1.

The factors of a^2-b^2-4c^2+4d^2-4(ad-bc) are

Answer»

Given \:a^{2} - b^{2}-4c^{2} + 4d^{2} - 4(ad-bc)

= \:a^{2} - b^{2}-4c^{2} + 4d^{2} - 4ad+ 4bc

/* REARRANGING the TERMS, we GET */

= a^{2} + 4d^{2} - 4ad - b^{2} - 4c^{2} + 4bc

= (a^{2} + 4d^{2} - 4ad) - (b^{2} +4c^{2} -4bc )

= [a^{2} + (2d)^{2} - 2\times a \times (2d)] - [b^{2} +(2c)^{2} - 2\times b \times (2c )]

= ( a - 2d )^{2} - ( b - 2c)^{2}

= [ (a-2d) + ( b - 2c ) ] [ (a-2d) - ( b - 2c ) ] \\= (a-2d+b-2c)(a-2d-b+2c) \\= ( a+b-2c-2d)(a-b+2c-2d)

THEREFORE.,

\red{a^{2} - b^{2}-4c^{2} + 4d^{2} - 4(ad-bc) }\\\green { = ( a+b-2c-2d)(a-b+2c-2d) }

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