Factorize the following :

$28 a^{2} + 14 a^{2} b^{2} - 21 a^{4}$

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Solution

$28 a^{2} + 14 a^{2} b^{2} - 21 a^{4}$

HCF of 28, 14, 21 = 7

HCF of $a^{2}, a^{2}, a^{4} = a^{2}$

HCF of $1, b^{2}, 1 = 1$

$∴ 28 a^{2} + 14 a^{2} b^{2} - 21 a^{4} = 7 a^{2}$

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

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