Question 5: Do the calculation: \( \left( {x – \frac{1}{2}} \right)\left( {x + \frac{1}{2}} \right)\left( { 4x – 1} \right)\)

We have:

\(\begin{array}{l} \left( {x – \frac{1}{2}} \right)\left( {x + \frac{1}{2}} \right)\left( { 4x – 1} \right) = \left( {xx + x.\frac{1}{2} – \frac{1}{2}.x – \frac{1}{2}.\frac{1} {2}} \right).\left( {4x – 1} \right)\\ = \left( {{x^2} + \frac{1}{2}x – \frac{1}{2} x – \frac{1}{4}} \right)\left( {4x – 1} \right) = \left( {{x^2} – \frac{1}{4}} \right)\left ( {4x – 1} \right) = {x^2}.4x + {x^2}.\left( { – 1} \right) + \left( { – \frac{1}{4}} \ right).4x + \left( { – \frac{1}{4}} \right).\left( { – 1} \right)\\ = 4{x^3} – {x^2} – x + \frac{1}{4} \end{array}\)