Chapter 2, Problem 59RE

Solution

a.

To find the : The zeros of a the function . $f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ .

The zeros of the given function are $x=\frac{5}{2},x=-1,x=1,x=-\frac{2}{3}$ .

Given information:

The given function is $f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ .

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 $f\left(x\right)=f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ .

By using the rational zero theorem $x=-1,1$ are the zero of the function

The leading coefficient here is $6$ and constant term is $10$ .

The integer factor of $6$ are $\pm 1,\pm 2,\pm 3,\pm 6$ .

The integer factors of $12$ are $$

  $\pm 1,\pm 2,\pm 5,\pm 10$ .

Hence the potential rational zeros are : $$

  $\frac{factorsof5}{factorsof10}:\frac{\pm 1,\pm 2,\pm 4,\pm 6}{\pm 1,\pm 2,\pm 5,\pm 10}$

. Then $\left(x-1\right),\left(x+1\right)$ are the two factors and by using synthetic division will get $\left(-\frac{2}{3}\right),\left(\frac{5}{2}\right)$ are the zeros of the given function

So zeros of the given function are $x=\frac{5}{2},x=-1,x=1,x=-\frac{2}{3}$ .

b.

To find : the linear factors of the function $f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ and the zeros are $x=\frac{5}{2},x=-1,x=1,x=-\frac{2}{3}$ .

The linear factorization of the function is $f\left(x\right)=\left(2x+5\right)\left(x-1\right)\left(x+1\right)\left(3x+5\right)$ .

Given information:

The given function $f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ where zeros are $x=\frac{5}{2},x=-1,x=1,x=-\frac{2}{3}$ .

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 $f\left(x\right)=6x^4+11x^3-16x^2-11x+10$ where zeros are $x=\frac{5}{2},x=-1,x=1,x=-\frac{2}{3}$ .

Then based on the linear factorization theorem the function can expressed as $f\left(x\right)=\left(2x+5\right)\left(x-1\right)\left(x+1\right)\left(3x+5\right)$ .

Hence the linear factorization of the function is $f\left(x\right)=\left(2x+5\right)\left(x-1\right)\left(x+1\right)\left(3x+5\right)$ .