How do you write $6 x^{2} + 4 x$ $6x^2 + 4x$ in factored form?

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How do you write $6 x^{2} + 4 x$ $6x^2 + 4x$ in factored form?

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Answer 1 · 2018-03-21T09:59:06

$\Rightarrow {(\sqrt{6} x + \frac{2}{\sqrt{6}})}^{2} - \frac{2}{3}$ $color(green)(=> (sqrt6 x + 2 / sqrt6)^2 - 2/3$

$\Rightarrow (\sqrt{6} x + \frac{2}{\sqrt{6}} + \sqrt{\frac{2}{3}}) \cdot (\sqrt{6} x + \frac{2}{\sqrt{6}} - \sqrt{\frac{2}{3}})$ $=> color(red)((sqrt6 x + 2/sqrt6 + sqrt(2/3)) * (sqrt6 x + 2/sqrt6 - sqrt(2/3))$

Explanation:

$6 x^{2} + 4 x = 6 x \cdot 2 + 4 x + \frac{4}{6} - \frac{4}{6}, Add & Subtract 4/6$ $6x^2 + 4x = 6x*2 + 4x + 4 / 6 - 4/6, " Add & Subtract 4/6"$

$\Rightarrow {(\sqrt{6} x)}^{2} + 2 \cdot (\frac{1}{\sqrt{6}}) \cdot 2 \cdot x + {(\frac{2}{\sqrt{6}})}^{2} - \frac{2}{3}$ $=> (sqrt6x)^2 + 2 * (1/sqrt6) * 2 * x + (2/sqrt6)^2 - 2/3$

It is in the form $(a 2 + 2 a b + b^{2}) where a = \sqrt{6} . b = \frac{2}{\sqrt{6}}$ $(a2 + 2ab + b^2) " where " a = sqrt6. b = 2 / sqrt6$

Hence $\Rightarrow {(\sqrt{6} x + \frac{2}{\sqrt{6}})}^{2} - {(\sqrt{\frac{2}{3}})}^{2}$ $=> (sqrt6 x + 2 / sqrt6)^2 - (sqrt(2/3))^2$

It is again in the for $a^{2} - b^{2}$ $a^2 - b^2$

$\Rightarrow (\sqrt{6} x + \frac{2}{\sqrt{6}} + \sqrt{\frac{2}{3}}) \cdot (\sqrt{6} x + \frac{2}{\sqrt{6}} - \sqrt{\frac{2}{3}})$ $=> (sqrt6 x + 2/sqrt6 + sqrt(2/3)) * (sqrt6 x + 2/sqrt6 - sqrt(2/3))$

Answer 2 · 2018-03-21T10:09:10

$2 x (3 x + 2)$ $2x(3x+2)$

Explanation:

$take out a common factor 2 x$ $"take out a "color(blue)"common factor "2x$

$\Rightarrow 6 x^{2} + 4 x = 2 x (3 x + 2)$ $rArr6x^2+4x=2x(3x+2)$

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How do you write $6 x^{2} + 4 x$ $6x^2 + 4x$ in factored form?

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