Example Question - mathematics binomial theorem

Simplifying an Algebraic Expression with Binomial Expansion

The image shows an algebraic expression to simplify: \[ 2^2 \cdot (2^{-\frac{1}{2}} + 1)^2 - (2^{\frac{3}{2}} + 1)^2 \] Let's simplify step by step: 1. Simplify the terms inside the parentheses first: \[ (2^{-\frac{1}{2}} + 1)^2 = ( \frac{1}{\sqrt{2}} + 1 )^2 \] \[ (2^{\frac{3}{2}} + 1)^2 = ( \sqrt{8} + 1 )^2 = (2\sqrt{2} + 1)^2 \] 2. Apply the binomial expansion to the squared terms: \[ ( \frac{1}{\sqrt{2}} + 1 )^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + 2 \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot 1 + 1^2 \] \[ (2\sqrt{2} + 1)^2 = (2\sqrt{2})^2 + 2 \cdot (2\sqrt{2}) \cdot 1 + 1^2 \] 3. Simplify each term: \[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \] \[ 2 \cdot \left(\frac{1}{\sqrt{2}}\right) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \] \[ (2\sqrt{2})^2 = 8 \] \[ 2 \cdot (2\sqrt{2}) = 4\sqrt{2} \] 4. Substitute back into the equation: \[ ( \frac{1}{2} + \sqrt{2} + 1 )^2 = ( \frac{3}{2} + \sqrt{2} )^2 \] \[ (8 + 4\sqrt{2} + 1 )^2 = (9 + 4\sqrt{2})^2 \] 5. Now multiply by the factor outside the parentheses in the original expression, which is \(2^2 = 4\), and subtract the squared terms: \[ 4 \cdot ( \frac{3}{2} + \sqrt{2} )^2 - (9 + 4\sqrt{2})^2 \] \[ = 4 \cdot (\frac{9}{4} + 3\sqrt{2} + 2) - (81 + 8\sqrt{2} + 16) \] \[ = (9 + 12\sqrt{2} + 8) - (81 + 8\sqrt{2} + 16) \] 6. Combine like terms: \[ 9 + 12\sqrt{2} + 8 - 81 - 8\sqrt{2} - 16 \] \[ = (9 + 8 - 81 - 16) + (12\sqrt{2} - 8\sqrt{2}) \] \[ = (-70) + 4\sqrt{2} \] So the simplified expression is: \[ -70 + 4\sqrt{2} \]