Factorise : $25 (2 x + y)^{2} - 16 (x - y)^{2}$

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Solution

$25 (2 x + y)^{2} - 16 (x - y)^{2} = [5 (2 x + y)]^{2} - [4 (x - y)]^{2} = (10 x + 5 y)^{2} - (4 x - 4 y)^{2} = (10 x + 5 y + 4 x - 4 y) (10 x + 5 y - 4 x + 4 y) = (14 x + y) (6 x + 9 y) = (14 x + y) 3 (2 x + 3 y) = 3 (14 x + y) (2 x + 3 y)$

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Similar questions

Q. Factorise by using formula:

25 (2 x + y)^{2} - 16 (x - y)^{2}

Q. Solve:

25 {(2 x + y)}^{2} - 16 {(x - y)}^{2}

Factorise :

$(i) a^{2} - (b - c)^{2} (i i) 25 (2 x - y)^{2} - 16 (x - 2 y)^{2} (i i i) 16 (5 x + 4)^{2} - 9 (3 x - 2)^{2} (i v) 9 x^{2} - \frac{1}{16} (v) 25 (x - 2 y)^{2} - 4$

Q. State whether True or False:

Factorization of

25 (2 x - y)^{2} - 16 (x - 2 y)^{2}

3 (14 x - 13 y) (2 x + y)

Q. Factorise:

16 {(2 x - y)}^{2} - 24 (4 x^{2} - y^{2}) + {9 (2 x + y)}^{2}

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Factorise : 25(2x+y)2−16(x−y)2

25(2x+y)2−16(x−y)2=[5(2x+y)]2−[4(x−y)]2=(10x+5y)2−(4x−4y)2=(10x+5y+4x−4y)(10x+5y−4x+4y)=(14x+y)(6x+9y)=(14x+y)3(2x+3y)=3(14x+y)(2x+3y)

Factorise : $25 (2 x + y)^{2} - 16 (x - y)^{2}$

$25 (2 x + y)^{2} - 16 (x - y)^{2} = [5 (2 x + y)]^{2} - [4 (x - y)]^{2} = (10 x + 5 y)^{2} - (4 x - 4 y)^{2} = (10 x + 5 y + 4 x - 4 y) (10 x + 5 y - 4 x + 4 y) = (14 x + y) (6 x + 9 y) = (14 x + y) 3 (2 x + 3 y) = 3 (14 x + y) (2 x + 3 y)$