factorise
(3x-y)^2 -(y+2x)^2
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how to find its zeroes by factorising
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Given (3x - y)^2 - (y + 2x)^2. ------- (1)
We know that (a-b)^2 = a^2 + b^2 - 2ab.
Then (3x - y)^2 = (3x)^2 + y^2 - 2 * 3xy
= 9x^2 + y^2 - 6xy. --------- (2)
We know that (a+b)^2 = a^2 + b^2 + 2ab
Then (y + 2x)^2 = y^2 + (2x)^2 + 2 * y * 2x
= y^2 + 4x^2 + 4xy. ----- (3)
Substitute (2) and (3) in (1), we get
(3x - y)^2 - (y + 2x)^2 = (9x^2 + y^2 - 6xy) - (4x^2 + y^2 + 4xy)
= 9x^2 + y^2 - 6xy - 4x^2 - y^2 - 4xy
= 5x^2 - 10xy.
Therefore (3x^2 - y)^2 - (y + 2x)^2 = 5x^2 - 10xy.
Hope this helps!
We know that (a-b)^2 = a^2 + b^2 - 2ab.
Then (3x - y)^2 = (3x)^2 + y^2 - 2 * 3xy
= 9x^2 + y^2 - 6xy. --------- (2)
We know that (a+b)^2 = a^2 + b^2 + 2ab
Then (y + 2x)^2 = y^2 + (2x)^2 + 2 * y * 2x
= y^2 + 4x^2 + 4xy. ----- (3)
Substitute (2) and (3) in (1), we get
(3x - y)^2 - (y + 2x)^2 = (9x^2 + y^2 - 6xy) - (4x^2 + y^2 + 4xy)
= 9x^2 + y^2 - 6xy - 4x^2 - y^2 - 4xy
= 5x^2 - 10xy.
Therefore (3x^2 - y)^2 - (y + 2x)^2 = 5x^2 - 10xy.
Hope this helps!
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