Question

solve the following
6x^3+12x^2-18x/3x​

Answer 1

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Question :

Solve the following

☑ 6x³+12x²-18/3x

$\: \dagger \large\boxed {\displaystyle \bf{ \color{cyan} \: answer}}$

$\: \implies \displaystyle \sf{ \color{red} \frac{ {6x}^{3} + {12x}^{2} - 18x }{3x} }$

$\: \: \implies \displaystyle \red { \sf{ \frac{x(6 {x}^{2} + 12x - 18 )}{3x} }}$

$\: \implies \displaystyle \red { \sf{ \cancel \frac{x}{x} \frac{6 {x}^{2} + 12 x - 18 }{3} }}$

$\: \: \implies \displaystyle \red{ \sf{ \frac{6 {x}^{2} + 12x - 18 }{3} }}$

$\: \implies \displaystyle \red{ \sf{ \: \frac{3( {2x}^{2} + 4x - 6) }{3} }}$

$\: \: \implies \displaystyle \red{ \sf{ \cancel \frac{3}{3} \color{orange} \: \: {2x}^{2} + 4x - 6}}$

$\: \: \implies \displaystyle \red{ \sf{ {2x}^{2} + 4x - 6 = 0}}$

$\: \: \\ \implies \displaystyle \red{ \sf{ {x}^{2} + 2x - 3 = 0}}$

$\: \: \\ \implies \displaystyle \red{ \sf{ {x}^{2} \: + 3x -x - 3 = 0}}$

$\: \\ \implies \displaystyle \red{ \sf{x(x + 3) - 1(x + 3) = 0}}$

$\: \\ \implies \displaystyle \red{ \sf{(x - 1) = 0 \: and \: (x + 3) = 0}}$

$\: \implies \displaystyle \red{ \sf{x = 1 \: and \: x = - 3}}$

$\: \: \red\bigstar \underline{ \boxed{ \sf{x = 1 \: and \: x = - 3 \: are \: the \: roots \: of \: this \: equation}}}$