Chapter 4.2, Problem 39AYU

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

51. $f\left(x\right)\text{ = }{x}^3\text{ + 2}{x}^2\text{ - 5}x\text{ - 6}$

To determine

To find: The real zeros of $f$ and to write the polynomial function in the factored form.

Answer to Problem 39AYU

The real zeros $=-3,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$.

Explanation of Solution

Given:

$f\left(x\right)=x^3+2x^2-5x-6$

The degree of the polynomial is 5. Therefore the number of real zeros by real zero theorem can be at most $=5$.

$f\left(x\right)=x^3+2x^2-5x-6$

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 $a_n$.

$f\left(x\right)=x^3+2x^2-5x-6$

Here $a_0=\text{ −3}$ and $a_n=4$.

Zeros of 4, $q=\pm 1,\pm 2,\pm 4$.

Zeros of $-3$, $p=\pm 1,\pm 3$.

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

Use repeated Synthetic division.

$f\left(x\right)=x^3+2x^2-5x-6$

The coefficients are 4, 12, 0, 0, $-1$ and $-3$.

The real zero $=-3$.

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

From factorizing further,

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

$(2x^2+1)$ has no real zero.

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

The real zeros $=-3,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$.