Chapter 4.2, Problem 43AYU

In Problems 51-68, find the real zeros of f. Use the real zeros to factor f

55. $f\left(x\right)\text{ = 3}{x}^3\text{ + 4}{x}^2\text{ + 4}x\text{ + 1}$

To determine

To find: The real zeros of $f$ and to use the real zeros to factor $f$.

Answer to Problem 43AYU

$\text{The real zeros }=-\frac{1}{3}$.

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

Explanation of Solution

Given:

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

$f$ is a polynomial function.

The degree of the polynomial is 3. Therefore the number of real zeros by real zero theorem can be at most $=3$.

From Descartes’ rule of signs,

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

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

$-$ to $+$, $+$ to $-$

There will be 3 or 1 negative real zeros.

Rational zeros theorem provides information about the potential rational zeros of a polynomial function with integer coefficients.

If $\frac{p}{q}$ in its lowest terms is a rational zero of $f$, then $p$ is a factor of $a_0$ and $q$ is the factor of $a_n$.

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

Here $a_0=\text{ 1}$ and $a_n=3$.

Zeros of 3, $q=\pm 1,\pm 3$.

Zeros of 1, $p=\pm 1$.

The potential rational zeros of $f=\pm 1,\pm 3,\pm \frac{1}{3}$.

From the graph of $f$, we have 1 real zero near $-\frac{1}{3}$.

$x=-\frac{1}{3}$ is zero of $f$.

Use long division or Synthetic division of $3x^3+4x^2+4x+1$.

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

The depressed equation $=\left(3x^2+3x+3\right)$ which is a quadratic equation.

From factorizing further,

$\left(3x^2+3x+3\right)$

$a=3$, $b=3$, $c=3$.

By quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4.a.c}}{2.a}=\frac{-3\pm \sqrt{3^2-\mathrm{4.3.3}}}{2.3}=\frac{-3\pm \sqrt{-27}}{2.3}=\frac{-1\pm i\sqrt{3}}{2}$

The real zeros $=-\frac{1}{3}$.

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