Question

Factorise the following by regrouping the terms.
a) 6xy-y^2+12xz-2yz
b) b^2-ab(1-a)-a^3
c) a^2-a(x+2y)+2xy
d) x^3-2x^2y+3xy^2-6y^3
e) xy(a^2+1)-a(x^2+y^2)


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Answer 1

$\huge \sf {\orange{\underline{\purple{\underline{Question :-}}}}}$

Factorise the following by regrouping the terms.

a) 6xy  - y² + 12xz - 2yz

b) b² - ab(1 - a) - a³

c) a² - a(x + 2y) + 2xy

d) x³ - 2x²y + 3xy² - 6y³

e) xy(a² + 1)-a(x² + y²)

$\huge \sf {\orange {\underline {\pink{\underline{Solution :-}}}}}$

a) $\sf{6xy-y^2+12xz-2yz}$

$\sf{= y(6x-y)+2z(6x-y) }$

$\sf{= (6x-y)(y+2z)}$

◘ The factorised form is (6x - y)(y + 2z).

_________________

b) $\sf{ b^2-ab(1-a)-a^3}$

$\sf{=b^2-ab+a^2b-a^3 }$

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

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

◘ The factorised form is (b - a)(b + a²).

_________________

c) $\sf{a^2-a(x+2y)+2xy}$

$\sf{=a^2-ax-2ay+2xy }$

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

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

◘ The factorised form is (a - x)(a - 2y).

_________________

d) $\sf{x^3-2x^2y+3xy^2-6y^3}$

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

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

◘ The factorised form is (x - 2y)(x² + 3y²).

_________________

e) $\sf{xy(a^2+1)-a(x^2+y^2)}$

$\sf{= a^2xy+xy-ax^2-ay^2}$

$\sf{=a^2xy - ax^2 + xy - ay^2}$

$\sf{=ax(ay-x)-y(ay-x)}$

$\sf{=( ax - y )( ay - x )}$

◘ The factorised form is (ax - y)(ay - x).

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Answer 2

Answer:

Solution :

[ a ]

6xy - y² + 12xz - 2yz

=> y ( 6x - y ) - 12z ( 6x - y )

=> ( 6x - y )( y - 12z )

[ b ]

b² - ab ( 1 - a ) - a³

=> b² - ab + a²b - a³

=> b ( b - a ) + a²( b - a )

=> ( a² + b )( b - a )

[ c ]

a² - a ( x + 2y ) + 2xy

=> a² - ax - 2ay + 2xy

=> a( a - x ) - 2y ( a - x )

=> ( a - x )( a - 2y )

[ d ]

x³ - 2x²y + 3xy² - 6y³

=> x² ( x - 2y ) + 3y² ( x - 2y )

=> ( x² + 3y² )( x - 2y )

[ e ]

xy ( a² + 1 ) - a ( x² + y² )

=> a²xy + xy - ax² - ay²

=> a²xy - ax² + xy - ay²

=> ax ( ay - x ) - y ( ay - x )

=> ( ax - y )( ay - x ) .

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Additional Information -

( a + b )² = a² + 2ab + b²

( a - b )² = a² - 2ab + b²

( a + b )( a - b ) = a² - b²

( a + b )³ = a³ + 3ab ( a + b ) + b³

( a - b )³ = a³ - 3ab ( a + b ) - b³

( a + b + c )³ = a³ + b³ + c³ + 3 ( a + b )( b + c )( c + a )

a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² - ab - bc - ca )

When a + b + c = 0 ,

a³ + b³ + c³ = 3abc .

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