Question

Factorise : 49a2 + 7oab + 25 b2

Answer 1

Answer:

$({7a})^{2} + 2(7a)(5b) + ({5b})^{2} \\ ={(7a + 5b)}^{2}$

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Answer 2

$\mathbf\red{Question}$

Factorise:-

$49 {a}^{2} + 70ab + 25 {b}^{2}$

$\mathbf\green{Solution}$

Given:- $49 {a}^{2} + 70ab + 25 {b}^{2}$

$49 = {7}^{2} \: and \: \: 25 = {b}^{2} \: and \: 70 = 2 \times 5 \times 7$

Expand the whole equation:-

$7 \times 7 \times a \times a + 2 \times 7 \times 5 \times a \times b + 5 \times 5 \times b \times b$

It can also be written as:-

${(7a)}^{2} + 2 \times 7 \times 5 \times a \times b + {(5b)}^{2}$

It is in the form :-

${(a + b)}^{2} = {a}^{2} + 2 \times a \times b + {b}^{2}$

Here, a is 7a and b is 25 b

So we get,

${(7a + 5b)}^{2}$

= (7a+5b)(7a+5b)

Therefore,the factors are:-

(7a+5b) and (7a+5b)