Find roots of polynomial $$ p(x) = 400p-8p^2 $$

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

$$ \begin{aligned}p_1 &= 0\\[1 em]p_2 &= 50 \end{aligned} $$

Step 1:

Write polynomial in descending order

$$ \begin{aligned} 400p-8p^2 & = 0\\[1 em] -8p^2+400p & = 0 \end{aligned} $$

Step 2:

Factor out $ \color{blue}{ -8p }$ from $ -8p^2+400p $ and solve two separate equations:

$$ \begin{aligned} -8p^2+400p & = 0\\[1 em] \color{blue}{ -8p }\cdot ( p-50 ) & = 0 \\[1 em] \color{blue}{ -8p = 0} ~~ \text{or} ~~ p-50 & = 0 \end{aligned} $$

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

Step 3:

To find the second zero, solve equation $ p-50 = 0 $

$$ \begin{aligned} p-50 & = 0 \\[1 em] p & = 50 \end{aligned} $$

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