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We are given a polynomial function,

$f\left(x\right)=3x^3-2x^2+x+4$

We have to find the bounds to the zeros of the function and using it we need to obtain its graph.

Let f denote a polynomial function whose leading coefficient is $1$.

$f\left(x\right)=x^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$

A bound M on the real zeros of f is the smaller of the two numbers

$Max\left\{1,\left|a_0\right|+\left|a_1\right|+...+\left|a_{n-1}\right|\right\},1+Max\left\{\left|a_0\right|,\left|a_1\right|,...,\left|a_{n-1}\right|\right\}$

where Max means 鈥渃hoose the largest entry in .鈥�

The leading coefficient of the given polynomial is $3$. So dividing the function by $3$, we get

$f\left(x\right)=x^3-\frac{2}{3}{x}^2+\frac{1}{3}x+\frac{4}{3}$

On comparing with standard form we get,

$$\begin{gathered} a_2=-\frac{2}{3} \\ {a}_1=\frac{1}{3} \\ {a}_0=\frac{4}{3} \end{gathered}$$

Using the formula the two numbers are found as,

$$\begin{gathered} Max\left\{1,\left|a_0\right|+\left|a_1\right|+鈥�+\left|a_{n-1}\right|\right\}=Max\left\{1,\left|-\frac{2}{3}\right|+\left|\frac{1}{3}\right|+\left|\frac{4}{3}\right|\right\} \\ =Max\left\{1,\frac{7}{3}\right\} \\ =\frac{7}{3} \end{gathered}$$

And,

$$\begin{gathered} 1+Max\left\{\left|a_0\right|,\left|a_1\right|,鈥�,\left|a_{n-1}\right|\right\}=1+Max\left\{\left|-\frac{2}{3}\right|,\left|\frac{1}{3}\right|,\left|\frac{4}{3}\right|\right\} \\ =1+\frac{4}{3} \\ =\frac{3+4}{3} \\ =\frac{7}{3} \end{gathered}$$

So, $\frac{7}{3}$ is the bound for this graph and hence every zero of the given polynomial will lie between$-\frac{7}{3}$and$\frac{7}{3}$.

Using the bound the graph of the function is as follows,