# What is the coefficient of x^4 in the expansion of (2+4x)^5(2-x)^4 ?

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Feb 13, 2018

$- 12768$ is the coefficient of ${x}^{4}$

#### Explanation:

${\left(2 + 4 x\right)}^{5} {\left(2 - x\right)}^{4}$
$\left({\left(2 + 4 x\right)}^{5}\right) {\left(2 - x\right)}^{4} = \left(\left(5 C 0\right) {2}^{5} {\left(4 x\right)}^{0} + \left(5 C 1\right) {2}^{4} {\left(4 x\right)}^{1} + \left(5 C 2\right) {2}^{3} {\left(4 x\right)}^{2} + \left(5 C 3\right) {2}^{2} {\left(4 x\right)}^{3} + \left(5 C 4\right) {2}^{1} {\left(4 x\right)}^{4} + \left(5 C 5\right) {2}^{0} {\left(4 x\right)}^{5}\right) \left(\left(4 c 0\right) {2}^{4} {x}^{0} - \left(4 c 1\right) {2}^{3} {x}^{1} + \left(4 c 2\right) {2}^{2} {x}^{2} - \left(4 c 3\right) {2}^{1} {x}^{3} + \left(4 c 4\right) {2}^{0} {x}^{4}\right)$
$\left(\left(1\right) \left(32\right) \left(1\right) + \left(5\right) \left(16\right) \left(4 x\right) + \left(10\right) \left(8\right) \left(16 {x}^{2}\right) + \left(10\right) \left(4\right) \left(64 {x}^{3}\right) + \left(5\right) \left(2\right) \left(256 {x}^{4}\right) + \left(1\right) \left(1\right) \left(1024 {x}^{5}\right)\right) \left(\left(1\right) \left(16\right) {\left(x\right)}^{0} - \left(4\right) \left(8\right) {\left(x\right)}^{1} + \left(6\right) \left(4\right) {\left(x\right)}^{2} - \left(4\right) \left(2\right) {\left(x\right)}^{3} + \left(1\right) \left(1\right) {\left(x\right)}^{4}\right)$
$\left(32 + 320 x + 1280 {x}^{2} + 2560 {x}^{3} + 2560 {x}^{4} + 1024 {x}^{5}\right) \left(16 - 32 x + 24 {x}^{2} - 8 {x}^{3} + {x}^{4}\right)$

Tabulating the products of coefficients
x 16 -32 24 -8 1
32 512 -1024 768 -256 32
320 5120 -10240 7680 -2560 320
1280 20480 -40960 30720 -10240 1280
2560 40960 -81920 61440 -20480 2560
2560 40960 -81920 61440 -20480 2560
1024 16384 -32768 24576 -8192 1024

The highlighted numbers are the coefficients of ${x}^{4}$
$- 12768$ is the coefficient of ${x}^{4}$

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