Question
Factorise: $9 a^2-b^2+4 b-4$
✓
Answer
$9 a^2-b^2+4 b-4$
$=9 a^2-\left(b^2-4 b+4\right)$
$=(3 a)^2-[(b)^2-2 \times b \times 2+(2)^2$
$=(3 a)^2-(b-2)^2.$
$\{\because a^2- 2ab + b^2= (a - b)^2\}$
$= (3a + b - 2)(3a - b + 2)$
$\{\because a^2- b^2= (a + b)(a - b)\}$
$=9 a^2-\left(b^2-4 b+4\right)$
$=(3 a)^2-[(b)^2-2 \times b \times 2+(2)^2$
$=(3 a)^2-(b-2)^2.$
$\{\because a^2- 2ab + b^2= (a - b)^2\}$
$= (3a + b - 2)(3a - b + 2)$
$\{\because a^2- b^2= (a + b)(a - b)\}$
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