Question:
Factorise:
9a2 − b2 + 4b − 4
Solution:
We have:
$9 a^{2}-b^{2}+4 b-4=9 a^{2}-\left(b^{2}-4 b+4\right)$
$=(3 a)^{2}-(b-2)^{2}$
$=\{3 a+(b-2)\}\{3 a-(b-2)\}$
$=(3 a+b-2)(3 a-b+2)$
$\therefore 9 a^{2}-b^{2}+4 b-4=(3 a+b-2)(3 a-b+2)$
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