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Q5(i):
Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$

Solution :

Initial Setup & Structural Analysis

We are tasked with factorising the following algebraic expression of six terms:

$P(x, y, z) = 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

[Per the fundamental theorems of polynomial algebra], an expression containing three perfect square terms and three cross-product terms strongly indicates the expansion of a squared trinomial. The governing algebraic identity is:

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Step 1: Extraction of Base Variables (Magnitude)

We must map the first three terms of our polynomial to the squared terms of the identity ($a^2, b^2, c^2$) to find the magnitudes of $a, b,$ and $c$.

  • First term: $a^2 = 4x^2 \implies |a| = \sqrt{4x^2} = 2x$
  • Second term: $b^2 = 9y^2 \implies |b| = \sqrt{9y^2} = 3y$
  • Third term: $c^2 = 16z^2 \implies |c| = \sqrt{16z^2} = 4z$

Step 2: Sign Determination via Cross-Product Analysis

The signs of $a, b,$ and $c$ are determined by analyzing the signs of the cross-product terms ($2ab, 2bc, 2ca$).

The given cross-product terms are:

  • $2ab = +12xy$ (Positive)
  • $2bc = -24yz$ (Negative)
  • $2ca = -16xz$ (Negative)

[By the rules of integer multiplication], the product $2ab$ is positive, which dictates that $a$ and $b$ must share the same sign. Let us assume both $a$ and $b$ are positive:

$a = +2x$

$b = +3y$

The products $2bc$ and $2ca$ are both negative. Since $b$ is positive, for $2bc$ to be negative, $c$ must be negative. Similarly, since $a$ is positive, for $2ca$ to be negative, $c$ must be negative. This confirms that the negative sign originates exclusively from the $z$-term.

$c = -4z$

Step 3: Verification of the Identity

We substitute $a = 2x$, $b = 3y$, and $c = -4z$ back into the expanded identity to ensure absolute mathematical equivalence:

$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
$= 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

The expanded form perfectly matches the original polynomial. Therefore, the expression can be written as the square of the trinomial $(2x + 3y - 4z)$.

Visualizing the Expansion (Tabular Area Model)

To rigorously prove the distribution of terms, we can use a tabular area model. The sum of all cells in the $3 \times 3$ matrix equals the original polynomial, demonstrating how the cross-terms combine.

Tabular Expansion Matrix of (2x + 3y - 4z)² 2x 3y -4z 2x 3y -4z 4x² 6xy -8xz 6xy 9y² -12yz -8xz -12yz 16z²

Notice how the symmetric off-diagonal terms combine perfectly: $(6xy + 6xy = 12xy)$, $(-12yz - 12yz = -24yz)$, and $(-8xz - 8xz = -16xz)$.

Step 4: Final Synthesis

Having established the base terms and their respective signs, we write the expression in its fully factorised form. Since factorisation requires expressing the polynomial as a product of its irreducible factors, we write the squared binomial as the product of two identical brackets.

Final Solution: The factorised form of the given polynomial is $(2x + 3y - 4z)(2x + 3y - 4z)$, which can also be written as $(2x + 3y - 4z)^2$.


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