Factorise each of the following expressions:
(i) $x^{2} + 6 x + 8$ (ii) $x^{2} + 4 x - 21$
(iii) $x^{2} - 7 x + 12$

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Solution

In order to Factorise $x^{2} + 6 x + 8$ , we find two numbers p and q such that p + q = 6 and
pq = 8.
Clearly, $2 + 4 = 6 a n d 2 \times 4 = 8$
We now split the middle term 6x in the given quadratic as $2 x + 4 x$ .
$∴ x^{2} + 6 x + 8 = x^{2} + 2 x + 4 x + 8$
$= (x^{2} + 2 x) + (4 x + 8)$
$= x (x + 2) + 4 (x + 2)$
$= (x + 2) (x + 4)$
(ii) In order to Factorise $x^{2} + 4 x - 21$ , we have to find two numbers p and q such that p+q=4 and pq = -21
Clearly, $7 + (- 3) = 4 a n d 7 \times - 3 = 21$ .
We now split the middle term $4 x o f x^{2} + 4 x - 21 a s 7 x - 3 x$
$∴ x^{2} + 4 x - 21 = x^{2} + 7 x - 3 x - 21$
$= (x^{2} + 7 x) - (3 x + 21)$
$= x (x + 7) - 3 (x + 7)$
$= (x + 7) (x - 3)$
(iii) In order to Factorise $x^{2} - 7 x + 12$ we have to find two numbers p and q such that p + q = -7 and pq = 12.
Clearly, -3 - 4 = -7
and $- 3 \times - 4 = 12$ .
We now split the middle term -7x of the given quadratic as -3x - 4x.
$∴ x^{2} - 7 x + 12 = x^{2} - 3 x - 4 x + 12$
$= (x^{2} - 3 x) - (4 x - 12)$
$= x (x - 3) - 4 (x - 3)$
$= (x - 3) (x - 4)$

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