Question

Factorisation using identities
$(3 x + 2y {)}^{2}$
$(xy - 2 {)}^{2}$
$(x - 7 {)}^{2}$
$49 - {x}^{2}$
$121 {x}^{2} - 81$


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Answer 1

ANSWER✔

$\large {\star}\:\: (3x+2y)^2 \\ \implies (3x)^2+(2y)^2+2(3x)(2y)\:---\boxed{a^2+b^2+2ab} \\ \implies 9x^2+4y^2+12xy$

$\large{\boxed{\bf{9x^2+4y^2+12xy }}}$

$\large {\star}\:\: (xy-2)^2\\ \implies (xy)^2+(2)^2-2(xy)(2)\:--\boxed{a^2+b^2-2ab}\\ \implies x^2y^2+4-4xy$

$\large{\boxed{\bf{x^2y^2+4-4xy }}}$

$\large {\star}\:\: (x-7)^2\\ \implies (x)^2+(7)^2-2(x)(7)\:---\boxed{a^2+b^2-2ab} \\ \implies x^2+49-14x$

$\large{\boxed{\bf{ x^2-14x+49}}}$

$\large {\star}\:\: 49-x^2 \\ \implies 7^2-x^2\\ \implies (7+x)(7-x)\:--\boxed{a^2-b^2= (a+b)(a-b)}$

$\large{\boxed{\bf{ (x+7)(x-7)}}}$

$\large {\star}\:\: 121x^2-81 \\ \implies (11x)^2-(9)^2\\ \implies (11x+9)(11x-9) \:--\boxed{ a^2-b^2= (a+b)(a-b)}$

$\large{\boxed{\bf{ (11x+9)(11x-9)}}}$

___________

Answer 2

(3x+2y)^2 (3x)^2+(2y)^2+2(3x)(2y){a^2+b^2+2ab} 9x^2+4y^2+12xy

⋆(3x+2y)

2

⟹(3x)

2

+(2y)

2

+2(3x)(2y)−−−

a

2

+b

2

+2ab

⟹9x

2

+4y

2

+12xy

9x

2

+4y

2

+12xy

(xy-2)^2(xy)^2+(2)^2-2(xy)(2)a^2+b^2-2ab x^2y^2+4-4xy

⋆(xy−2)

2

⟹(xy)

2

+(2)

2

−2(xy)(2)−−

a

2

+b

2

−2ab

⟹x

2

y

2

+4−4xy

x

2

y

2

+4−4xy

(x-7)^2(x)^2+(7)^2-2(x)(7)a^2+b^2-2ab x^2+49-14x\

⋆(x−7)

2

⟹(x)

2

+(7)

2

−2(x)(7)−−−

a

2

+b

2

−2ab

⟹x

2

+49−14x

x

2

−14x+49

49-x^2 7^2-x^2

(7+x)(7-x)

a^2-b^2= (a+b)(a-b)

49−x

2

⟹7

2

−x

2

⟹(7+x)(7−x)−−

a

2

−b

2

=(a+b)(a−b)

(x+7)(x-7)

(x+7)(x−7)

121x^2-81

(11x)^2-(9)^2

(11x+9)(11x-9) { a^2-b^2= (a+b)(a-b)}

⋆121x

2

−81

⟹(11x)

2

−(9)

2

⟹(11x+9)(11x−9)−−

a

2

−b

2

=(a+b)(a−b)

(11x+9)(11x-9)

(11x+9)(11x−9)