[5 marks sum]

Question 15 Marks

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

Answer

$\left(a^2-3 a\right)\left(a^2-3 a+7\right)+10$ Let us assume, $a^2-3 a=x$
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, $a^2-3 a+5$ 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 25 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 [5 marks sum]

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