Find roots of polynomial $$ p(x) = 12x^2-18x $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 3 }{ 2 } \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ 6x }$ from $ 12x^2-18x $ and solve two separate equations:

$$ \begin{aligned} 12x^2-18x & = 0\\[1 em] \color{blue}{ 6x }\cdot ( 2x-3 ) & = 0 \\[1 em] \color{blue}{ 6x = 0} ~~ \text{or} ~~ 2x-3 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

To find the second zero, solve equation $ 2x-3 = 0 $

$$ \begin{aligned} 2x-3 & = 0 \\[1 em] 2 \cdot x & = 3 \\[1 em] x & = \frac{ 3 }{ 2 } \end{aligned} $$

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