Chapter 4.2, Problem 42AYU

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

54. $f\left(x\right)\text{ = 2}{x}^3\text{ - 5}{x}^2\text{ - 4}x\text{ + 12}$

To determine

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

Answer to Problem 42AYU

$\text{The real zeros }=2\text{ multiplicity }2\text{ and }-\frac{3}{2}$.

$f\left(x\right)=2x^3-5x^2-4x+12=\left(x-2\right)\left(x-2\right)\left(2x+3\right)$

Explanation of Solution

Given:

$f\left(x\right)=2x^3-5x^2-4x+12$

$f$ 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,

$f\left(x\right)=2x^3-5x^2-4x+12$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right)=-2x^3-5x^2+4x+12$

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 $a_n$.

$f\left(x\right)=2x^3-5x^2-4x+12$

Here $a_0=\text{ 12}$ and $a_n=2$.

Zeros of 2, $q=\pm 1,\pm 2$.

Zeros of 12, $p=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$.

The potential rational zeros of $f=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2}$.

From the graph of $f$, we have 2 real zeros near $-\frac{3}{2}$ and 2.

$x\text{ = 2}$ is zero of $f$.

Use long division or Synthetic division of $2x^3-5x^2-4x+12$.

$f\left(x\right)=\left(x-2\right)\left(2x^2-x-6\right)$

The depressed equation $=\left(2x^2-x-6\right)$ which is a quadratic equation.

From factorizing further,

$\left(2x^2-x-6\right)=2x^2-4x+3x-6=2x\left(x-2\right)+3\left(x-2\right)=\left(2x+3\right)\left(x-2\right)$

The real zeros $=2$ multiplicity 2 and $-\frac{3}{2}$.

$f\left(x\right)=2x^3-5x^2-4x+12=\left(x-2\right)\left(x-2\right)\left(2x+3\right)$