Factorise: 4x^{2} + 9y^{2} + 16z^{2} + 12xy - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given expression is $4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz$.We use the algebraic identity: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} +

Vedclass · Accepted Answer

$(2x + 3y - 4z)^{2}$

Vedclass · Answer

(A) The given expression is $4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz$.We use the algebraic identity: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca$.Comparing the given expression with the identity:$4x^{2} = (2x)^{2}$$9y^{2} = (3y)^{2}$$16z^{2} = (-4z)^{2}$ (Since the terms $-24yz$ and $-16xz$ are negative,the $z$ term must be negative).Now,rewrite the expression:$= (2x)^{2} + (3y)^{2} + (-4z)^{2} + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$Using the identity,this simplifies to:$= (2x + 3y - 4z)^{2}$Thus,the factorised form is $(2x + 3y - 4z)(2x + 3y - 4z)$.

Factorise: $4x^{2} + 9y^{2} + 16z^{2} + 12xy - 24yz - 16xz$

Similar Questions

Find the following products using appropriate identities:
$(i) (x + 3) (x + 3)$
$(ii) (x - 3) (x + 5)$

Evaluate using suitable identities: $(104)^{3}$

Evaluate $(215)^{2}$.

Factorise $9x^{2} + 42x + 49$.

Multiply $x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z$ by $(-z+x-2 y)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module