Question

Factorise the following expression:
1) y square+y-132
2) x square-11x-42
3) z square-12z+27

This question is from Algebraic Identities....


Plz solve these 3 questions!​

Answer 1

Solutions :-

$i) \: {y}^{2} + y - 132 \\ \\ : \implies {y}^{2} + 12y - 11y - 132 \\ \\ : \implies y(y + 12) - 11(y + 12) \\ \\ : \implies (y - 11)(y + 12) \\ \\ : \implies y = 11 \: and \: - 12$

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$2) \: {x}^{2} - 11x - 42 \\ \\ : \implies {x}^{2} - 14x + 3x - 42 \\ \\ : \implies x(x - 14) + 3(x - 14) \\ \\ : \implies (x + 3)(x - 14) \\ \\ : \implies x = 14 \: and \: - 3$

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$3) \: {z}^{2} - 12z + 27 \\ \\ : \implies {z}^{2} - 9z - 3z + 27 \\ \\ : \implies z(z - 9) - 3(z - 9) \\ \\ : \implies (z - 9)(z - 3) \\ \\ : \implies z = 9 \: and \: 3$

Answer 2

Solution :-

Case 1 :-

⇒ y² + y - 132

⇒ y² + 12y - 11y - 132

⇒ y ( y + 12 ) - 11 ( y + 12 )

⇒ ( y - 11 ) ( y + 12 )

⇒ y = 11 , - 12

Case 2 :-

⇒ x² - 11x - 42

⇒ x² - 14x + 3x - 42

⇒ x ( x - 14 ) 3 ( x - 14 )

⇒ ( x + 3 ) ( x - 14 )

⇒ x = - 3 , 14

Case 3 :-

⇒ z² - 12z + 27

⇒ z² - 9z - 3z + 27

⇒ z ( z - 9 ) - 3 ( z - 9 )

⇒ ( z - 3 ) ( z - 9 )

⇒ z = 3 , 9