Question 13 Marks
Factorise $: x^2+\frac{1}{x^2}-3$
Answer$ x^2+\frac{1}{x^2}-3$
$ =x^2+\frac{1}{x^2}-2 \times x \times \frac{1}{x}-1$
$ =\left(x-\frac{1}{x}\right)^2-1$
$ =\left(x-\frac{1}{x}\right)^2-(1)^2$
$ =\left(x-\frac{1}{x}-1\right)\left(x-\frac{1}{x}+1\right)\left[\because a^2-b^2=(a+b)(a-b)\right]$
View full question & answer→Question 23 Marks
Factorise $:4x^4+ 9y^4+ 11x^2y^2$
Answer$4x^4+ 9y^4+ 11x^2y^{2}$
$= (2x^2)^2 + (3y^2)^2 + 12x^2y^2 - x^2y^2$
$= (2x^2 + 3y^2)^2 - x^2y^2$
$= (2x^2 + 3y^2)^2 - (xy)^2$
$= ( 2x^2 + 3y^2 - xy )( 2x^2 + 3y^2 + xy)$
$ [ \because a^2 - b^2 = ( a + b )( a - b )]$
View full question & answer→Question 33 Marks
Factorise$ : x^4+y^4-27 x^2 y^2$
Answer$ x^4+y^4-27 x^2 y^2$
$ =\left(x^2\right)^2+\left(y^2\right)^2-2 x^2 y^2-25 x^2 y^2$
$ =\left(x^2-y^2\right)^2-25 x^2 y^2$
$ =\left(x^2-y^2\right)^2-(5 x y)^2 \quad[\because \mathrm{a} 2-\mathrm{b} 2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})]$
$ =\left[\left(x^2-y^2\right)+5 x y\right]\left[\left(x^2-y^2\right)-5 x y\right]$
$ =\left[x^2+5 x y-y^2\right]\left[x^2-5 x y-y^2\right]$
View full question & answer→Question 43 Marks
Factorise $: (9 a)^2+\frac{1}{(9 a)^2}-2-12 a+\frac{4}{3 a}$
Answer$ (9 a)^2+\frac{1}{(9 a)^2}-2-12 a+\frac{4}{3 a}$
$ =(3 a)^2+\frac{1}{(3 a)^2}-2 \times 3 a \times \frac{1}{3 a}-4\left(3 a-\frac{1}{3 a}\right)$
$ =\left(3 a-\frac{1}{3 a}\right)^2-4\left(3 a-\frac{1}{3 a}\right)$
$ =\left(3 a-\frac{1}{3 a}\right)\left[\left(3 a-\frac{1}{3 a}\right)-4\right]$
$ =\left(3 a-\frac{1}{3 a}\right)\left(3 a-4-\frac{1}{3 a}\right)$
View full question & answer→Question 53 Marks
Factorise $: x^2+\frac{1}{(4 x)^2}+1-7 x-\frac{7}{2 x}$
Answer$ x^2+\frac{1}{(4 x)^2}+1-7 x-\frac{7}{2 x}$
$ =(x)^2+\frac{1}{(2 x)^2}+2 \times x \times \frac{1}{2 x}-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)^2-7\left(x+\frac{1}{2 x}\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x+\frac{1}{2 x}-7\right)$
$ =\left(x+\frac{1}{2 x}\right)\left(x-7+\frac{1}{2 x}\right)$
View full question & answer→Question 63 Marks
Factorise : $2(ab+cd) - a^2- b^2+ c^2+ d^2$
Answer$2(ab+cd) - a^2- b^2+ c^2+ d^2$
$= 2ab + 2cd - a^2 - b^2 + c^2 + d^2$
$= c^2+ d^2 + 2cd - a^2 - b^2 + 2ab$
$= ( c^2 + d^2 + 2cd ) - ( a^2 + b^2 - 2ab )$
$= ( c + d )^2 - ( a - b )^2$
$= ( c + d + a - b )( c + d - a + b )$
View full question & answer→Question 73 Marks
Factorise : $3 - 5x + 5y - 12(x - y)^2$
Answer$3 - 5x + 5y - 12(x - y)^2= 3 - 5( x - y ) - 12(x - y)^2$
Let us assume that $x - y = a$
Then the given expression is
$3 - 5x + 5y - 12(x - y)^2$
$= 3 - 5a - 12a^2$
$= 3 - 9a + 4a - 12a^2$
$= 3( 1 - 3a ) + 4a( 1 - 3a )$
$= ( 3 + 4a )( 1 - 3a ) [ $resubstitute the value of $a ]$
$= [ 3 + 4( x - y )][ 1 - 3( x - y )]$
$= ( 3 + 4x - 4y )( 1 - 3x + 3y )$
View full question & answer→Question 83 Marks
Factorise $:4(2x - 3y)^2- 8x+12y - 3$
Answer$4(2x - 3y)^2- 8x+12y - 3$
$= 4(2x - 3y)^2- 4(2x - 3y) - 3$
Let us assume that $ 2x - 3y = a$
Then the given expression is
$4(2x - 3y)^2 - 8x+12y - 3$
$= 4a^2 - 4a - 3$
$= 4a^2 - 6a + 2a - 3$
$= 2a( 2a - 3 ) + 1( 2a - 3)$
$= ( 2a - 3 )( 2a + 1 )$
$= [ 2( 2x - 3y ) - 3 ][ 2( 2x - 3y ) + 1 ]$
$= ( 4x - 6y - 3 )( 4x - 6y + 1 )$
View full question & answer→Question 93 Marks
Factorise : $\mathrm{a}^3-\frac{27}{a^3}$
Answer$ a^3-\frac{27}{a^3}$
$ =(a)^3-\left(\frac{3}{a}\right)^3$
$ =\left(a-\frac{3}{a}\right)\left[a^2+a \times \frac{3}{a}+\left(\frac{3}{a}\right)^2\right]$
$\left[\because a^3+b^3=(a b)\left(a^2\right.\right. \left.\left.+a b+b^2\right)\right]$
$ =\left(a-\frac{3}{a}\right)\left(a^2+3+\frac{9}{a^2}\right)$
View full question & answer→Question 103 Marks
Factorise : $3x^7y - 81x^4y^4$
Answer$3x^7y - 81x^4y^4$
$= 3xy( x^6 - 27x^3y^3 )$
$= 3xy[ (x^2)^3 - ( 3xy )^3 ]$
$= 3xy( x^2 - 3xy )[ (x^2)^2 + x^2 \times 3xy + (3xy)^2 ] [ \because a^3 - b^3 = ( a -b )( a^2 + ab + b^2 )]$
$= 3xy( x^2 - 3xy )[ x^4 + 3x^3y + 9x^2y^2 ]$
$= 3xy [ x( x + 3y) x^2( x^2 + 3xy + 9y^2 ) ]$
$= 3x^4y( x - 3y )( x^2 + 3xy + 9y^2 )$
View full question & answer→Question 113 Marks
Factorise $:a^3- 27b^3+ 2a^2b - 6ab^2$
Answer$a^3- 27b^3+ 2a^2b - 6ab^2$
We know that,
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(1)$
$a^3 - 27b^3 + 2a^2b - 6ab^2$
$= (a)^3 - (3b)^3 + 2ab( a - 3b )$
$= ( a - 3b )[ a^2 + a \times 3b + (3b)^2 ] + 2ab( a - 3b ) [$From$(1)]$
$= ( a - 3b )[ a^2 + 3ab + 9b^2 ] + 2ab( a - 3b )$
$= ( a - 3b )[ a^2 + 3ab + 9b^2 + 2ab ]$
$= ( a - 3b )[ a^2 + 5ab + 9b^2 ]$
View full question & answer→Question 123 Marks
Factorise :$a^6- 7a^3- 8$
AnswerWe know that,
$a^3 + b^3 = ( a + b )( a^2 - ab + b^2 ) ....(1)$
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(2)$
$a^6- 7a^3 - 8$
$= a^6 - 8a^3 + a^3 - 8$
$= a^3( a^3 - 8) + 1( a^3 - 8 )$
$= ( a^3 + 1 )( a^3 - 8 )$
$= ( a^3 + 1^3 )( a^3 - 2^3 )$
$= ( a + 1 )( a^2 - a + 1 )( a - 2 )( a^2 + 2a + 4 )$
$[$ From$(1)$ and $(2) ]$
$= ( a + 1 )( a - 2)( a^2 - a + 1 )( a^2 + 2a + 4 )$
View full question & answer→Question 133 Marks
Factorise :$a^6- b^6$
AnswerWe know that,
$a^3 + b^3 = ( a + b )( a^2 - ab + b^2 ) ....(1)$
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(2)$
$a^6 - b^6$
$= ( a^3)^2 - (b^3)^2$
$= ( a^3 + b^3 )( a^3 - b^3 )$
$= ( a + b )( a^2 - ab + b^2 )( a - b )( a^2 + ab + b^2 )$
$[$ From$(1)$ and $(2) ]$
$= ( a + b )( a - b )( a^2 - ab + b^2 )( a^2 + ab + b^2 )$
View full question & answer→Question 143 Marks
Factorise : $\frac{(8 a)^3}{27}-\frac{b^3}{8}$
Answer$ \frac{(8 a)^3}{27}-\frac{b^3}{8}$
$ =\left(\frac{2 a}{3}\right)^3-\left(\frac{b}{2}\right)^3$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\left(\frac{2 a}{3}\right)^2+\frac{2 a}{3} \times \frac{b}{2}+\left(\frac{b}{2}\right)^2\right]$
$ {\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right] }$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\frac{4 a^2}{9}+\frac{a b}{3}+\frac{b^2}{4}\right]$
View full question & answer→Question 153 Marks
Factorise : $9(a - 2)^2- 16(a + 2)^2$
Answer$9(a - 2)^2- 16(a + 2)^2$
$= [ 3( a - 2 )]^2 - [4( a + 2 )]^2$
$= [ 3( a - 2 ) - 4( a + 2 )][ 3( a - 2) + 4( a + 2 )]$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 3a - 6 - 4a - 8 ][ 3a - 6 + 4a + 8 ]$
$= ( - a - 14 )( 7a + 2 )$
$= - ( a + 14 )( 7a + 2 )$
View full question & answer→Question 163 Marks
Factorise : $25(2a - b)^2- 81b^2$
Answer$25(2a - b)^2- 81b^2$
$= [ 5( 2a - b )]^2 - (9b)^2$
$= [ 5( 2a - b ) - 9b ][ 5( 2a - b ) + 9b ]$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 10a - 5b - 9b ][ 10a - 5b + 9b ]$
$= [ 10a - 14b ][ 10a + 4b ]$
$= 2 \times ( 5a - 7b ) \times 2 \times ( 5a + 2b )$
$= 4( 5a - 7b )( 5a + 2b )$
View full question & answer→Question 173 Marks
Factorise : $4x^4- x^2- 12x - 36$
AnswerFactorise : $4x^4- x^2- 12x - 36$
$= 4x^4 - ( x^2 + 12x + 36 )$
$= ( 2x^2)^2 - ( x^2 + 2x \times 6x + 6^2 )$
$= ( 2x^2)^2 - ( x + 6 )^2$
$= ( 2x^2 + x + 6 )( 2x^2 - x - 6 )$
$= ( 2x^2 + x + 6 )( 2x^2 - 4x + 3x - 6 )$
$= ( 2x^2 + x + 6 )[ 2x( x - 2 ) + 3( x - 2 )]$
$= ( 2x^2 + x + 6 )[ ( x - 2)( 2x + 3 )]$
$= ( 2x^2 + x + 6 )( x - 2 )( 2x + 3 )$
View full question & answer→Question 183 Marks
Factorize : $9a^2- (a^2- 4)^2$
Answer$9a^2- (a^2 - 4)^2$
$= (3a)^2 - (a^2 - 4)^2$
$= [ 3a + (a^2 - 4)][ 3a - (a^2 - 4)]$
$= [ 3a + a^2 - 4 ][ 3a - a^2 + 4 ]$
$= [ a^2 + 3a - 4 ][- a^2 + 3a + 4]$
$= (a + 4)(a - 1)(-a^2+ 3a + 4)$
$=(a + 4)(a - 1)[-(a^2- 3a - 4)]$
$= - (a + 4)(a - 1)(a - 4)(a + 1)$
$= (a + 4)(a - 1)(a + 1)(4 - a).$
View full question & answer→Question 193 Marks
Factorise : $(x^2+ 4y^2- 9z^2)^2- 16x^2y^2$
Answer$(x^2+ 4y^2- 9z^2)^2- 16x^2y^2= (x^2+ 4y^2- 9z^2)^2 - ( 4xy )^2$
$= ( x^2 + 4y^2 - 9z^2 - 4xy )( x^2 + 4y^2 - 9z^2 + 4xy ) [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( x^2 + 4y^2 - 4xy - 9z^2 )( x^2 + 4y^2 + 4xy - 9z^2 )$
$= [( x - 2y )^2 - (3z)^2 ][ ( x + 2y )^2 - (3z)^2]$
$= [( x - 2y ) - 3z ][( x - 2y ) + 3z ][( x + 2y ) - 3z ][( x + 2y ) + 3z ]$
$= [ x - 2y - 3z ][ x - 2y + 3z ][ x + 2y - 3z ][ x + 2y + 3z ]$
View full question & answer→Question 203 Marks
Factorise$ : (a^2+ b^2- 4c^2)^2- 4a^2b^2$
Answer$\left(a^2+b^2-4 c^2\right)^2-4 a^2 b^2$
$=\left(a^2+b^2-4 c^2\right)^2-(2 a b)^2$
$=\left(a^2+b^2-4 c^2-2 a b\right)\left(a^2+b^2-4 c^2+2 a b\right) $
$\left[\because a^2-b^2=(a\right.+b)(a-b)]$
$=\left(a^2+b^2-2 a b-4 c^2\right)\left(a^2+b^2+2 a b-4 c^2\right)$
$=\left[(a-b)^2-(2 c)^2\right]\left[(a+b)^2-(2 c)^2\right]$
$=(a-b+2 c)(a-b-2 c)(a+b+2 c)(a+b-2 c)$
View full question & answer→Question 213 Marks
Factorise : $(a^2- 1) (b^2- 1) + 4ab$
Answer$(a^2- 1) (b^2- 1) + 4ab$
$= a^2b^2 - a^2 - b^2 + 1 + 4ab$
$= a^2b^2 + 1 + 2ab - a^2 - b^2 + 2ab$
$= ( a^2b^2 + 1 + 2ab ) - ( a^2 + b^2 - 2ab )$
$= ( ab + 1)^2 - ( a - b )^2$
$= [( ab + 1 ) - ( a - b )][( ab + 1 ) + ( a - b )] [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ ab + 1 - a + b ][ ab + 1 + a - b ]$
View full question & answer→Question 223 Marks
Factorise : $4x^2- 12ax - y^2- z^2- 2yz + 9a^2$
Answer$4x^2- 12ax - y^2- z^2- 2yz + 9a^2$
$= 4x^2 + 9a^2 - 12ax - y^2 - z^2 - 2yz$
$= ( 2x )^2 + ( 3a )^2 - 12ax - ( y^2 + z^2 + 2yz )$
$= ( 2x - 3a )^2 - ( y + z )^2$
$= [( 2x - 3a ) - ( y + z )][( 2x - 3a ) + ( y + z )]$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 2x - 3a - y - z ][ 2x - 3a + y + z ]$
View full question & answer→Question 233 Marks
Factorise : $a^2+ b^2- c^2- d^2+ 2ab - 2cd$
Answer$a^2+ b^2- c^2- d^2+ 2ab - 2cd$
$= ( a^2 + b^2 + 2ab ) - ( c^2 + d^2 + 2cd )$
$= ( a + b )^2 - ( c + d )^2$
$= [( a + b ) - ( c + d )][( a + b ) + ( c + d )] [\because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a + b - c - d )( a + b + c + d )$
View full question & answer→Question 243 Marks
Factorise : $4xy - x^2- 4y^2+ z^2$
Answer$4xy - x^2- 4y^2+ z^2$
$= z^2- ( x^2+ 4y^2- 4xy )$
$= z^2- ( x - 2y )^2$
$= [ z - ( x - 2y )][ z + ( x - 2y )] [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ z - x + 2y ][ z + x - 2y ]$
View full question & answer→Question 253 Marks
Factorise : $4a^2- 12a + 9 - 49b^2$
Answer$4a^2 - 12a + 9 - 49b^2= (2a)^2 - 12a + (3)^2 - 49b^2$
$= (2a)^2 - 2(2a)(3) + (3)^2 - (7b)^2 ...[\because (a - b)^2= a^2-2ab + b^2]$
$= (2a - 3)^2 - (7b)^2$
$= (2a - 3 - 7b)(2a - 3 + 7b) ...[\because a^2 - b^2 = (a + b)(a - b)]$
View full question & answer→Question 263 Marks
Factorise : $9a^2+ 3a - 8b - 64b^2$
Answer$9a^2+ 3a - 8b - 64b^2$
$= 9a^2- 64b^2+ 3a - 8b$
$= ( 3a )^2 - ( 8b )^2 + 3a - 8b$
$= ( 3a - 8b )( 3a + 8b ) + ( 3a - 8b )$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( 3a - 8b )( 3a + 8b + 1 )$
View full question & answer→Question 273 Marks
Factorise : $(a + b)^3- a - b$
Answer$(a + b)^3 - a - b$
$= ( a + b )^3 - ( a + b )$
$= ( a + b )[ ( a + b )^2 - 1 ]$
$= ( a + b )[ ( a + b )^2 - (1)^2 ]$
$= ( a + b )[( a + b ) + 1 ][( a + b ) - 1] [ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( a + b )( a + b + 1 )( a + b - 1 )$
View full question & answer→Question 283 Marks
Find trinomial $($quadratic expression$),$ given below, find whether it is factorisable or not. Factorise, if possible.$x(2x - 1) - 1$
AnswerGiven expression :$ x(2x - 1) - 1$
Now , $x(2x - 1) - 1 = 2x^2 - x - 1$
Comparing with $ax^2 + bx + c$, we get $a = 2, b = - 1$, and $c = - 1$
$\therefore b^2 - 4ac = (- 1)^2 - 4(2)(-1) = 1 + 8 = 9$, which is a perfect square.
$\therefore 2x^2 - x - 1$ is factorisable.
Now, $2x^2 - x - 1 = 2x2 - 2x + x - 1$
$= 2x( x - 1 ) + 1( x - 1 )$
$= ( 2x + 1 )( x - 1 )$
View full question & answer→Question 293 Marks
Factories: $(x^2- 3x)(x^2- 3x - 1) - 20.$
Answer$(x^2- 3x)(x^2- 3x - 1) - 20$
$= (x^2- 3x)[(x^2- 3x) - 1] - 20$
$= a[a - 1] - 20 ….( $Taking $x^2 - 3x = a )$
$= a^2- a - 20$
$= a^2- 5a + 4a - 20$
$= a(a - 5) + 4(a - 5)$
$= (a - 5)(a + 4)$
$= (x^2- 3x - 5)(x^2- 3x + 4)$
View full question & answer→Question 303 Marks
Factorise : $\frac{1}{35}+\frac{12}{35} a+a^2$
Answer$\frac{1}{35}+\frac{12}{35} a+a^2 $
$=\frac{1}{35}+\left(\frac{1}{5}+\frac{1}{7}\right) a+a^2$
$ =\frac{1}{35}+\frac{1}{5} a+\frac{1}{7} a+a^2$
$=\frac{1}{5}\left(\frac{1}{7}+a\right)+a\left(\frac{1}{7}+a\right) $
$ =\left(\frac{1}{7}+a\right)\left(\frac{1}{5}+a\right)$
View full question & answer→Question 313 Marks
Factorise : $5 - (3a^2- 2a) (6 - 3a^2+ 2a)$
Answer$5 - (3a^2- 2a) (6 - 3a^2+ 2a)$
$= 5 - ( 3a^2 - 2a )[ 6 - ( 3a^2 - 2a )]$
Assume that $3a^2 - 2a = x$
Therefore,
$5 - ( 3a^2 - 2a )( 6 - 3a^2 + 2a )$
$= 5 - x( 6 - x )$
$= 5 - 6x + x^2$
$= 5( 1 - x ) - x( 1 - x )$
$= ( 5 - x )( 1 - x )$
$= ( x - 5 )( x - 1 )$
$= ( 3a^2 - 2a - 5 )( 3a^2 - 2a - 1 )$
$= ( 3a^2 - 5a + 3a -5 )(3a^2 - 3a + a - 1 )$
$= [ a( 3a - 5 ) + 1( 3a - 5)][3a( a - 1) + 1( a - 1)]$
$= ( 3a - 5 )( a + 1 )( 3a + 1 )( a - 1 )$
View full question & answer→Question 323 Marks
Factorise : $1 - 2a - 2b - 3 (a + b)^2$
Answer$1 - 2a - 2b - 3 (a + b)^21 - 2(a + b) - 3(a + b)^2$
Let $a + b = x$
$= 1 - 2x - 3x^2$
$= -3x^2 - 2x + 1$
$= - [3x^2 + 2x - 1]$
$= - [3x^2 + (3 - 1)x - 1]$
$= - [3x^2 + 3x - 1x - 1]$
$= - [3x + (x + 1) - 1(x + 1]$
$= - (x + 1) (3x - 1)$
$= (x + 1) (1 - 3x)$
$= (a + b + 1) [1 - 3(a + b)]$
$= (a + b + 1) (1 - 3a - 3b)$
View full question & answer→Question 333 Marks
Factorise : $(2a + b)^2- 6a - 3b - 4$
Answer$(2a + b)^2- 6a - 3b - 4$
$= ( 2a + b )^2 - 3( 2a + b ) - 4$
Assume that, $2a + b = x$
Therefore,
$(2a + b)^2- 6a - 3b - 4$
$= x2 - 3x - 4$
$= x2 - 4x + x - 4$
$= 1( x - 4 ) + x( x - 4 )$
$= ( x + 1 )( x - 4 )$
$= ( 2a + b + 1 )( 2a + b - 4 )$
$[$ resubstitute the value of $x ]$
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