Find the real zeros of f. Use the real zeros to factor f.

$f\left(x\right)=x^3-8x^2+17x-6$

The given function $f\left(x\right)=x^3-8x^2+17x-6$is of degree three, so it has at most three real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $-6$are

role="math" localid="1646111952464" $p:\pm 1,\pm 2,\pm 3,\pm 6$

The factors of the leading coefficient $1$are

$q:\pm 1$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 6$

The graph of the function is given as

So the graph has two turning points and thus has three real roots.

From the graph, it appears that $3$is a zero of the function. On performing synthetic division we get

As the remainder is zero so $3$is a zero of the function.

And the quotient $x^2-5x+2$is the depressed polynomial. So the given function is factored as

$f\left(x\right)=(x-3)(x^2-5x+2)$

Equating the depressing polynomial with zero we get$x^2-5x+2=0$. Now solving it using the quadratic formula we get

$$\begin{gathered} x=\frac{-(-5)\pm \sqrt{{(-5)}^2-4\times 1\times 2}}{2\times 1} \\ x=\frac{5\pm \sqrt{25-8}}{2} \\ x=\frac{5\pm \sqrt{17}}{2} \end{gathered}$$

So the other factors are $(x-\frac{5+\sqrt{17}}{2}),(x-\frac{5-\sqrt{17}}{2})$

Thus the factored form isrole="math" localid="1646112488264" $f\left(x\right)=(x-3)(x-\frac{5+\sqrt{17}}{2})(x-\frac{5-\sqrt{17}}{2})$