Question

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

$f\left(x\right)=3x^3-7x^2+12x-28$

Short answer

The only real zero of the function is $\frac{7}{3}$.

The function can be written in factored form as $f\left(x\right)=(3x-7)(x^2+4)$.

Step-by-step solution

Step 1. Find the possible number of zeros     

The given function $f\left(x\right)=3x^3-7x^2+12x-28$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 $-28$are

role="math" localid="1646111022250" $p:\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28$

The factors of the leading coefficient $3$are

$q:\pm 1,\pm 3$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{7}{3},\pm \frac{14}{3},\pm \frac{28}{3}$

Step 3. Graph the function

The graph of the function is given as

The graph resembles a cubic function and from the graph it can be seen that it has a real root which is greater than and near to the$2$.

Step 4. Find the first root

The number $\frac{7}{3}$is close to $2$.

So finding $f\left(\frac{7}{3}\right)$we get

$$\begin{gathered} f\left(\frac{7}{3}\right)=3·{\left(\frac{7}{3}\right)}^3-7·{\left(\frac{7}{3}\right)}^2+12·\frac{7}{3}-28 \\ f\left(\frac{7}{3}\right)=\frac{343}{9}-\frac{343}{9}+28-28 \\ f\left(\frac{7}{3}\right)=0 \end{gathered}$$

As the function value is zero, so $\frac{7}{3}$is a zero of the function and $3x-7$is a factor.

Step 5. Find the depressed polynomial

Using synthetic division divide $3x^3-7x^2+12x-28$by$3x+7$

So the quotient $x^2+4$is the depressed polynomial and it is a prime polynomial, so it cannot be factored further. Thus the function has only one real zero and itis written in factored form as

$f\left(x\right)=(3x+7)(x^2+4)$