Question

factorise using suitable identity 9 a square - 4 Y square - 6a + 1

Answer 1

9a^2 - 4y^2 - 6a + 1
By reordering the terms we get,
9a^2 - 6a + 1 - 4y^2
By using the identity a^2 - 2ab + b^2 we get,

(3a)^2 - 2(3a)(1) + (1)^2 - 4y^2
(3a - 1)^2 - 4y^2
Now, 4y^2 = (2y)^2
so we have,
(3a -1)^2 - (2y)^2
By using the identity a^2 - b^2 = (a + b)(a - b), we get

(3a - 1 + 2y)(3a - 1 - 2y)

Hope this will help you buddy

Answer 2

9a^2 - 4b^2 - 6a + 1
By reordering the terms we get,
9a^2 - 6a + 1 - 4b^2
By using the identity a^2 - 2ab + b^2 we get,

(3a)^2 - 2(3a)(1) + (1)^2 - 4b^2
(3a - 1)^2 - 4b^2
Now, 4b^2 = (2y)^2
so we have,
(3a -1)^2 - (2b)^2
By using the identity a^2 - b^2 = (a + b)(a - b), we get

(3a - 1 + 2b)(3a - 1 - 2b)

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