Answer
$f\left( x \right)=2{{x}^{4}}-4{{x}^{2}}+1$ has a zero between $-1$ and 0.
Work Step by Step
Fnd $f\left( -1 \right)$ and $f\left( 0 \right)$ and if the sign changes in value then there must be at least one zero in between -1 and 0.
$\begin{align}
& f\left( x \right)=2{{x}^{4}}-4{{x}^{2}}+1 \\
& f\left( -1 \right)=2\times 1-4\times 1+1 \\
& f\left( -1 \right)=2-4+1 \\
& f\left( -1 \right)=-1 \\
\end{align}$
Which is negative.
And,
$\begin{align}
& f\left( x \right)=2{{x}^{4}}-4{{x}^{2}}+1 \\
& f\left( 0 \right)=2\times 0-4\times 0+1 \\
& f\left( 0 \right)=1 \\
\end{align}$
Which is positive.
Therefore, there is sign change in the value of $f\left( -1 \right)$ and $f\left( 0 \right)$ and thus there is at least one zero between -1 and 0.
