[4 marks sum]

Question 14 Marks

Factorise :$12(3x - 2y)^2- 3x + 2y - 1$

Answer

$12(3x - 2y)^2 - 3x + 2y - 1 = 12( 3x - 2y )^2 - ( 3x - 2y ) - 1$
Let us assume that
$3x - 2y = a$
Then the given expression is
$3 x-2 y=a$
$12 a^2-a-1$
$12 a^2-(4-3) a-1$
$12 a^2-4 a+3 a-1$
$4 a(3 a-1)+1(3 a-1)$
$(3 a-1)(4 a+1)$
${[3(3 x 2 y)-1][4 x(3 x-2 y+1)]}$
$(9 x-6 y-1)(12 x-8 y+1)$

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Question 24 Marks

Factorise :$5a^2- b^2- 4ab+ 7a - 7b$

Answer

$5a^2- b^2- 4ab+ 7a - 7b$
$= 4a^2 + a^2 - b^2 - 4ab + 7a - 7b$
$= a^2 - b^2 + 4a^2 - 4ab + 7a - 7b$
$= ( a^2 - b^2 ) + 4a( a - b ) + 7( a - b )$
$= ( a - b )( a + b ) + 4a( a - b ) + 7( a - b ) [ \because a^2 - b^2 = ( a + b )( a - b ) ]$
$= ( a - b )[ ( a + b ) + 4a + 7 ]$
$= ( a - b )[ ( a + b ) + 4a + 7 ]$
$= ( a - b )[ 5a + b + 7 ]$

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Question 34 Marks

Factorise : $(a^2- a) (4a^2- 4a - 5) - 6$

Answer

Let us assume,
$a^2 - a = x$
Then the given expression is,
$(a^2- a) (4a^2- 4a - 5) - 6$
$= x( 4x - 5 ) - 6$
$= 4x^2 - 5x - 6$
$= 4x^2 - 8x + 3x - 6$
$= 4x( x - 2 ) + 3( x - 2 )$
$= ( 4x + 3 )( x - 2 )$
$= [ 4( a^2 - a ) + 3 ]( a^2 - a - 2 ) [$ resubstitute the value of $x ]$
$= [ 4a^2 - 4a + 3 ]( a^2 - a - 2 )$
$=[ 4a^2 - 4a + 3 ]( a^2 - 2a + a - 2 )$
$= [ 4a^2 - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]$
$= [ 4a^2 - 4a + 3 ]( a - 2 )( a + 1 )$

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Question 44 Marks

Evaluate:$\frac{5.67 \times 5.67 \times 5.67+4.33 \times 4.33 \times 4.33}{5.67 \times 5.67-5.67 \times 4.33+4.33 \times 4.33}$

Answer

Let $a=5.67$ and $b=4.33$
Then,
$\frac{5.67 \times 5.67 \times 5.67+4.33 \times 4.33 \times 4.33}{5.67 \times 5.67-5.67 \times 4.33+4.33 \times 4.33}$
$ =\frac{a \times a \times a+b \times b \times b}{a \times a-a \times b+b \times b}$
$ =\frac{a^3+b^3}{a^2-a b+b^2}$
$ =\frac{(a+b)\left(a^2-a b+b^2\right)}{a^2-a b+b^2}$
$ =a+b$
$ =5.67+4.33$
$ =10$

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Question 54 Marks

Factorise : $(a^2- 3a) (a^2 - 3a + 7) + 10$

Answer

$(a^2- 3a) (a^2 - 3a + 7) + 10$
Let us assume $, a^2 - 3a = x$
Then, our polynomial becomes,
$( a^2 - 3a )( a^2 - 3a + 7 ) + 10$
$= x( x + 7 ) + 10$
$= x^2 + 7x + 10$
$= x^2 + 5x + 2x + 10$
$= x( x + 5 ) + 2 ( x + 5 )$
$= ( x + 5 )( x + 2 )$
By resubstituting the value of $x,$
$= (a^2 - 3a + 5)( a^2 - 3a + 2 )$
Now$, a^2 - 3a + 5$ will have no factor as discriminant is $-11$ that is less than $0$.
And,
$\therefore a^2 - 3a + 2 = a^2 - 2a - a + 2 = a( a - 2) - 1(a - 2) = (a - 1)(a - 2)$
So, factor of given polynomial are,
$a^2 - 3a + 2 = a^2 - 2a - a + 2$
$= (a^2 - 3a + 5)(a - 1)(a - 2)$

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Question 64 Marks

Factorise : $\frac{1}{4}(a+b)^2-\frac{9}{16}(2 a-b)^2$

Answer

$ \frac{1}{4}(a+b)^2-\frac{9}{16}(2 a-b)^2$
$ =\frac{1}{4}\left[(a+b)^2-\frac{9}{4}(2 a-b)^2\right]$
$ =\frac{1}{4}\left[(a+b)^2-\left[\frac{3}{2}(2 a-b)^2\right]\right]$
$ =\frac{1}{4}\left[\left(a+b+\frac{3}{2}(2 a-b)\right)\left(a+b-\frac{3}{2}(2 a-b)\right)\right]$
$ =\frac{1}{4}\left[\left(a+b+3 a-\frac{3 b}{2}\right)\left(a+b-3 a+\frac{3 b}{2}\right)\right]$
$ =\frac{1}{4}\left[\left(4 a-\frac{b}{2}\right)\left(\frac{5 b}{2}-2 a\right)\right]$
$ =\frac{1}{4}\left[\left(\frac{8 a-b}{2}\right)\left(\frac{5 b-4 a}{2}\right)\right]$
$ =\frac{1}{4}\left[\frac{1}{4}(8 a-b)(5 b-4 a)\right]$
$ =\frac{1}{16}(8 a-b)(5 b-4 a)$

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MATHEMATICS [4 marks sum]

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