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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.
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 $(2x + 3y - 4z)^2$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
CBSE Solutions for Class 9 Mathematics Polynomials
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