Question

factorise:49a^2+70ab+25b^2

Answer 1

49a² + 70ab + 25b²

[(a+b)² = a²+2ab+b²]

= (7a)² + 2(7ax5b) + (5b)²

=(7a+5b)² = (7a+5b)(7a+5b)

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Answer 2

Answer:

The factors of $49a^{2}+70ab+25b^{2}$ is (7a+5b)and (7a+5b)

Solution:

Given  $49a^{2}+70ab+25b^{2}$,

$49=7^{2}$ and $25=5^{2}$

On expanding the equation we get,

$\Rightarrow 7\times7\times a^{2}+70ab+ 5\times5\times b^{2}$

$\Rightarrow (7a)^{2}+70ab+(5b)^{2}$

70ab can be written as $2\times5\times7\times ab$

Therefore, $\Rightarrow (7a)^{2}+2\times5\times7\times ab+(5b)^{2}$

It is in the form of $a^{2}+2ab+b^{2} = (a+b)^{2}$

Here, a is 7a and b is 5b

On substituting the values we get,

$(7a\times5b)^{2}$ or $(7a\times5b)(7a\times5b)$

Hence, the factors are (7a+5b)and (7a+5b).