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Q5(ii):
Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$

Solution :

Given Polynomial & Algebraic Foundation

We are tasked with factorising the following multivariable polynomial of degree 2:

$P(x,y,z) = 2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$

The structure of this expression—comprising three squared terms and three cross-product terms—directly corresponds to the algebraic identity for the square of a trinomial. [Per the standard algebraic expansion theorem]:

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

Geometric Area Model of a Trinomial Square A square divided into 9 regions demonstrating the expansion of (a+b+c) squared. ab ac ab bc ac bc a b c a b c Geometric Area Model: (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac

Step 1: Extracting the Base Magnitudes

We begin by equating the pure squared terms from the given polynomial to the squared terms in the identity to find the absolute values of $a$, $b$, and $c$.

  • $a^2 = 2x^2 \implies |a| = \sqrt{2}x$
  • $b^2 = y^2 \implies |b| = y$
  • $c^2 = 8z^2 \implies |c| = \sqrt{8}z = 2\sqrt{2}z$ [Since $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$]

Step 2: Determining the Algebraic Signs

To assign the correct positive ($+$) or negative ($-$) signs to $a$, $b$, and $c$, we must analyze the signs of the cross-product terms in the given polynomial:

Cross-Product Term Given Value Sign Analysis
$2ab$ $-2\sqrt{2}xy$ Negative. [Implies $a$ and $b$ have opposite signs].
$2bc$ $+4\sqrt{2}yz$ Positive. [Implies $b$ and $c$ have the same sign].
$2ca$ $-8xz$ Negative. [Implies $c$ and $a$ have opposite signs].

From the analysis above, $b$ and $c$ share the same sign, while $a$ has the opposite sign to both. We can conventionally set $b$ and $c$ as positive, which forces $a$ to be negative. Therefore, we define our terms as:

  • $a = -\sqrt{2}x$
  • $b = y$
  • $c = 2\sqrt{2}z$

Note: Alternatively, setting $a$ as positive forces $b$ and $c$ to be negative. Both conventions are mathematically equivalent.

Step 3: Verifying the Cross-Product Terms

Before finalizing the factorization, we must rigorously verify that our chosen terms reconstruct the original polynomial exactly.

  • Check $2ab$: $2(-\sqrt{2}x)(y) = -2\sqrt{2}xy$ (Matches the given term)
  • Check $2bc$: $2(y)(2\sqrt{2}z) = 4\sqrt{2}yz$ (Matches the given term)
  • Check $2ca$: $2(2\sqrt{2}z)(-\sqrt{2}x) = -4(\sqrt{2} \cdot \sqrt{2})xz = -4(2)xz = -8xz$ (Matches the given term)

Step 4: Constructing the Factored Form

Since all terms perfectly align with the identity $(a + b + c)^2$, we substitute our derived values of $a$, $b$, and $c$ into the factored form.

$(a + b + c)^2 = (-\sqrt{2}x + y + 2\sqrt{2}z)^2$

To express this as a product of its factors, we write the squared binomial as the product of two identical trinomials.

Final Solution: The factorised form of the polynomial is $(-\sqrt{2}x + y + 2\sqrt{2}z)(-\sqrt{2}x + y + 2\sqrt{2}z)$.

(Note: Factoring out a $-1$ yields the equally valid alternative form: $(\sqrt{2}x - y - 2\sqrt{2}z)(\sqrt{2}x - y - 2\sqrt{2}z)$).


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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