Chapter 2, Problem 54RE

Solution

a.

To find the : The real zeros of a the function . $k\left(t\right)=t^4-7t^2+12$ .

The real zeros are $x=2,x=-2,x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}}$ .

Given information:

The given function is $k\left(t\right)=t^4-7t^2+12$ .

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

  $k\left(t\right)=t^4-7t^2+12$ by using ration function theorem can find the zeros.

The leading coefficient here is $1$ and constant term is $12$ .

The integer factor of 1 are $\pm 1$ .

The integer factors of 12 are $$

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ .

Hence the potential rational zeros are : $$

  $\frac{factorsof1}{factorsof12}:\frac{\pm 1}{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12}$

Now take x=2 is a zero then by synthetic division check $x=2$ is a real zero or not.

  $\begin{array}{l}\overline{)2}\overline{)\begin{array}{l}10-7012\\ 24-612\end{array}}\\ 12-3-60\end{array}$

since the remainder is n equal to zero when $x-2$ is the factor then the other factor is $x^3+2x^2-3x-6$ .Now again synthetic division can find the other factor .Take $x=-2$ as a zero then by synthetic division can check which is a zero or not.

  $\begin{array}{l}\overline{)-2}\overline{)\begin{array}{l}12-3-6\\ -206\end{array}}\\ 10\text{}\text{}\text{}\text{}\text{}\text{}-30\end{array}$

The remainder is zero and then $x=-2$ is the zero of the given polynomial .Now the factors are $\left(x-2\right),\left(x+2\right),\left(x^2-3\right)$ ,

By using the quadratic formula the factors are:

  $\begin{array}{l}{x}^2-3=0\\ \Rightarrow x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}}\end{array}$

So the zeros are $x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}},x=-2,x=2$

b.

To identify : the zeros are rational, irrational or complex zeros of the function . $k\left(t\right)=t^4-7t^2+12$ and the zeros are $x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}},x=-2,x=2s$ .

The real zeros $2,-2$ are rational zeros and $x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}$ are the irrational zeros and there is no complex zeros.

Given information:

The given function is $k\left(t\right)=t^4-7t^2+12$ and the the zeros are $x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}},x=-2,x=2$ .

Concept used:

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

Rational number can be written in the format $\frac{p}{q}$ ,and An irrational number is a real number that cannot be expressed as a ratio of integers.

Calculations:

The function $k\left(t\right)=t^4-7t^2+12$

by using ration function theorem can find the zeros and the zeros are $x=\frac{1}{\sqrt{3}},x=\frac{-1}{\sqrt{3}},x=-2,x=2$ .

Here real zeros $2,-2$ are rational zeros and $x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}$ are the irrational zeros and there is no complex zeros.