How do you evaluate (2x + 16) ( \frac { 1} { 2} x - 8)?

Mar 3, 2018

$= {x}^{2} - 8 x - 128$

Explanation:

$\left(2 x + 16\right) \left(\frac{1}{2} x - 8\right)$

Distribute

$\left(2 x\right) \left(\frac{1}{2} x\right) + \left(2 x\right) \left(- 8\right) + \left(16\right) \left(\frac{1}{2} x\right) + \left(16\right) \left(- 8\right)$

$\frac{2 {x}^{2}}{2} - 16 x + \left(\frac{16}{2} x\right) - 128$

${x}^{2} - 16 x + 8 x - 128$

$= {x}^{2} - 8 x - 128$

Mar 3, 2018

Expand brackets.

Explanation:

This method helps me when it comes to expansions I can't do off the top of my head:

$2 x \left(\frac{1}{2} x - 8\right) + 16 \left(\frac{1}{2} x - 8\right)$

Then just expand as normal:

$\left(2 x\right) \left(\frac{1}{2} x\right) - \left(2 x\right) \left(8\right) + \left(16\right) \left(\frac{1}{2} x\right) - \left(16\right) \left(8\right)$

$= {x}^{2} - 16 x + 8 x - 128$

$= {x}^{2} - 8 x - 128$