Chapter 2, Problem 58RE

Solution

a.

To find the : The zeros of a the function . $f\left(x\right)=5x^3-24x^2+x+12$ .

The zeros of the given function are $x=\frac{4}{5},x=2+\sqrt{7},x=2-\sqrt{7}$ .

Given information:

The given function is $f\left(x\right)=5x^3-24x^2+x+12$ .

Concept used:

Factor theorem: A polynomial function has a factor $x-k$ if only if $f\left(x\right)=0$.

Rational zero theorem: Suppose f is a function of degree $n\ge 1$ of the form

  $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{......}+a_0$ with every coefficient an integer and $a_0\ne 0.$ If $x=\frac{p}{q}$ is a rational zero of $f$ where $p$ and $q$ have no common integers other than $\pm 1$ then $p$ is an integer factor of the constant $a_0$ and $q$ is an integer factor of the leading coefficient $a_0$ .

Calculations:

The function $f\left(x\right)=5x^3-24x^2+x+12$ .

By using the rational zero theorem $x=\frac{4}{5}$ is a zero of the function. Then $\left(x-\frac{4}{5}\right)\Rightarrow \left(5x-4\right)$ is factor.

Then the expression can write as $\left(5x-4\right)\left(x^2-4x-3\right)$ .Then by quadratic formula the roots of the equation

  $\begin{array}{c}\left(x^2-4x-3\right)=\frac{4\pm \sqrt{16+12}}{2}\\ =2\pm \sqrt{7}\end{array}$

So zeros of the given function are $x=\frac{4}{5},x=2+\sqrt{7},x=2-\sqrt{7}$ .

b.

To find : the linear factors of the function $f\left(x\right)=5x^3-24x^2+x+12$ where zeros are $x=\frac{4}{5},x=2+\sqrt{7},x=2-\sqrt{7}$ .

The linear factorization of the function is $f\left(x\right)=\left(5x+4\right)\left(x-\left(2+\sqrt{7}\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)$ .

Given information:

The given function $f\left(x\right)=5x^3-24x^2+x+12$ where zeros are $x=\frac{4}{5},x=2+\sqrt{7},x=2-\sqrt{7}$ .

Concept used:

If $f\left(x\right)$ is a polynomial of degree $n>0$ then $f\left(x\right)$ has $n$ linear factor and $f\left(x\right)=a\left(x-z_1\right)\left(x-z_2\right)\mathrm{.....}\left(x-z_n\right)$ where $a$ is the leading coefficient.

Calculations:

The function $f\left(x\right)=5x^3-24x^2+x+12$ where zeros are $x=\frac{4}{5},x=2+\sqrt{7},x=2-\sqrt{7}$ .

Then based on the linear factorization theorem the function becomes $f\left(x\right)=\left(5x+4\right)\left(x-\left(2+\sqrt{7}\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)$ .

Hence the linear factorization of the function is $f\left(x\right)=\left(5x+4\right)\left(x-\left(2+\sqrt{7}\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)$ .