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Factorise By HCF \( 2(2 x+y)-\left(6 x^{2}-3 x y\right) \)

Ask by Donnelly Wagner. in South Africa
Feb 25,2025

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Answer

The expression \(2(2x + y) - (6x^2 - 3xy)\) simplifies to \(-6x^2 + 3xy + 4x + 2y\) and can be factored as \(-3x(2x - y) + 2(2x + y)\).

Solution

Simplify the expression by following steps: - step0: Subtract the terms: \(2\left(2x+y\right)-\left(6x^{2}-3xy\right)\) - step1: Remove the parentheses: \(2\left(2x+y\right)-6x^{2}+3xy\) - step2: Expand the expression: \(4x+2y-6x^{2}+3xy\) Let's factor the expression step by step. 1. **Start with the original expression:** \[ 2(2x + y) - (6x^2 - 3xy) \] 2. **Expand the expression:** \[ 2(2x + y) = 4x + 2y \] \[ -(6x^2 - 3xy) = -6x^2 + 3xy \] Combining these gives: \[ 4x + 2y - 6x^2 + 3xy \] Rearranging the terms: \[ -6x^2 + 3xy + 4x + 2y \] 3. **Group the terms:** We can group the terms to factor by grouping: \[ (-6x^2 + 3xy) + (4x + 2y) \] 4. **Factor out the common factors in each group:** - From the first group \(-6x^2 + 3xy\), we can factor out \(-3x\): \[ -3x(2x - y) \] - From the second group \(4x + 2y\), we can factor out \(2\): \[ 2(2x + y) \] 5. **Combine the factored groups:** Now we have: \[ -3x(2x - y) + 2(2x + y) \] 6. **Final factorization:** The expression can be factored further if we find a common factor. However, it appears that the expression does not have a straightforward common factor across both groups. Thus, the expression \(2(2x + y) - (6x^2 - 3xy)\) simplifies to: \[ -6x^2 + 3xy + 4x + 2y \] And the factored form is: \[ -3x(2x - y) + 2(2x + y) \] If you need a specific form or further simplification, please let me know!

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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!

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