Question
Find roots of polynomial $$ p(x) = x+\frac{1}{100}x^2 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -100 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 100 } $.
$$ \begin{aligned} x+\frac{1}{100}x^2 & = 0 ~~~ / \cdot \color{blue}{ 100 } \\[1 em] 100x+x^2 & = 0 \end{aligned} $$Step 2:
Write polynomial in descending order
$$ \begin{aligned} 100x+x^2 & = 0\\[1 em] x^2+100x & = 0 \end{aligned} $$Step 3:
Factor out $ \color{blue}{ x }$ from $ x^2+100x $ and solve two separate equations:
$$ \begin{aligned} x^2+100x & = 0\\[1 em] \color{blue}{ x }\cdot ( x+100 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x+100 & = 0 \end{aligned} $$One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.
Step 4:
To find the second zero, solve equation $ x+100 = 0 $
$$ \begin{aligned} x+100 & = 0 \\[1 em] x & = -100 \end{aligned} $$This page was created using
Polynomial Roots Calculator