Question: User: (1/3-2x)+(6x-1)(x+1)\n(3x-2)ex2+(2x+9)(4-6x) \nfactorise
1. **State the problem:** Factorise the expressions \(\left(\frac{1}{3}-2x\right)+(6x-1)(x+1)\) and \((3x-2)ex2+(2x+9)(4-6x)\).\n\n2. **First expression:** \(\left(\frac{1}{3}-2x\right)+(6x-1)(x+1)\)\n- Expand \((6x-1)(x+1)\):\n$$ (6x-1)(x+1) = 6x \cdot x + 6x \cdot 1 - 1 \cdot x - 1 \cdot 1 = 6x^2 + 6x - x - 1 = 6x^2 + 5x - 1 $$\n- Add \(\frac{1}{3} - 2x\):\n$$ \frac{1}{3} - 2x + 6x^2 + 5x - 1 = 6x^2 + (5x - 2x) + \left(\frac{1}{3} - 1\right) = 6x^2 + 3x - \frac{2}{3} $$\n\n3. **Factorise the quadratic:**\nWe want to factor \(6x^2 + 3x - \frac{2}{3}\). Multiply entire expression by 3 to clear fraction:\n$$ 3 \times \left(6x^2 + 3x - \frac{2}{3}\right) = 18x^2 + 9x - 2 $$\nFind two numbers that multiply to \(18 \times (-2) = -36\) and add to \(9\): these are \(12\) and \(-3\).\n\nRewrite middle term:\n$$ 18x^2 + 12x - 3x - 2 $$\nGroup terms:\n$$ (18x^2 + 12x) - (3x + 2) $$\nFactor each group:\n$$ 6x(3x + 2) - 1(3x + 2) $$\nFactor out common binomial:\n$$ (3x + 2)(6x - 1) $$\n\nSince we multiplied by 3 earlier, the original expression is:\n$$ \frac{(3x + 2)(6x - 1)}{3} $$\n\n4. **Second expression:** \((3x-2)ex2+(2x+9)(4-6x)\)\nAssuming \(ex2\) means \(\times 2\), rewrite as:\n$$ 2(3x - 2) + (2x + 9)(4 - 6x) $$\nExpand:\n$$ 6x - 4 + (2x + 9)(4 - 6x) $$\nExpand second product:\n$$ 2x \cdot 4 + 2x \cdot (-6x) + 9 \cdot 4 + 9 \cdot (-6x) = 8x - 12x^2 + 36 - 54x $$\nCombine all terms:\n$$ 6x - 4 + 8x - 12x^2 + 36 - 54x = -12x^2 + (6x + 8x - 54x) + (-4 + 36) = -12x^2 - 40x + 32 $$\n\n5. **Factorise \(-12x^2 - 40x + 32\):**\nFactor out \(-4\):\n$$ -4(3x^2 + 10x - 8) $$\nFind two numbers that multiply to \(3 \times (-8) = -24\) and add to \(10\): these are \(12\) and \(-2\).\nRewrite middle term:\n$$ 3x^2 + 12x - 2x - 8 $$\nGroup terms:\n$$ (3x^2 + 12x) - (2x + 8) $$\nFactor each group:\n$$ 3x(x + 4) - 2(x + 4) $$\nFactor out common binomial:\n$$ (x + 4)(3x - 2) $$\n\n6. **Final factorised form:**\n$$ -4(x + 4)(3x - 2) $$