Chapter 4.2, Problem 45AYU

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

57. $f\left(x\right)\text{ = }{x}^3\text{ - 10}{x}^2\text{ + 28}x\text{ - 16}$

To determine

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

Answer to Problem 45AYU

$\text{The real zeros }=4,3\pm \sqrt{5}$.

$f\left(x\right)=x^3-10x^2+28x-16=\left(x-4\right)\left[x-\left(3+\sqrt{5}\right)\right]\left[x-\left( 3-\sqrt{5}\right)\right]$

Explanation of Solution

Given:

$f\left(x\right)=x^3-10x^2+28x-16$

$f$ 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,

$f\left(x\right)=x^3-10x^2+28x-16$

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

There will be 3 or 1 positive real zeros.

$f\left(-x\right)=-x^3-10x^2-28x-16$

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 $a_n$.

$f\left(x\right)=x^3-10x^2+28x-16$

Here $a_0=\text{ −16}$ and $a_n=1$.

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

Zeros of $-16$, $p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$.

The potential rational zeros of $f=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$.

From the graph of $f$, we have 3 real zero near 1, 4 and 5.

$x=4$ is zero of $f$.

Use long division or Synthetic division of $x^3-10x^2+28x-16$.

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

The depressed equation $=\left(x^2-6x+4\right)$ which is a quadratic equation.

From factorizing further,

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

$a=1$, $b=-6$, $c=4$.

By quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4.a.c}}{2.a}=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-\mathrm{4.1.4}}}{2.1}=\frac{6\pm 2\sqrt{5}}{2}=3\pm \sqrt{5}$

The real zeros $=4,3\pm \sqrt{5}$.

$f\left(x\right)=x^3-10x^2+28x-16=\left(x-4\right)\left[x-\left(3+\sqrt{5}\right)\right]\left[x-\left( 3-\sqrt{5}\right)\right]$