Factorize:

$25 (a - 5 b)^{2} - 4 (a - 3 b)^{2}$

$(3 a - 19 b) . (7 a + 31 b)$

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$(7 a - 19 b) . (3 a - 31 b)$

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$(3 a + 19 b) . (7 a + 31 b)$

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$(3 a - 19 b) . (7 a - 31 b)$

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Solution

The correct option is D
$(3 a - 19 b) . (7 a - 31 b)$

$\Rightarrow 25 (a - 5 b)^{2} - 4 (a - 3 b)^{2}$

$\Rightarrow [5 (a - 5 b)]^{2} - [2 (a - 3 b)]^{2}$

$\Rightarrow [5 (a - 5 b) - 2 (a - 3 b)] . [5 (a - 5 b) + 2 (a - 3 b)]$

$\Rightarrow [5 a - 25 b - 2 a + 6 b)] . [5 a - 25 b + 2 a - 6 b)]$

$\Rightarrow [3 a - 19 b] . [7 a - 31 b]$

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Similar questions

Q. 25(a-5b)²-4(a-3b)²

Q. State whether True or False:

Factorization of

25 (a - 5 b)^{2} - 4 (a - 3 b)^{2}

(7 a - 31 b) (3 a - 19 b)

Factories :

$(i) (a - 3 b)^{2} - 36 b^{2} (i i) 25 (a - 5 b)^{2} - 4 (a - 3 b)^{2} (i i i) a^{2} - 0.36 b^{2} (i v) a^{4} - 625 (v) x^{4} - 5 x^{2} - 36 (v i) 15 (2 x - y)^{2} - 16 (2 x - y) - 15$

Q. Solve

6 a^{2} - 3 b^{2} + 2 a^{2} + 5 b^{2} - 4 a^{2}

(4a+3b)² -(4a-3b)²+48ab

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Factorize: 25(a−5b)2−4(a−3b)2

The correct option is D (3a−19b).(7a−31b) ⇒25(a−5b)2−4(a−3b)2 ⇒[5(a−5b)]2−[2(a−3b)]2 ⇒[5(a−5b)−2(a−3b)].[5(a−5b)+2(a−3b)] ⇒[5a−25b−2a+6b)].[5a−25b+2a−6b)] ⇒[3a−19b].[7a−31b]

Factorize:

$25 (a - 5 b)^{2} - 4 (a - 3 b)^{2}$