Chapter 4, Problem 57RE

To determine

To calculate:

All the real zeroes of the following functions of polynomial with the help of Rational zeroes theorem and the factor $f$ over the real zeroes:

  $f\left(x\right)=x^4-4x^3+9x^2-20x+20$

Answer to Problem 57RE

The real zero of the given polynomial function are $f\left(x\right)=x^4-4x^3+9x^2-20x+20$ is $2$

Explanation of Solution

Given information:

  $f\left(x\right)=x^4-4x^3+9x^2-20x+20$

Formula Used:

The potential rational zeroes is given by the following expression:

  $x=\pm \frac{p}{q}$

Where,

Quotients of the factors of the last term: $p$

The factors of the leading coefficient: $q$

Calculation:

The given function is:

  $f\left(x\right)=x^4-4x^3+9x^2-20x+20$

The degree of the given polynomial function is $4$ which shows that $f$ must have at most $4$ real zeroes.

Now, the values of $p,q$ are defined as:

  $\begin{array}{l}p=20\\ q=1\end{array}$

Factors of $p$ are $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$ and factors of $q$ is. $\pm 1$

So the potential rational zeroes are defined as:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

Now test the potential rational zeroes with the help of the synthetic division:

If the value of remainder is zero $f\left(1\right)=0$ then $1$ is zero of $f$ and $x-1$ is a factor.

  $\begin{array}{cccccc}1& 1& -4& 9& -20& 20\\ & & 1& -3& 6& -14\\ & 1& -3& 6& -14& 6\end{array}$

Here, the remainder is $f\left(1\right)=6$ . That’s why $1$ is not zero of $q_1$ .

Now test for $-1$ by using the synthetic division:

  $\begin{array}{cccccc}1& 1& -4& 9& -20& 20\\ & & -1& 5& -14& 34\\ & 1& -5& 14& -34& 54\end{array}$

Here, the remainder is $f\left(-1\right)=54$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $2$ by using the synthetic division:

  $\begin{array}{cccccc}1& 1& -4& 9& -20& 20\\ & & 2& -4& 10& -20\\ & 1& -2& 5& -10& 0\end{array}$

Here, the remainder is $f\left(2\right)=0$ . That’s why $2$ is zero of $q_1$ and $x-2$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=x^4-4x^3+9x^2-20x+20\\ f\left(x\right)=\left(x-2\right)\left(x^3-2x^2+5x-10\right)\end{array}$

The solution of the given polynomial equation $f\left(x\right)=x^4-4x^3+9x^2-20x+20$ are the zeroes of the $q_1\left(x\right)=x^3-2x^2+5x-10$ .

The potential rational zeroes of $q_1\left(x\right)=x^3-2x^2+5x-10$ :

  $\begin{array}{l}p=10\\ q=1\\ \frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10\end{array}$

Now test the potential rational zeroes with the help of the synthetic division:

Now test for $1$ : :

  $\begin{array}{ccccc}1& 1& -2& 5& -10\\ & & 1& -1& 4\\ & 1& -1& 4& -6\end{array}$

Here, the remainder is $f\left(1\right)=-6$ . That’s why $1$ is not zero of $q_1$ .

Now test for $-1$ by using the synthetic division:

  $\begin{array}{ccccc}1& 1& -2& 5& -10\\ & & -1& 3& -8\\ & 1& -3& 8& -18\end{array}$

Here, the remainder is $f\left(-1\right)=18$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $2$ by using the synthetic division:

  $\begin{array}{ccccc}2& 1& -2& 5& -10\\ & & 2& 0& 10\\ & 1& 0& 5& 0\end{array}$

Here, the remainder is $f\left(2\right)=0$ . That’s why $2$ is zero of $q_1$ and $x-2$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=x^4-4x^3+9x^2-20x+20\\ f\left(x\right)=\left(x-2\right)\left(x^3-2x^2+5x-10\right)\\ f\left(x\right)=\left(x-2\right)\left(x-2\right)\left(x^2+5\right)\end{array}$

As the $x^2+5=0$ has no real solution, $2$ is the only zeroes of $f$ .

When factored, $f$ is

  $\begin{array}{l}f\left(x\right)=x^4-4x^3+9x^2-20x+20\\ f\left(x\right)=\left(x-2\right)\left(x^3-2x^2+5x-10\right)\\ f\left(x\right)=\left(x-2\right)\left(x-2\right)\left(x^2+5\right)\\ f\left(x\right)={\left(x-2\right)}^2\left(x^2+5\right)\end{array}$

Hence, real zero is $2$ for the given polynomial function.