Q5(iii):

Factorise : (iii) $x^3 + 13x^2 + 32x + 20$

Initial Setup & Given Polynomial

We are tasked with completely factorising the cubic polynomial:

$P(x) = x^3 + 13x^2 + 32x + 20$

Since the highest degree of the polynomial is $3$, it will have at most three linear factors. We will determine these factors systematically using the Factor Theorem and algebraic manipulation.

Step 1: Applying the Rational Root Theorem & Factor Theorem

To find the first linear factor, we look for an integer root by testing the factors of the constant term. The constant term is $20$.

The possible integer roots are the divisors of $20$: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$.

[Logical Deduction: Notice that all coefficients in $P(x)$ are positive. Therefore, substituting any positive value for $x$ will result in a positive sum, meaning $P(x)$ cannot equal zero. Thus, we only need to test the negative divisors.]

Let us test $x = -1$:

$P(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$

$P(-1) = -1 + 13(1) - 32 + 20$

$P(-1) = -1 + 13 - 32 + 20$

$P(-1) = 12 - 32 + 20 = -20 + 20 = 0$

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

Step 2: Factoring out $(x + 1)$ via Algebraic Splitting

Now that we know $(x + 1)$ is a factor, we can rewrite the terms of the original polynomial to factor out $(x + 1)$ by grouping. We split the $x^2$ and $x$ terms to match the coefficients required to pull out $(x + 1)$.

$P(x) = x^3 + 13x^2 + 32x + 20$

• Split $13x^2$ into $x^2 + 12x^2$
• Split $32x$ into $12x + 20x$

Substituting these into the polynomial:

$P(x) = x^3 + x^2 + 12x^2 + 12x + 20x + 20$

Now, group the terms in pairs:

$P(x) = (x^3 + x^2) + (12x^2 + 12x) + (20x + 20)$

Factor out the greatest common monomial from each pair:

$P(x) = x^2(x + 1) + 12x(x + 1) + 20(x + 1)$

Factor out the common binomial $(x + 1)$:

$P(x) = (x + 1)(x^2 + 12x + 20)$

Step 3: Factorising the Quadratic Quotient

We are left with a quadratic polynomial: $Q(x) = x^2 + 12x + 20$. We will factorise this by splitting the middle term.

We need to find two numbers, $a$ and $b$, such that:

• Sum: $a + b = 12$ (the coefficient of $x$)
• Product: $a \cdot b = 20$ (the constant term)

The factors of $20$ are $(1, 20)$, $(2, 10)$, and $(4, 5)$. The pair that adds up to $12$ is $10$ and $2$.

Rewrite the middle term $12x$ as $10x + 2x$:

$Q(x) = x^2 + 10x + 2x + 20$

Group the terms and factorise:

$Q(x) = x(x + 10) + 2(x + 10)$

Factor out the common binomial $(x + 10)$:

$Q(x) = (x + 2)(x + 10)$

Step 4: Final Synthesis & Geometric Interpretation

Combining the linear factor from Step 1 with the two linear factors from Step 3, we get the complete factorisation of the cubic polynomial:

$P(x) = (x + 1)(x + 2)(x + 10)$

[Geometric Justification: The roots of the polynomial are the $x$-intercepts of its graph. Setting $P(x) = 0$ yields $x = -1$, $x = -2$, and $x = -10$. The graph below accurately plots $y = x^3 + 13x^2 + 32x + 20$, demonstrating the curve crossing the $x$-axis exactly at these three calculated roots.]

Final Solution: The complete factorisation of the polynomial is $(x + 1)(x + 2)(x + 10)$.