Chapter 2.5, Problem 53E

a

To determine

To find: the polynomial function by using the product of factors that are irreducible over the rational

Answer to Problem 53E

The zeros of the given polynomial function $f$ are $\left(x\right)=1+\sqrt{3},1-\sqrt{3}$

Explanation of Solution

Given information:

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

One factor is $x^2-2x-2$

Concept used:

Factors of a polynomial;

Every polynomial of degree $n>0$ with real coefficient can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

We have a factor $x^2-2x-2$

This polynomial function can be written as,

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

Now, simplify as quadratic equation.

  $\begin{array}{c}f\left(x\right)=\left(x^2-2x+3\right)\left(x^2-2x-2+1-1\right)\\ =\left(x^2-2x+3\right)\left(x^2-2x+1-3\right)\\ =\left(x^2-2x+3\right)\left({\left(x-1\right)}^2-3\right)\end{array}$

Now, solving the value of $x$

  $\begin{array}{l}{\left(x-1\right)}^2-3=0\\ {\left(x-1\right)}^2=3\\ x=1\pm \sqrt{3}\end{array}$

Following the above polynomial function as quadratic factors $\left(1+\sqrt{3},1-\sqrt{3}\right)$ is irreducible over the rational.

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

b.

To determine

To find: the polynomial function by using the product of factors that are irreducible over the rational

Answer to Problem 53E

The zeros of the given polynomial function $f$ are $\left(x\right)=1\pm \sqrt{2i}$

Explanation of Solution

Given information:

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

One factor is $x^2-2x-2$

Concept used:

Factors of a polynomial;

Every polynomial of degree $n>0$ with real coefficient can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

One factor is $x^2-2x-2$

This polynomial function can be written as,

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

Now, simplify as quadratic equation to get the value of $x$

Take one factor $\left(x^2-2x+3\right)=0$

Put $+1$ and $-1$ to both side

  $\begin{array}{c}\left(x^2-2x+3+1-1\right)=0+1-1\\ \left(x^2-2x+1+2\right)=0\\ \left({\left(x-1\right)}^2+2\right)=0\end{array}$

Now, solving the value of $x$

  $\begin{array}{c}{\left(x-1\right)}^2+2=0\\ {\left(x-1\right)}^2=-2\\ x=1\pm \sqrt{2i}\end{array}$

Therefore the above polynomial function as quadratic factors $\left(1+\sqrt{2i},1-\sqrt{2i}\right)$ is irreducible over the rational.

  $f\left(x\right)=\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)\left(x^2-2x+3\right)$

c

To determine

To find: the polynomial function by using the product of factors that are irreducible over the rational in completely factored form.

Answer to Problem 53E

The zeros of the given polynomial function $f$ are $\left(x\right)=1\pm \sqrt{3},1\pm \sqrt{2i}$

Explanation of Solution

Given information:

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

One factor is $x^2-2x-2$

Concept used:

Factors of a polynomial;

Every polynomial of degree $n>0$ with real coefficient can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros

The given polynomial function is,

  $f\left(x\right)=x^4-4x^3+5x^2-2x-6$

One factor is $x^2-2x-2$

This polynomial function can be written as,

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

Now, simplify as quadratic equation.

  $\begin{array}{c}f\left(x\right)=\left(x^2-2x+3\right)\left(x^2-2x-2+1-1\right)\\ =\left(x^2-2x+3\right)\left(x^2-2x+1-3\right)\\ =\left(x^2-2x+3\right)\left({\left(x-1\right)}^2-3\right)\end{array}$

Now, solving the value of $x$

  $\begin{array}{l}\left(x^2-2x-2\right)=0\\ \left({\left(x-1\right)}^2-3\right)=0\\ {\left(x-1\right)}^2=0\\ x=1\pm \sqrt{3}\end{array}$

Following the above polynomial function as quadratic factors $\left(1+\sqrt{3},1-\sqrt{3}\right)$ is irreducible over the rational.

Also, $\left(x^2-2x+3=0\right)$ is a factor of this polynomial function.

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

Also, solving the value of $x$

  $\begin{array}{l}\left(x^2-2x+3\right)=0\\ {\left(x-1\right)}^2+2=0\\ x-1=\pm \sqrt{2i}\\ x=1\pm \sqrt{2i}\end{array}$

Therefore, we have the completely factored form of the polynomial function are.

  $f\left(x\right)=\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{2i}\right)\left(x-1-\sqrt{2i}\right)$

Hence, we can conclude that the zeros of $f$ are $x=1\pm \sqrt{3},1\pm \sqrt{2i}$