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

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Answer 1 · 2018-03-03T17:27:57

$=x^2-8x-128$

$(2x+16)(1/2x-8)$

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$(2x)(1/2x)+(2x)(-8)+(16)(1/2x)+(16)(-8)$

$(2x^2)/2 -16x+(16/2x)-128$

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

$=x^2-8x-128$

Answer 2 · 2018-03-03T17:28:21

Expand brackets.

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

$2x(1/2x - 8) + 16(1/2x - 8)$

Then just expand as normal:

$(2x)(1/2x) - (2x)(8) + (16)(1/2x) - (16)(8)$

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

$= x^2 - 8x -128$

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