Question

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

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

Short answer

The real zeros of the function are $3,\frac{5+\sqrt{17}}{2},\frac{5-\sqrt{17}}{2}$.

And its factored form is $f\left(x\right)=(x-3)(x-\frac{5+\sqrt{17}}{2})(x-\frac{5-\sqrt{17}}{2})$.

Step-by-step solution

Step 1. Find the possible number of zeros    

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.

Step 2. Use the rational zero theorem    

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

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

Step 3. Graph the function

The graph of the function is given as

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

Step 4. Find the first factor   

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

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

Step 5. Solve the depressed polynomial

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" $f\left(x\right)=(x-3)(x-\frac{5+\sqrt{17}}{2})(x-\frac{5-\sqrt{17}}{2})$