x - 5x - 5$

[Group the terms into pairs]
$P(x) = (x^3 + x^2) - (4x^2 + 4x) - (5x + 5)$

[Factor out the greatest common monomial from each group]
$P(x) = x^2(x + 1) - 4x(x + 1) - 5(x + 1)$

[Factor out the common binomial $(x + 1)$]
$P(x) = (x + 1)(x^2 - 4x - 5)$

Step 3: Factorising the Quadratic Quotient

We are now left with a linear factor and a quadratic factor: $Q(x) = x^2 - 4x - 5$. We must factorise this quadratic expression by splitting the middle term.

We seek two numbers that multiply to the constant term ($-5$) and add up to the coefficient of the middle term ($-4$). These numbers are $-5$ and $1$.

$x^2 - 4x - 5$
$= x^2 - 5x + 1x - 5$
$= x(x - 5) + 1(x - 5)$
$= (x - 5)(x + 1)$

Step 4: Final Synthesis of Factors

Substitute the fully factorised quadratic back into the equation from Step 2:

$P(x) = (x + 1) \cdot [(x - 5)(x + 1)]$
$P(x) = (x + 1)(x + 1)(x - 5)$
$P(x) = (x + 1)^2(x - 5)$

Visual Verification: Graph of the Polynomial

The graph below illustrates the function $y = x^3 - 3x^2 - 9x - 5$. Notice that the curve is tangent to the x-axis at $x = -1$ (indicating a root with a multiplicity of 2, hence $(x+1)^2$) and intersects the x-axis at $x = 5$ (indicating a root with a multiplicity of 1, hence $(x-5)$).

Final Solution: The completely factorised form of the polynomial $x^3 - 3x^2 - 9x - 5$ is $(x + 1)^2(x - 5)$.