Question

Factorise 25a2-4b2+28bc-49c2

Answer 1

Answer:

(5a+2b-7c)(5a-2b+7c)

Explanation:

25a²-4b²+28bc-49c²

= 25a²-(4b²-28bc+49c²)

= (5a)²- [ (2b)²-2*2b*7c+(7c)²]

= (5a)² - (2b-7c)²

[tex]\boxed {x^{2}-y^{2}=(x+y)\times(x-y)}[\tex]

= [5a+(2b-7c)][5a-(2b-7c)]

= (5a+2b-7c)(5a-2b+7c)

••••

Answer 2

$\bf\huge\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Factorise = 25a^2 - 4b^2 + 28bc - 49c^2

= 5a² - 4b² + 28bc - 49c²

= 25a² - (4b² - 28bc + 49c²)

Here we get

= (5a)²- (2b)²- 2 × 2b × 7c + (7c)²

= (5a)² - (2b - 7c)²

Using Identity we get

= x^2 - y^2 = (x + y)(x - y)

= 5a + (2b - 7c)(5a - (2b - 7c)

$\bf\huge\bf\huge{\boxed{\bigstar{{(5a+2b-7c)(5a-2b+7c)}}}}$