Chapter 4, Problem 6CRT

a.

To determine

To find: All zeros of $Q(x)=x^5-3x^4+3x^3+x^2-4x+2$ for real and complex.

Answer to Problem 6CRT

  $x=1,-1,(1-i),(1+i)$

Explanation of Solution

Formula used:

Use of quadratic formula

  $\begin{array}{l}\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

Calculation:

Given expression is $Q(x)=x^5-3x^4+3x^3+x^2-4x+2$

  $\begin{array}{l}{x}^5-3x^4+3x^3+x^2-4x+2=0\\ x^5-x^4-2x^4+2x^3+x^3-x^2+2x^2-2x-2x+2=0\\ x^4(x-1)-2x^3(x-1)+x^2(x-1)+2x(x-1)-2(x-1)=0\\ (x^4-2x^3+x^2+2x-2)(x-1)=0\\ (x^3(x-1)-x^2(x-1)+2(x-1))(x-1)=0\\ (x^3-x^2+2)(x-1)(x-1)=0\\ $ (x^2(x+1)-2x(x+1)+2(x+1)){(x-1)}^2=0$$ (x^2-2x+2)(x+1){(x-1)}^2=\mathrm{0............}eqn(1)$\end{array}$$$

Recall that the quadratic equation formula $\begin{array}{l}\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

Where a,b,c is a real number

So the

  $x^2-2x+2$ has $a=1,b=-2,c=2$

  $\begin{array}{l}x=\frac{2\pm \sqrt{4-8}}{2}\\ x=\frac{2\pm 2i}{2}\\ x=\frac{1\pm i}{1}\\ x=1\pm i\end{array}$

Therefore the eqn(01) has

  $x=1\pm i,1,-1$

So the real values are

  $x=1andx=-1$

And the imaginary values are

  $1+iand1-i$

b.

To determine

To Factor the given expression completely.

Answer to Problem 6CRT

The complete factorization are

  ${(x-1)}^2(x+1)(x^2-2x+2)$

Explanation of Solution

Given information:

Expression given in the form of 5^th degree form

Calculation:

  $\begin{array}{l}{x}^5-3x^4+3x^3+x^2-4x+2=0\\ x^5-x^4-2x^4+2x^3+x^3-x^2+2x^2-2x-2x+2=0\\ x^4(x-1)-2x^3(x-1)+x^2(x-1)+2x(x-1)-2(x-1)=0\\ (x^4-2x^3+x^2+2x-2)(x-1)=0\\ (x^3(x-1)-x^2(x-1)+2(x-1))(x-1)=0\\ (x^3-x^2+2)(x-1)(x-1)=0\\ $ (x^2(x+1)-2x(x+1)+2(x+1)){(x-1)}^2=0$$ (x^2-2x+2)(x+1){(x-1)}^2=0$\end{array}$

Therefore, the complete factorization is

  ${(x-1)}^2(x+1)(x^2-2x+2)$