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Mathematics

Evaluate :

(3x+12)\Big(3x +\dfrac{1}{2}\Big) (2x+13)\Big(2x +\dfrac{1}{3}\Big)

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Answer

(3x+12)(2x+13)=3x(2x+13)+12(2x+13)=3x×2x+3x×13+12×2x+12×13=6x2+3x3+2x2+1×12×3=6x2+x+x+16=6x2+2x+16\Big(3x +\dfrac{1}{2}\Big) \Big(2x +\dfrac{1}{3}\Big) \\[1em] = 3x \Big(2x +\dfrac{1}{3}\Big) + \dfrac{1}{2} \Big(2x +\dfrac{1}{3}\Big) \\[1em] = 3x \times 2x + 3x \times \dfrac{1}{3} + \dfrac{1}{2} \times 2x + \dfrac{1}{2} \times \dfrac{1}{3} \\[1em] = 6x^2 + \dfrac{3x}{3} + \dfrac{2x}{2} + \dfrac{1 \times 1}{2 \times 3} \\[1em] = 6x^2 + x + x + \dfrac{1}{6} \\[1em] = 6x^2 + 2x + \dfrac{1}{6}

Hence, (3x+12)(2x+13)=6x2+2x+16\Big(3x +\dfrac{1}{2}\Big) \Big(2x +\dfrac{1}{3}\Big) = 6x^2 + 2x + \dfrac{1}{6}

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