Chapter 4, Problem 58RE

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:

  $f\left(x\right)=x^4+6x^3+11x^2+12x+18$

Answer to Problem 58RE

The real zero of the given polynomial function are $f\left(x\right)=x^4+6x^3+11x^2+12x+18$ is $-3$

Explanation of Solution

Given information:

  $f\left(x\right)=x^4+6x^3+11x^2+12x+18$

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:

  $f\left(x\right)=x^4+6x^3+11x^2+12x+18$

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:

  $\begin{array}{l}p=18\\ q=1\end{array}$

Factors of $p$ are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$ and factors of $q$ is $\pm 1$ .

So the potential rational zeroes are defined as:

  $\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

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.

  $\begin{array}{cccccc}1& 1& 6& 11& 12& 18\\ & & 1& 7& 18& 30\\ & 1& 7& 18& 30& 48\end{array}$

Here, the remainder is $f\left(1\right)=48$ . That’s why $1$ is not zero of $q_1$ .

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

  $\begin{array}{cccccc}-1& 1& 6& 11& 12& 18\\ & & -1& -6& -5& -7\\ & 1& 6& 5& 7& 11\end{array}$

Here, the remainder is $f\left(-1\right)=11$ . That’s why $-1$ is not zero of $q_1$ .

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

  $\begin{array}{cccccc}2& 1& 6& 11& 12& 18\\ & & 2& 16& 54& 132\\ & 1& 8& 27& 66& 150\end{array}$

Here, the remainder is $f\left(2\right)=150$ . That’s why $2$ is not zero of $q_1$ .

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

  $\begin{array}{cccccc}-2& 1& 6& 11& 12& 18\\ & & -2& -8& -6& -12\\ & 1& 4& 3& 6& 6\end{array}$

Here, the remainder is $f\left(-2\right)=6$ . That’s why $-2$ is not zero of $q_1$ .

Now test for $3$ by using the synthetic division:

  $\begin{array}{cccccc}3& 1& 6& 11& 12& 18\\ & & 3& 27& 114& 378\\ & 1& 9& 38& 126& 396\end{array}$

Here, the remainder is $f\left(3\right)=396$ . That’s why $3$ is not zero of $q_1$ .

Now test for $-3$ by using the synthetic division:

  $\begin{array}{cccccc}-3& 1& 6& 11& 12& 18\\ & & -3& -9& -6& -18\\ & 1& 3& 2& 6& 0\end{array}$

Here, the remainder is $f\left(-3\right)=0$ . That’s why $-3$ is zero of $q_1$ and $x+3$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=x^4+6x^3+11x^2+12x+18\\ f\left(x\right)=\left(x+3\right)\left(x^3+3x^2+2x+6\right)\end{array}$

The solution of the given polynomial equation $f\left(x\right)=x^4-4x^3+9x^2-20x+20$ are the zeroes of the $q_1\left(x\right)=x^3+3x^2+2x+6$ .

The potential rational zeroes of $q_1\left(x\right)=x^3+3x^2+2x+6$ :

  $\begin{array}{l}p=6\\ q=1\\ \frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6\end{array}$

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

Now test for $1$ : :

  $\begin{array}{ccccc}1& 1& 3& 2& 6\\ & & 1& 4& 6\\ & 1& 4& 6& 12\end{array}$

Here, the remainder is $f\left(1\right)=12$ . That’s why $1$ is not zero of $q_1$ .

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

  $\begin{array}{ccccc}-1& 1& 3& 2& 6\\ & & -1& -2& 0\\ & 1& 2& 0& 6\end{array}$

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

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

  $\begin{array}{ccccc}2& 1& 3& 2& 6\\ & & 2& 10& 40\\ & 1& 5& 20& 46\end{array}$

Here, the remainder is $f\left(2\right)=46$ . That’s why $2$ is not zero of $q_1$ .

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

  $\begin{array}{ccccc}-2& 1& 3& 2& 6\\ & & -2& -2& 0\\ & 1& 1& 0& 6\end{array}$

Here, the remainder is $f\left(-2\right)=6$ . That’s why $-2$ is not zero of $q_1$ .

Now test for $3$ by using the synthetic division:

  $\begin{array}{ccccc}3& 1& 3& 2& 6\\ & & 3& 18& 40\\ & 1& 6& 20& 46\end{array}$

Here, the remainder is $f\left(3\right)=46$ . That’s why $3$ is not zero of $q_1$ .

Now test for $3$ by using the synthetic division:

  $\begin{array}{ccccc}-3& 1& 3& 2& 6\\ & & -3& 0& -6\\ & 1& 0& 2& 0\end{array}$

Here, the remainder is $f\left(-3\right)=0$ . That’s why $-3$ is zero of $q_1$ and $x+3$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=x^4+6x^3+11x^2+12x+18\\ f\left(x\right)=\left(x+3\right)\left(x^3+3x^2+2x+6\right)\\ f\left(x\right)=\left(x+3\right)\left(x+3\right)\left(x^2+2\right)\end{array}$

As the $x^2+2=0$ has no real solution, $-3$ is the only zeroes of $f$ .

When factored, $f$ is

  $\begin{array}{l}f\left(x\right)=x^4+6x^3+11x^2+12x+18\\ f\left(x\right)=\left(x+3\right)\left(x^3+3x^2+2x+6\right)\\ f\left(x\right)=\left(x+3\right)\left(x+3\right)\left(x^2+2\right)\\ f\left(x\right)={\left(x+3\right)}^2\left(x^2+2\right)\end{array}$

Hence, real zero is $-3$ for the given polynomial function.