Chapter 4.2, Problem 46AYU

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

58. $f\left(x\right)\text{ = }{x}^3\text{ + 6}{x}^2\text{ + 6}x\text{ - 4}$

To determine

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

Answer to Problem 46AYU

$\text{The real zeros }=-2,-\text{2 }\pm \sqrt{6}$.

$f\left(x\right)=x^3+6x^2+6x-4=\left(x+2\right)\left[x-\left(2+\sqrt{6}\right)\right]\left[x-\left(2-\sqrt{6}\right)\right]$

Explanation of Solution

Given:

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

$f$ 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,

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

$+$ to $-$

There will be 1 positive real zero.

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

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

There will be 2 or 0 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 $a_n$.

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

Here $a_0=\text{ −4}$ and $a_n=1$.

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

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

The potential rational zeros of $f=\pm 1,\pm 2,\pm 4$.

From the graph of $f$, we have 3 real zeros near $-2$, $-4$ and 1.

$x=-2$ is zero of $f$.

Use long division or Synthetic division of $x^3+6x^2+6x-4$.

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

The depressed equation $=\left(x^2+4x-2\right)$ which is a quadratic equation.

From factorizing further,

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

$a=1$, $b=4$, $c=-2$.

By quadratic formula,

$x=\frac{-b\pm \sqrt[]{b^2-4.a.c}}{2.a}=\frac{-4\pm \sqrt[]{4^2-4.1.\left(-2\right)}}{2.1}=\frac{-4\pm 2\sqrt[]{6}}{2}=-2\pm \sqrt[]{6}$

The real zeros $=-2,-\text{2 }\pm \sqrt{6}$.

$f\left(x\right)=x^3+6x^2+6x-4=\left(x+2\right)\left[x-\left(2+\sqrt{6}\right)\right]\left[x-\left(2-\sqrt{6}\right)\right]$