Question
Factorise:
$25 a^2-4 b^2+28 b c-49 c^2$
$25 a^2-4 b^2+28 b c-49 c^2$
✓
Answer
$25 a^2-4 b^2+28 b c-49 c^2$
$=25-\left[4 b^2-28 b c+49 c^2\right]$
${\left[\because a^2-2 a b+b^2=(a-b)^2\right]}$
$=(5 a)^2-\left[(2 b)^2-2 \times 2 b \times 7 c+(7 c)^2\right] $
$ =(5 a)^2-(2 b-7 c)^2 $
$\left\{\because\left(a^2-b^2=(a+b)(a-b)\right\}\right.$
$ = (5a + 2b - 7c)(5a - 2b + 7c)$
$=25-\left[4 b^2-28 b c+49 c^2\right]$
${\left[\because a^2-2 a b+b^2=(a-b)^2\right]}$
$=(5 a)^2-\left[(2 b)^2-2 \times 2 b \times 7 c+(7 c)^2\right] $
$ =(5 a)^2-(2 b-7 c)^2 $
$\left\{\because\left(a^2-b^2=(a+b)(a-b)\right\}\right.$
$ = (5a + 2b - 7c)(5a - 2b + 7c)$
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