Chapter 2.5, Problem 31E

To determine

To find : the real solution of the polynomial function

Answer to Problem 31E

The real solution of the polynomial function is: $\boxed{\boxed{-3;1;-2+\sqrt{6};-2-\sqrt{6}}}$

Explanation of Solution

Given information:

  $x^4+6x^3+3x^2-16x+6=0$

Concept Involved:

The Rational Zero Test : The Rational Zero Test relates the possible rational zeros of a polynomial (havinginteger coefficients) to the leading coefficient and to the constant term of the polynomial.

If the polynomial $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_2{x}^2+a_{1}x+a_0$ has integer coefficients, then every rational zero of $f$ has the formRational zero = $p/q$ where p and q have no common factors other than 1, $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$ .

To use the Rational Zero Test, you should first list all rational numbers whosenumerators are factors of the constant term and whose denominators are factors of theleading coefficient.

Possible rational zeros: $\frac{Factorsof\text{constant}term}{Factorsofleadingcoefficient}$

Synthetic Division (for a Cubic Polynomial):To divide $a{x}^3+b{x}^2+cx+d$ by $x-k$ , use this pattern.

$\begin{array}{l}\begin{array}{l}k\\ \end{array}|\stackrel{Coefficientofdividend}{\overbrace{\underset{\_}{\begin{array}{cccc}\boxed{a}& b& c& d\\ \downarrow & \underset{\downarrow }{\boxed{ka}}& \underset{\downarrow }{\boxed{}}& \underset{\downarrow }{\boxed{}}\end{array}}}}\\ \begin{array}{ccccc}& \nearrow & \nearrow & \nearrow & \end{array}\\ \underset{Coefficientofquotient}{\underbrace{\begin{array}{ccc}\boxed{a}& \boxed{}& \boxed{}\end{array}}}\begin{array}{cc}\boxed{r}& \leftarrow \text{Remainder}\end{array}\end{array}$

In case when we have a polynomial with a missing term, insert placeholders with zero coefficients for missing powers of the variable. Vertical pattern: Add terms in columns Diagonal pattern: Multiply results by k. This algorithm for synthetic division works only for divisors of the form x - k. Remember that $x+k=x-\left(-k\right)$ .

The Division Algorithm: If $f\left(x\right)$ and $d\left(x\right)$ are polynomials such that $d\left(x\right)\ne 0$ , and the degree of $d\left(x\right)$ isless than or equal to the degree of $f\left(x\right)$ , then there exist unique polynomials $q\left(x\right)$ and $r\left(x\right)$ such that $\underset{\stackrel{\uparrow }{{\displaystyle Dividend}}}{{\displaystyle \text{f (x)}}}\text{ = }\underset{\stackrel{\uparrow }{{\displaystyle Divisor}}}{{\displaystyle \text{d(x)}}}\cdot \underset{\stackrel{\uparrow }{{\displaystyle Quotient}}}{{\displaystyle \text{q(x)}}}+\underset{\stackrel{\uparrow }{{\displaystyle Remainder}}}{{\displaystyle \text{ r (x)}}}$ where $r\left(x\right)=0$ or the degree of $r\left(x\right)$ is less than the degree of $d\left(x\right)$ . If theremainder $r\left(x\right)$ is zero, then $d\left(x\right)$ divides evenly into $f\left(x\right)$ and k can be called as zero of the polynomial.

Graph:

  

Interpretation:

From the graph of the function we can pick possible zeros of the function as

  $x=-2$ and check using synthetic division whether they are actual zeros.

Calculation:

Identify the polynomial by bringing all the terms to left side of the equation and write it in the form $f\left(x\right)=0$ . For our problem its already in that form.

  $f\left(x\right)=x^4+6x^3+3x^2-16x+6$

Identifying the constant and leading coefficient of the given function

Constant is $6$

Leading coefficient is $1$

Listing the factors of constant $-6$ : $\pm 1,\pm 2,\pm 3,\pm 6$

Listing the factors of Leading Coefficient 8: $\pm 1$

The possible rational zeros of the function are: $\pm 1,\pm 2,\pm 3,\pm 6$

From the graph let us use $x=-3$ and perform synthetic division

  $\begin{array}{l}\begin{array}{l}-3\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}6\\ -3\end{array}\begin{array}{c}3\\ -9\end{array}\begin{array}{c}-16\\ 18\end{array}\begin{array}{c}6\\ -6\end{array}}\\ 13-62\overline{)0}\leftarrow Remainder\end{array}$

If $k=-3$ is zero, then $x-k$ that is $x+3$ will be the divisor. The numbers in the last row make up your coefficients of the quotient as well as the remainder. The final value on the right is the remainder. Working right to left, the next number is your constant, the next is the coefficient for x , the next is the coefficient for x squared, and so on.

  $\left(x^4+6x^3+3x^2-16x+6\right)÷\left(x+3\right)=x^3+3x^2-6x+2$

From the graph let us use $x=1$ and perform synthetic division further

  $\begin{array}{l}\begin{array}{l}1\\ \end{array}|\underset{\_}{\begin{array}{c}1\\ \end{array}\begin{array}{c}3\\ 1\end{array}\begin{array}{c}-6\\ 4\end{array}\begin{array}{c}2\\ -2\end{array}}\\ 14-2\overline{)0}\leftarrow Remainder\end{array}$

If $k=1$ is zero, then $x-k$ that is $x-1$ will be the divisor. The numbers in the last row make up your coefficients of the quotient as well as the remainder. The final value on the right is the remainder. Working right to left, the next number is your constant, the next is the coefficient for x , the next is the coefficient for x squared, and so on.

  $\left(x^3+3x^2-6x+2\right)÷\left(x-1\right)=x^2+4x-2$

If we divide original function by both

  $\left(x+3\right)\&\left(x-1\right)$ we end with a quotient $x^2+4x-2$

  $\frac{x^4+6x^3+3x^2-16x+6}{\left(x+3\right)\left(x-1\right)}=x^2+4x-2$

To find other zeros of the polynomial

  $f\left(x\right)=x^4+6x^3+3x^2-16x+6$ , set quotient of synthetic division $x^2+4x-2$ to zero and solve for x .

  $x^2+4x-2=0$

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $x=\frac{-4\pm \sqrt{4^2-4\left(1\right)\left(-2\right)}}{2\left(1\right)}$

Simplify the expression inside the square root

  $\begin{array}{l}x=\frac{-4\pm \sqrt{16+8}}{2}\\ =\frac{-4\pm \sqrt{24}}{2}\end{array}$

Simplify the radical by splitting 20 as product of two numbers out of one is square number.

  $\begin{array}{l}x=\frac{-4\pm \sqrt{4}\cdot \sqrt{6}}{2}\\ =\frac{-4\pm 2\sqrt{6}}{2}\end{array}$

Cancel the common factor in the numerator and denominator to get the final answer

  $x=-2+\sqrt{6};x=-2-\sqrt{6}$

Conclusion:

The possible rational zeros of the function are: $\pm 1,\pm 2,\pm 3,\pm 6$

The actual rational zeros of the function are: $-3,1$

The real solution of the polynomial function is: $-3;1;-2+\sqrt{6};-2-\sqrt{6}$