Chapter 2, Problem 60RE

Solution

a.

To find the : The zeros of a the function . $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ and one of the zero is $1+2i$

The zeros of the given function are $x=3+i,3-i,1+2i,1-2i$ .

Given information:

The given function is $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ .

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 $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ .Since $$

  $1+2i$ is a zero then its complex conjugate also is the zero of the given function .Hence by using the synthetic division can find the other factors. The quadratic factor is $x^2-6x+10$

By using the quadratic formula the zeros are

  $\begin{array}{c}x=\frac{6\pm \sqrt{36-40}}{2}\\ =3\pm i\end{array}$

So the zeros of this functions are $x=3+i,3-i,1+2i,1-2i$

b.

To find : the linear factors of the function $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ where zeros are $x=3+i,x=3-i,x=1+2i,x=1-2i$ .

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

Given information:

The given function $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ where zeros are $x=3+i,x=3-i,x=1+2i,x=1-2i$ .

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 $f\left(x\right)=x^4-8x^3+27x^2-50x+50$ and the zeros are $x=3+i,x=3-i,x=1+2i,x=1-2i$ .

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

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