Answer
a) Integer root of the equation is $2$.
b) The other two roots of the equation are $\frac{1}{2},\frac{2}{3}$.
Work Step by Step
(a)
As observed from the given graph, the function crosses the x-axis at three places, one at $x=2$ and twice in between $x=0$ and $x=1$.
Therefore, the equation has three roots, $x=2$ and two non-integral roots between 0 and 1.
Thus, the integer root of the equation is 2.
Hence, the integer root of the equation $f\left( x \right)=6{{x}^{3}}-19{{x}^{2}}+16x-4$ is 2.
(b)
From part (a), the integral root of the function $f\left( x \right)=6{{x}^{3}}-19{{x}^{2}}+16x-4$ is 2. So $\left( x-2 \right)$ is a factor of the function. Hence, the division of the given equation by $\left( x-2 \right)$ can be obtained using synthetic division as,
$\begin{matrix}
\left. {\underline {\,
2 \,}}\! \right| & 6 & -19 & 16 & -4 \\
{} & {} & 12 & -14 & 4 \\
{} & 6 & -7 & 2 & 0 \\
\end{matrix}$
Remainder 0 shows that $\left( x-2 \right)$ is a factor of the equation.
Factorizing the function $6{{x}^{3}}-19{{x}^{2}}+16x-4$:
$\begin{align}
& 6{{x}^{3}}-19{{x}^{2}}+16x-4=\left( x-2 \right)\left( 6{{x}^{2}}-7x+2 \right) \\
& =\left( x-2 \right)\left( 6{{x}^{2}}-4x-3x+2 \right) \\
& =\left( x-2 \right)\left( 2x\left( 3x-2 \right)-1\left( 3x-2 \right) \right) \\
& =\left( x-2 \right)\left( 3x-2 \right)\left( 2x-1 \right)
\end{align}$
Now, to get roots of the equation, put $f\left( x \right)=0.$
$\begin{align}
& 6{{x}^{3}}-19{{x}^{2}}+16x-4=0 \\
& \left( x-2 \right)\left( 3x-2 \right)\left( 2x-1 \right)=0 \\
& x=2,x=\frac{2}{3},x=\frac{1}{2}
\end{align}$
So, the other two roots are $\frac{1}{2},\frac{2}{3}.$
