Question

4 x^2 y - 9 y^3 factorise​

Answer 1

Answer:

$\blue{\boxed{\large{y \: ( \: 2x + 3y \: ) \: \: ( \: 2x - 3y \: )}}}$

Step-by-step explanation:

$\underline{\bold{To \: find}} = > factors \: of \: \: (4 {x}^{2} y - 9 {y}^{3} )$

$\underline{\bold{Concept \: used}} = > \\ 1) \: 4 = 2 \times 2 = {2}^{2}$

$2) \: 9 = 3 \times 3 \: = {3}^{2}$

$3) \: {a}^{2} - {b}^{2} \: = ( \: a + b \: ) \: ( \: a - b \: )$

$\underline{\bold{Solution}} = > \\ 4 {x}^{2} y - 9 {y}^{3} \\ taking \: y \: common \: from \: both \: terms \\ = y \: ( \: 4 {x}^{2} \: - 9 {y}^{2} \: ) \\ = y \: ( {2}^{2} \: {x}^{2} - {3}^{2} \: {y}^{2} ) \\ = y \: ( \: {( \: 2x \: )}^{2} - {( \: 3y \: )}^{2} ) \\ now \: applying \: ( {a}^{2} - {b}^{2} ) = (a + b) \: (a - b) \\ = y \: (2x + 3y) \: (2x - 3y)$

$\underline{\bold{Additional \: information}} = > \\ 1)\blue{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab} \\ 2) \red{{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab} \\ 3)\green{( {a}^{3} + {b}^{3} )= (a + b) \: ( {a}^{2} + {b}^{2} - ab) }\\ 4)\red{( {a}^{3} - {b}^{3}) = (a - b) \: ( {a}^{2} + {b}^{2} + ab \: )}$

Answer 2

Answer:

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