Question

$9x {}^{2} - 30x + 16 - 4y { }^{2} + 12y$
​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: ANSWER \: \: } \mid}}}}}$

$\: \: \: \: \: \: \tt{ \red{9x {}^{2} - 30x + 16 - 4y { }^{2} + 12y }}\\ \\ \tt{ \blue{= 9x {}^{2} - 30x + (25 - 9) - 4y {}^{2} + 12y}} \\ \\ \tt{ \pink{=9x {}^{2} - 30x + 25 - 9 - 4y {}^{2} + 12y }}\\ \\ \tt{ \green{=9x {}^{2} - 30x + 25 - 4y {}^{2} + 12y - 9}} \\ \\ \tt{ \purple{= (9x {}^{2} - 30x + 25) - (4y {}^{2} - 12y + 9)}} \\ \\ \tt{=\{(3x) {}^{2} - 2 \times 3x \times 5 + (5) {}^{2} \} - }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{\{ (2y) {}^{2} - 2 \times 2y \times 3 +( 3) {}^{2} \} }\\ \\ \tt{ \purple{=\{ (3x - 5) {}^{2} \} - \{ (2y - 3) {}^{2} \} }}\\ \\ \tt{ \green{= \{(3x - 5) + (2y - 3) \} \{(3x - 5) - (2y - 3) \} }}\\ \\ \tt{ \pink{= \{3x - 5 + 2y - 3 \} \{3x - 5 - 2y + 3 \}}} \\ \\ \tt{ \blue{=(3x + 2y - 5 - 3)(3x - 2y - 5 + 3)}} \\ \\ \tt{ \red{=\underline{\: (3y - 2y - 8)(3y - 2y - 2) \: }}}$

$\star \: \: \: \: \underline{ \purple{ \text{ \: FORMULA \: \: USED : \: }}}$

$\bold{1. \: (a + b) {}^{2} = a {}^{2} + 2ab + b {}^{2}} \\ \\ \bold{2. \: (a - b) {}^{2} = a {}^{2} - 2ab + b {}^{2}} \\ \\ \bold{3. \: a {}^{2} - b {}^{2} =( a + b)(a - b)}$

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathcal{ \pink{ \: \:Solution \: \: } \mid}}}}} </p><p></p><p>$

$\: \: \: \: \: \: \bf{ 9x {}^{2} - 30x + 16 - 4y { }^{2} + 12y }\\ \\ \bf{= 9x {}^{2} - 30x + (25 - 9) - 4y {}^{2} + 12y} \\ \\ \bf{=9x {}^{2} - 30x + 25 - 9 - 4y {}^{2} + 12y }\\ \\ \bf{=9x {}^{2} - 30x + 25 - 4y {}^{2} + 12y - 9} \\ \\ \bf{= (9x {}^{2} - 30x + 25) - (4y {}^{2} - 12y + 9)} \\ \\ \bf{=\{(3x) {}^{2} - 2 \times 3x \times 5 + (5) {}^{2} \} - }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{\{ (2y) {}^{2} - 2 \times 2y \times 3 +( 3) {}^{2} \} }\\ \\ \bf{=\{ (3x - 5) {}^{2} \} - \{ (2y - 3) {}^{2} \} }\\ \\ \bf{= \{(3x - 5) + (2y - 3) \} \{(3x - 5) - (2y - 3) \} }\\ \\ \bf{= \{3x - 5 + 2y - 3 \} \{3x - 5 - 2y + 3 \}} \\ \\ \bf{=\underline{\: (3x + 2y - 8)(3x - 2y - 2) \: }}$