Chapter 4.2, Problem 44AYU

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

56. $f\left(x\right)\text{ = 3}{x}^3\text{ - 7}{x}^2\text{ + 12}x\text{ - 28}$

To determine

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

Answer to Problem 44AYU

$\text{The real zeros }=\frac{7}{3}$.

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

Explanation of Solution

Given:

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

$f$ 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,

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

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

There will be 3 or 1 positive real zeros.

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

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 $a_n$.

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

Here $a_0=\text{ −28}$ and $a_n=3$.

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

Zeros of $-28$, $p=\pm 1,\pm 2,\pm 4,\pm 7,\pm 14,\pm 28$.

The potential rational zeros of $f=\pm 1,\pm 2,\pm 3,\pm 4,\pm 7,\pm 14,\pm 28,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{7}{3},\pm \frac{14}{3},\pm \frac{28}{3}$.

From the graph of $f$, we have 1 real zero near $\frac{7}{3}$.

$x=\frac{7}{3}$ is zero of $f$.

Use long division or Synthetic division of $3x^3-7x^2+12x-28$.

$f\left(x\right)=\left[\left(x-\frac{7}{3}\right)\right]\left(3x^2+\text{ 12}\right)$

The depressed equation $=\left(3x^2+\text{ 12}\right)$ which is a quadratic equation.

From factorizing further,

$\left(3x^2+12\right)=3\left(x^2+4\right)=x^2=-4$

The real zeros $=\frac{7}{3}$.

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