Question 12 Marks
Factorise $:(2a - 3)^2- 2 (2a - 3) (a - 1) + (a - 1)^2$
Answer$(2a - 3)^2- 2 (2a - 3) (a - 1) + (a - 1)^2$
$= [( 2a - 3 ) - ( a - 1 )]^2$
$= [ 2a - 3 - a + 1 ]^2$
$= ( a - 2 )^2$
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Factorise$ :a - b - 4a^2+ 4b^2$
Answer$a - b - 4a^2+ 4b^2$
$= ( a - b ) - 4( a^2 - b^2 )$
$= ( a - b ) - 4( a - b )( a + b ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a - b )[ 1 - 4( a + b )]$
$= ( a - b )[ 1 - 4a - 4b ]$
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Factorise $: x^2+\frac{a^2+1}{a} x+1$
Answer$ x^2+\frac{a^2+1}{a} x+1=0$
$ \therefore x^2+a x+\frac{1}{a} x+1=0$
$ \therefore x(x+a)+\frac{1}{a}(x+a)=0$
$ \therefore(x+a)\left(x+\frac{1}{a}\right)$
$=0$
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Find the value of $: \frac{(18.5)^2-(6.5)^2}{18.5+6.5}$
Answer$ \frac{(18.5)^2-(6.5)^2}{18.5+6.5}$
$= \frac{(18.5+6.5)(18.5-6.5)}{(18.5+6.5)}$
$= 12$
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Find the value of $: \frac{(6.7)^2-(3.3)^2}{6.7-3.3}$
Answer$ \frac{(6.7)^2-(3.3)^2}{6.7-3.3}$
$= \frac{(6.7+3.3)(6.7-3.3)}{(6.7-3.3)}$
$= 10$
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Find the value of $:( 67.8 )^2 - ( 32.2 )^2$
Answer$( 67.8 )^2 - ( 32.2 )^2$
$= ( 67.8 + 32.2 )( 67.8 - 32.2 )$
$= 100\times 35.6$
$= 3560$
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Find the value of $: ( 987 )^2 - (13)^2$
Answer$( 987 )^2 - (13)^2$
$= ( 987 + 13 )( 987 - 13)$
$= 1000 \times 974$
$= 974000$
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Factorise $:2\sqrt3x^2 + x - 5\sqrt3$
Answer$2\sqrt3x^2 + x - 5\sqrt3$
$= 2\sqrt3x^2+ 6x - 5x - 5\sqrt3$
$= 2\sqrt3x( x + \sqrt3 ) - 5( x + \sqrt3 )$
$= ( 2\sqrt3x - 5 )( x + \sqrt3 )$
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Factorise$ : 9x^2+ 3x - 8y - 64y^2$Factorise$ : 9x^2+ 3x - 8y - 64y^2$
Answer$9x^2+ 3x - 8y - 64y^2$
$= 9x^2 - 64y^2 + 3x - 8y$
$= [ (3x)^2 - (8y)^2 ] + ( 3x - 8y )$
$= [( 3x + 8y )( 3x - 8y )] + ( 3x - 8y )$
$= ( 3x - 8y )( 3x + 8y + 1 )$
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Factorise $:x^4+ y^4- 3x^2y^2$
Answer$x^4+ y^4- 3x^2y^2$
$= x^4 + y^4 - 2x^2y^2 - x^2y^2$
$= (x^2)^2 + (y^2)^2 - 2x^2y^2 - x^2y^2$
$= ( x^2 - y^2 )^2 - (xy)^2$
$= ( x^2 - y^2 - xy )( x^2 - y^2 + xy ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
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Factorise$: (x - y)^3- 8x^3$
Answer$( x - y )^3- 8x^3$
$= ( x - y )^3 - ( 2x )^3$
$= ( x - y - 2x )[ (x - y)^2+ 2x(x - y) + (2x)^2]$
[ Using identity $(a^3- b^3) = (a - b)(a^2+ab+ b^2) ]$
$= (- x - y ) [ x^2+ y^2- 2xy + 2x^2- 2xy + 4x^2]$
$=- ( x + y ) [ 7x^2- 4xy + y^2]$
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Factorise$ :a^4 - 343a$
Answer$a^4 - 343a$
$= a( a^3 - 7^3 )$
$= a( a - 7 )[(a)^2 + a x 7 + (7)^2 ] [ \because a^3 - b^3 = ( a - b )( a^2 + ab + b^2 )]$
$= a( a - 7 )( a^2 + 7a + 49 )$
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Factorise $:a^3+ 0.064$
Answer$a^3 + 0.064$
$= (a)^3 + (0.4)^3$
$= ( a + 0.4 )[ (a)^2 - a x 0.4 + (0.4)^2 ] [ \because a^3 + b^3 = ( a + b )( a^2 - ab + b^2) ]$
$ = ( a + 0.4 )( a^2 - 0.4a + 0.16 )$
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Factorise$ :a^6+ 27b^3$
Answer$a^6+ 27b^3$
$= ( a^2 )^3 + ( 3b )^3$
$= ( a^2 + 3b )[ (a^2)^2 - a^2 x 3b + (3b)^2 ] [ \because a^3 + b^3 = ( a + b )( a^2 - ab + b^2 )]$
$= ( a^2 + 3b )[ a^4 - 3a^2b + 9b^2 ]$
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Factorise $:64 - a^3b^3$
Answer$64 - a^3b^3$
$= (4)^3 - (ab)^3$
$= ( 4 - ab )[(4)^2 + 4 x ab + (ab)^2 ] [ \because a^3 - b^3 = ( a - b )( a^2 + ab + b^2 )]$
$= ( 4 - ab )( 16 + 4ab + a^2b^2 )$
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Factorise$ :1 - 8a^3$
Answer$1 - 8a^3$
$= (1)^3 - (2a)^3$
$= ( 1 - 2a )[ (1)^2 + 1 x 2a + (2a)^2 ]$
$[ \because a^3 - b^3 = ( a - b )( a^2 + ab + b^2 )]$
$= ( 1 - 2a )[ 1 + 2a + 4a^2 ]$
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Factorise$ :a^3- 27$
Answer$a^3 - 27$
$= ( a )^3 - ( 3 )^3$
$= ( a - 3 )[ (a)^2 + a x^3 + (3)^2 ] [ \because a^3 - b^3 = ( a - b )( a^2 + ab + b^2 )]$
$= ( a - 3 )[ a^2 + 3a + 9 ]$
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Show that $ :35^3+ 27^3$ is divisible by $62$
Answer$(35^3+ 27^3)$
[Using identity $(a^3+ b^3)=(a + b)(a^2-ab+ b^2)]$
$= ( 35 + 27 )( 35^2+ 35\times 27 + 27^2)$
$= 62 \times ( 35^2+ 35 \times 27 + 27^2)$
Therefore, the number is divisible by $62.$
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Show that $: 13^3- 5^3$ is divisible by $8$
Answer$( 13^3- 5^3)$
[ Using identity $(a^3- b^3) = (a - b)(a^2+ab+ b^2)]$
$= ( 13 - 5 )( 13^2+ 13 \times 5 + 5^2)$
$= 8( 169 + 65 + 25 )$
Therefore$,$ the number is divisible by $ 8.$
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Factorise $: 1029 - 3x^3$
Answer$1029 - 3x^3$
$= 3( 343 - x^3)$
$= 3( 7^3- x^3)$
$= 3( 7 - x )( 7^2+ 7x + x^2)$
$= 3( 7 - x )( 49 + 7x + x^2)$
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Factorise $: 2x^3+ 54y^3- 4x - 12y$
Answer$2x^3+ 54y^3- 4x - 12y$
$= 2 ( x^3+ 27y^3 - 2x - 6y )$
$= 2 [ { (x)^3+ (3y)^3}- 2(x+ 3y) ]$
$[$ Using identity $(a^3+b^3) = (a + b)(a^2-ab+ b^2) ]$
$= 2[ {(x + 3y)(x^2- 3xy + 9y^2)} - 2(x + 3y) ]$
$= 2(x + 3y)(x^2- 3xy + 9y^2- 2)$
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Factorise $:a - b - a^3+ b^3$
Answerwe know that,
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(1)$
$a - b - a^3+ b^3$
$= a - b - ( a^3 - b^3 )$
$= ( a - b ) - ( a - b )[ a^2 + ab + b^2 ] [ From (1) ]$
$= ( a - b )[ 1 - a^2 - ab - b^2 ]$
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Factorise $:8a^3- b^3- 4ax + 2bx$
AnswerWe know that,
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) .....(1)$
$8a^3- b^3- 4ax + 2bx$
$= [ (2a)^3 - (b)^3 ] - 2 \times ( 2a - b )$
$= ( 2a - b )[ (2a)^2 + 2a \times b + (b)^2 ] - 2 x ( 2a - b ) [$ From$(1) ]$
$= ( 2a - b )[ 4a^2 + 2ab + b^2 ] - 2 \times ( 2a - b )$
$= ( 2a - b )[ 4a^2 + 2ab + b^2 - 2x ]$
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Factorise $: a^4- 1$
Answer$a^4 - 1$
$= ( a^2 )^2 - ( 1 )^2$
$= ( a^2 + 1 )( a^2 - 1 ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a^2 + 1 )[ (a)^2 - (1)^2]$
$= ( a^2 + 1 )( a + 1 )( a - 1 )$
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Factorise $: 3a^5- 108a^3$
Answer$3a^5- 108a^3$
$= 3a^3( a^2 - 36 )$
$= 3a^3[( a )^2 - ( 6 )^2]$
$= 3a^3( a - 6 )( a + 6 ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
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Factorise$ : 4a^2b - 9b^3$
Answer$4a^2b - 9b^3$
$= b( 4a^2 - 9b^2 )$
$= b[ (2a)^2 - (3b)^2 ]$
$= b( 2a - 3b )( 2a + 3b ) [ \because a2 - b2 = ( a + b )( a - b )]$
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Factorise$ : 50a^3- 2a$
Answer$50a^3 - 2a$
$= 2a( 25a^2 - 1 )$
$= 2a[ (5a)^2 - (1)^2]$
$= 2a( 5a + 1 )( 5a - 1 ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
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Factorise $: a^2- 81 (b-c)^2$
Answer$a^2- 81 (b-c)^2$
$= ( a )^2 - [ 9( b - c ) ]^2$
$= [ a - ( 9b - 9c )][ a + ( 9b - 9c )] [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a - 9b + 9c )( a + 9b - 9c )$
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Factorise$ : a^2( b + c) - (b + c)^3$
Answer$a^2( b + c) - (b + c)^3$
$= ( b + c )[ a^2 - ( b + c )^2 ]$
$= ( b + c )[( a + b + c )( a - b - c )]$
$= ( b + c )( a + b + c )( a - b - c )$
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Factorise$ : a^2- (2a + 3b)^2$
Answer$a^2- (2a + 3b)^2$
$= a^2 - ( 2a + 3b )^2$
$= ( a - 2a - 3b )( a + 2a + 3b ) [ \because a^2 - b^2 = ( a + b )(a - b)]$
$= ( - a - 3b )( 3a + 3b )$
$= - 3( a + 3b )( a + b )$
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Factorise $: 4 x^2+\frac{1}{(4 x)^2}+1$
Answer$ 4 x^2+\frac{1}{(4 x)^2}+1$
$ =4 x^2+\frac{1}{(4 x)^2}+2-1$
$ =4 x^2+\frac{1}{(4 x)^2}+2 \times 2 x \times \frac{1}{2 x}-1$
$ =\left(2 x+\frac{1}{2 x}+1\right)^2-(1)^2$
$ =\left(2 x+\frac{1}{2 x}+1\right)\left(2 x+\frac{1}{2 x}-1\right)$
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Factorise$: x^2+\frac{1}{x^2}-11$
Answer$ x^2+\frac{1}{x^2}-11$
$ =x^2+\frac{1}{x^2}-2-9$
$ =x^2+\frac{1}{x^2}-2 \times x \times \frac{1}{x}-9$
$ =\left(x-\frac{1}{x}\right)^2-(3)^2$
$ =\left(x-\frac{1}{x}+3\right)\left(x-\frac{1}{x}-3\right)$
View full question & answer→Question 332 Marks
Factorise $: a^2- b^2- (a + b)^2$
Answer$a^2- b^2- (a + b)^2$
$= a^2 - b^2 - ( a^2 + 2ab + b^2 )$
$= a^2 - b^2 - a^2 - 2ab - b^2$
$= - 2ab - 2b^2$
$= - 2b( a + b )$
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Factorise $ :(a + b)^2- a^2+ b^2$
Answer$(a + b)^2- a^2+ b^2$
$= a^2 + 2ab + b^2 - a^2 + b^2$
$= 2ab + 2b^2$
$= 2b( a + b )$
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Factorise$ : x^4+ x^2+ 1$
Answer$x^4+ x^2+ 1$
$= x^4 + 2x^2 + 1 - x^2$
$= (x^2)^2 + 2x^2 + (1)^2 - x^2$
$= ( x^2 + 1 )^2 - (x)^2 [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( x^2 + 1 - x )( x^2 + 1 + x )$
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Factorise $: 25a^2- 9b^2$
Answer$25a^2- 9b^2$
$= ( 5a )^2 - ( 3b )^2$
$= ( 5a - 3b )( 5a + 3b )$
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Factorise$ : 4a^2- 49b^2+ 2a - 7b$
Answer$4a^2- 49b^2 + 2a - 7b$
$= [ ( 2a )^2 - ( 7b )^2] + [ 2a - 7b ]$
$= [ 2a - 7b ][ 2a + 7b ] + [ 2a - 7b ] [ \because a^2 - b^2 = ( a + b )( a - b ) ]$
$= [ 2a - 7b ][ 2a + 7b + 1 ]$
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Factorise $: 4a^2- (4b^2+ 4bc + c^2)$
Answer$4a^2- (4b^2+ 4bc + c^2)$
$= ( 2a )^2 - ( 2b + c )^2$
$= [ 2a - ( 2b + c )][ 2a + (2b + c )] [\because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 2a - 2b - c ][ 2a + 2b + c ]$
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Factorise $: a (a - 1) - b (b - 1)$
Answer$a (a - 1) - b (b - 1)$
$= a^2 - a - b^2 + b$
$= a^2 - b^2 - a + b$
$= ( a + b )( a - b ) - ( a - b ) [\because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a - b )[( a + b ) - 1]$
$= ( a - b )[ a + b - 1 ]$
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Factorise $: a^3+ 2a^2 - a - 2$
Answer$a^3+ 2a^2 - a - 2$
$= a^2( a + 2 ) - 1( a + 2 )$
$= ( a^2 - 1 )( a + 2 )$
$= ( a + 1 )( a - 1 )( a + 2 ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
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Factorise $: 3 - a (4 + 7a)$
Answer$3 - a (4 + 7a)$
$= 3 - 4a - 7a^2$
$= 3 - 7a + 3a - 7a^2$
$= 1( 3 - 7a ) + a( 3 - 7a )$
$= ( 3 - 7a )( a + 1 )$
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Factorise$ : a^2b^2+ 8ab - 9$
Answer$a^2b^2 + 8ab - 9$
$= a^2b^2 + 9ab - ab - 9$
$= ab( ab + 9 ) -1( ab + 9 )$
$= ( ab + 9 )( ab - 1 )$
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Factorise $: a(3a - 2) - 1$
Answer$a(3a - 2) - 1$
$= 3a^2 - 2a - 1$
$= 3a^2 - 3a + a - 1$
$= 3a( a - 1 ) + 1( a - 1 )$
$= ( 3a + 1 )( a - 1 )$
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Factorise $: 24a^3+ 37a^2- 5a$
Answer$24a^3+ 37a^2 - 5a$
$= a( 24a^2 + 37a - 5 )$
$= a( 24a^2 + 40a -3a - 5 )$
$= a \times [ 8a( 3a + 5 ) - 1( 3a + 5 )]$
$= a[( 8a - 1 )( 3a + 5 )]$
$= a( 8a - 1 )( 3a + 5 )$
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Factorise $: 6a^2 - a - 15$
Answer$6a^2 - a - 15$
$= 6a^2 - 10a + 9a - 15$
$= 2a( 3a - 5 ) + 3( 3a - 5 )$
$= ( 2a + 3 )( 3a - 5 )$
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Factorise $:x^2- 3ax - 88a^2$
Answer$x^2- 3ax - 88a^2$
$= x^2 - 11ax + 8ax - 88a^2$
$= x( x - 11a ) + 8a( x - 11a )$
$= ( x + 8a )( x - 11a )$
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Give possible expressions for the length and the breadth of the rectangle whose area is $12x^2- 35x + 25$
Answer$12x^2- 35x + 25$
$= 12x^2- 20x - 15x + 25$
$= 4x(3x - 5) - 5(3x - 5)$
$= (3x - 5)(4x - 5)$
Thus, Length $= (3x - 5)$ and breadth $= (4x - 5)$
OR Length $= (4x - 5)$ and breadth $= (3x - 5)$
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Factorise $: 7\sqrt2x^2 - 10x - 4\sqrt2$
Answer$7\sqrt2x^2 - 10x - 4\sqrt2$
$= 7\sqrt2x^2 - 14x + 4x - 4\sqrt2$
$= 7\sqrt2x( x - \sqrt2 ) + 4( x - \sqrt2 )$
$= ( x - \sqrt2 )( 7\sqrt2x + 4 )$
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Factorise $: 4\sqrt3x^2 + 5x - 2\sqrt3$
Answer$4\sqrt3x^2 + 5x - 2\sqrt3$
$= 4\sqrt3x^2+ 8x - 3x - 2\sqrt3$
$= 4x( \sqrt3x + 2 ) - \sqrt3( \sqrt3x + 2 )$
$= ( \sqrt3x + 2 )( 4x - \sqrt3 )$
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Find trinomial $($quadratic expression$),$ given below$,$ find whether it is factorisable or not. Factorise$,$ if possible.$3x^2+ 4x - 10$
AnswerGiven expression : $3x^2+ 4x - 10$
Comparing with $ax^2 + bx + c$, we get $a = 3, b = 4$, and $c = -10$
$\therefore b^2 - 4ac = (4)^2 - 4(3)(-10) = 16 + 120 = 136,$ which is not a perfect square.
$\therefore 3x^2 + 4x - 10$ is not factorisable.
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