SAT Mathematics Practice Test 3

The given equation can be rewritten as ${\mathit{x}}^5-5{\mathit{x}}^3+4\mathit{x}=0$.
Since the polynomial expression on the left has no constant term, it has x as a factor: $\mathit{x}\left({\mathit{x}}^4-5{\mathit{x}}^2+4\right)=0$.
The expression in parentheses is a quadratic equation in ${\mathit{x}}^2$ that can be factored, giving $\mathit{x}\left({\mathit{x}}^2-1\right)\left({\mathit{x}}^2-4\right)=0.$
This further factors as $\mathit{x}\left(\mathit{x}-1\right)\left(\mathit{x}+1\right)\left(\mathit{x}-2\right)\left(\mathit{x}+2\right)=0$.
The solutions for x are $\mathit{x}=0,\mathit{x}=1,\mathit{x}=-1,\mathit{x}=2,$ and $\mathit{x}=-2$.
Since it is given that $\mathit{x}>0$, the possible values of x are x = 1 and x = 2.
Either 1 or 2 may be gridded as the correct answer.