The exponential function h, represented in the table, can be written as h(x)=a\cdot b^x x x. h(x) 0 10
1. 4

The answer is h(x)=10⋅(2/5)^x

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Ответ:

ptrlvn01

30.01.2021

Mathematics

$=2x\left(2x-1\right)\left(x+3\right)$

Step-by-step explanation:

$4x^3+10x^2-6x\\\mathrm{Factor\:out\:common\:term\:}2x:\quad 2x\left(2x^2+5x-3\right)\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^2=xx\\x^3=x^2x\\=4x^2x+10xx-6x\\\mathrm{Rewrite\:}6\mathrm{\:as\:}2\cdot \:3\\\mathrm{Rewrite\:}10\mathrm{\:as\:}2\cdot \:5\\\mathrm{Rewrite\:}4\mathrm{\:as\:}2\cdot \:2\\=2\cdot \:2x^2x+2\cdot \:5xx-2\cdot \:3x\\\mathrm{Factor\:out\:common\:term\:}2x\\=2x\left(2x^2+5x-3\right)\\\mathrm{Factor}\:2x^2+5x-3:\quad \left(2x-1\right)\left(x+3\right)$

$2x^2+5x-3\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(2x^2-x\right)+\left(6x-3\right)\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-x\mathrm{:\quad }x\left(2x-1\right)\\2x^2-x\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^2=xx\\=2xx-x\\\mathrm{Factor\:out\:common\:term\:}x\\=x\left(2x-1\right)\\\mathrm{Factor\:out\:}3\mathrm{\:from\:}6x-3\mathrm{:\quad }3\left(2x-1\right)\\6x-3\\\mathrm{Rewrite\:}6\mathrm{\:as\:}3\cdot \:2\\=3\cdot \:2x-3$

$\mathrm{Factor\:out\:common\:term\:}3\\=3\left(2x-1\right)\\=x\left(2x-1\right)+3\left(2x-1\right)\\\mathrm{Factor\:out\:common\:term\:}2x-1\\=\left(2x-1\right)\left(x+3\right)\\=2x\left(2x-1\right)\left(x+3\right)$

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