Factorise By HCF \( 2(2 x+y)-\left(6 x^{2}-3 x y\right) \)
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To factorise the expression \( 2(2x+y)-\left(6x^{2}-3xy\right) \), we first rewrite it by distributing the minus sign in the second part: \[ 2(2x+y) - 6x^2 + 3xy \] Now let's gather like terms: \[ -6x^2 + 2(2x+y) + 3xy = -6x^2 + 4x + 2y + 3xy \] Next, we will look for the greatest common factor (HCF) in each term. A suitable HCF strategy is to group terms: \[ -6x^2 + (4x + 3xy) + 2y \] Notice that we can factor out \( x \) from the second group: \[ -6x^2 + x(4 + 3y) + 2y \] Now, we notice that we have a common factor among \( -6x^2 \), \( 4x \), and \( 3xy \), which is \( x \): \[ = x(-6x + 4 + 3y) + 2y \] Thus, we can write it as: \[ = -2(3x^2 - 2x - y + 2y) \] Finally, the full factorised form is: \[ = -2(3x^2 - 2x + 2y - y) \] So, the factorization of the original expression is: \[ = -2(3x^2 - 2x + (3y - 2y)) \] Or simplified in form, depending on how you want to express it!