Question

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

$f\left(x\right)=3x^3+4x^2+4x+1$

Short answer

The only real zero of the function is $-\frac{1}{3}$.

The function can be written in factored form as $f\left(x\right)=(3x+1)(x^2+x+1)$.

Step-by-step solution

Step 1. Find the possible number of zeros    

The given function $f\left(x\right)=3x^3+4x^2+4x+1$is of degree three, so it has at most three real zeros.

Step 2. Use the rational zero theorem   

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

The factors of the constant term $1$are

$p:\pm 1$

The factors of the leading coefficient $3$are

$q:\pm 1,\pm 3$

So the possible rational zeros are

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

Step 3. Find the function values

For the possible rational zeros find the function value for which the function is zero and we get

$$\begin{gathered} f(-1)=3·{(-1)}^3+4·{(-1)}^2+4·(-1)+1 \\ f(-1)=-3+4-4+1 \\ f(-1)=-3 \\ f(-\frac{1}{3})=3·{(-\frac{1}{3})}^3+4{(-\frac{1}{3})}^2+4(-\frac{1}{3})+1 \\ f(-\frac{1}{3})=-\frac{1}{9}+\frac{4}{9}-\frac{4}{3}+1 \\ f(-\frac{1}{3})=0 \end{gathered}$$

So for $x=-\frac{1}{3}$the function value is zero, so it is a real zero and$3x+1$is a factor of the function.

Step 4. Perform synthetic

Dividing the function $3x^3+4x^2+4x+1$by $3x+1$using synthetic division

So the quotient $x^2+x+1$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

$f\left(x\right)=(3x+1)(x^2+x+1)$