Question

1) P² + P - (a+1) (a+ 2)
2) x²+3x - a² - a +2​

Answer 1

Answer:

✨ Solution (1)

p² + p - ( a + 1 ) ( a + 2 )

=> p² + { ( a + 2 ) - ( a + 1 ) } p - ( a + 1 ) ( a + 2 )

=> p² + ( a + 2 ) p - ( a + 1 ) p - ( a + 1 ) ( a - 2 )

=> p ( p + a + 2 ) - ( a + 1 ) ( p + a + 2 )

=> (p + a + 2 ) { p - ( a + 1 ) }

=> ( p + a + 2 ) ( p - a - 1 )

✨ Solution (2)

x² + 3x - a² - a + 2

=> x² +3x - ( a² + a - 2 )

=> x² +3x - ( a² +2a - a -2 )

=> x²+3x - { a ( a + 2 ) - 1 ( a + 2 )}

=> x² +3x - ( a + 2 ) ( a - 1 )

=> x² + { ( a + 2 ) - ( a - 1 ) } x - ( a + 2 ) ( a - 1 )

=> x² + ( a + 2 ) x - ( a -1 ) x - ( a + 2 ) ( a - 1 )

=> x (x + a + 2 ) - ( a - 1 ) ( x + a + 2 )

=> ( x + a + 2 ) { x - ( a - 1 ) }

=> ( x + a + 2 ) ( x - a + 1 )

Answer 2

Required Solutions:-

1) p² + p - (a+1)(a+2)

By middle Term Factorising,

= $\sf{p^2 - p(a+1) + p(a+2) - (a+1)(a+2)}$

= $\sf{p^2 + p(a+2) - p(a+1) - (a+1)(a+2)}$

= $\sf{p[p + (a+2)] - (a+1)[p + (a+2)]}$

= $\sf{p(p+a+2) - (a+1)(p+a+2)}$

= $\sf{(p+a+2)[p-(a+1)]}$

= $\sf{(p+a+2)(p-a-1)}$

Hence Factorised.

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2) x² + 3x - a² - a + 2

= $\sf{x^2 + 3x - (a^2 + a - 2)}$

By middle term factorizing,

= $\sf{x^2 + 3x - (a^2 +2a - a -2)}$

= $\sf{x^2 + 3x - [a(a + 2) -1(a+2)]}$

= $\sf{x^2 + 3x - [(a+2)(a-1)]}$

= $\sf{x^2 + 3x - (a+2)(a-1)}$

= $\sf{x^2 + [(a+2)-(a-1)]x - (a+2)(a-1)}$

= $\sf{x^2 + (a+2)x - (a-1)x - (a+2)(a-1)}$

= $\sf{x[x + (a+2)] - (a-1)[x+(a+2)]}$

= $\sf{x[x + a + 2] - (a-1)[x+a+2]}$

= $\sf{(x+a+2)[x-(a-1)]}$

= $\sf{(x+a+2)(x-a+1)}$

Hence Factorized.

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