Chapter 4.5, Problem 68AYU

To determine

The solution of the equation $2x^4+x^3-24x^2+20x+16=0$ in the real number system.

Answer to Problem 68AYU

Solution:

The real solution of the equation $2x^4+x^3-24x^2+20x+16=0$ is $x=-4,-\frac{1}{2},2$

Explanation of Solution

Given Information:

The equation $2x^4+x^3-24x^2+20x+16=0$

Step-1: The degree of the polynomial function $f\left(x\right)=2x^4+x^3-24x^2+20x+16$ is 4; therefore, there are at the most 4 real zeros.

Step-2: Since there are 2 variations in the sign of the non-zero coefficients of $f\left(x\right)$ by using the Descartes’ Rule of Signs, there are either 2 or 0 positive real roots

$f\left(-x\right)=2{\left(-x\right)}^4+{\left(-x\right)}^3-24{\left(-x\right)}^2+20\left(-x\right)+16$

$\Rightarrow f\left(-x\right)=2x^4-x^3-24x^2-20x+16$.

There are 2 variations in the sign of the non-zero coefficients of $f\left(-x\right)$ by using the Descartes’ Rule of Signs, there are either 2 negative real zeros or 0 negative real zero.

Step-3: By the Rational Zero Theorem, all possible rational zeroes are of the form $x=\frac{p}{q}$, where $p$ means all factors of the constant term in the polynomial and $q$ means all factors of the leading terms of the polynomial.

In the polynomial function $f\left(x\right)=2x^4+x^3-24x^2+20x+16$, the constant term is $16$ and the leading term is 2

Factors of the constant term $16=p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$

Factors of the leading term $2=q=\pm 1,\pm 2$

Therefore, all the possible rational zeros are:

$\frac{p}{q}:$ $\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{2}{1},\pm \frac{2}{2},\pm \frac{4}{1},\pm \frac{4}{2},\pm \frac{8}{1},\pm \frac{8}{2},\pm \frac{16}{1},\pm \frac{16}{2}$.

Simplifies and reduces to

$\frac{p}{q}:\pm 1,\pm \frac{1}{2},\pm 2,\pm 4,\pm 8,\pm 16$

Thus, all the potential rational zeros of the polynomial function $f\left(x\right)=2x^4+x^3-24x^2+20x+16$ are $x=1,-1,\frac{1}{2},-\frac{1}{2},2,-2,4,-4,8,-8,16,-16$. $$

Now, test $x=2$ using synthetic division.

$\begin{array}{l}2\overline{)21-242016}\\ \underset{\_}{410-28-16}\\ 25-14-80\end{array}$

Here, since the remainder is 0, $x=2$ is a zero of $f$ and $\left(x-2\right)$ is a factor of $f$

To write the factors of $f$, use the entries in the bottom row of the above synthetic division.

$f\left(x\right)=2x^4+x^3-24x^2+20x+16=\left(x-2\right)\left(2x^3+5x^2-14x-8\right)$.

Here, $f_1\left(x\right)=2x^3+5x^2-14x-8=0$ is called a depressed equation

Any solution to this depressed equation is also a zero of $f$, so now find zeros of this depressed equation.

Repeat step-3 to find zeros of this depressed equation.

In the polynomial $f_1\left(x\right)=2x^3+5x^2-14x-8$, the constant term is $-8$ and the leading term is 2

Factors of the constant term $-8=p=\pm 1,\pm 2,\pm 4,\pm 8$

Factors of the leading term $2=q=\pm 1,\pm 2$

Therefore, all possible rational zeros are:

$\frac{p}{q}:$ $\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{2}{1},\pm \frac{2}{2},\pm \frac{4}{1},\pm \frac{4}{2},\pm \frac{8}{1},\pm \frac{8}{2}$.

Simplifies and reduces to

$\frac{p}{q}:\pm 1,\pm \frac{1}{2},\pm 2,\pm 4,\pm 8$.

Thus, all the potential rational zeros of the depressed equation $f_1\left(x\right)=2x^3+5x^2-14x-8$ are $x=1,-1,\frac{1}{2},-\frac{1}{2},2,-2,4,-4,8,-8$. $$

Now, test $x=1$ using synthetic division.

$\begin{array}{l}1\overline{)25-14-8}\\ \underset{\_}{27-7}\\ 27-7-15\end{array}$

Here, since the remainder is $-15$, $x=1$ is not a zero of $f$

Now, test $x=-\frac{1}{2}$ using synthetic division.

$\begin{array}{l}-\frac{1}{2}\overline{)25-14-8}\\ \underset{\_}{2-1-28}\\ 24-160\end{array}$

Here, since the remainder is 0, $x=-\frac{1}{2}$ is a zero of $f$ and $\left(x+\frac{1}{2}\right)$ is a factor of $f$

Use the entries in the bottom row of the synthetic division to continue the factoring of $f$

$f\left(x\right)=2x^4+x^3-24x^2+20x+16=\left(x-2\right)\left(x+\frac{1}{2}\right)\left(2x^2+4x-16\right)$.

$\Rightarrow f\left(x\right)=2x^4+x^3-24x^2+20x+16=2\left(x-2\right)\left(x+\frac{1}{2}\right)\left(x^2+2x-8\right)$.

Here, $f_2\left(x\right)=x^2+2x-8=0$ is called a depressed equation and it can be factored as follows:

$x^2+2x-8=\left(x-2\right)\left(x+4\right)=0$.

$\Rightarrow \left(x-2\right)\left(x+4\right)=0$

$\Rightarrow \left(x-2\right)=0$ or $\left(x+4\right)=0$

$\Rightarrow x=2$ or $x=-4$.

Thus, the real solution of the equation $2x^4+x^3-24x^2+20x+16=0$ is $x=-\frac{1}{2},-4,2$ with 2 being a repeated zero of multiplicity 2.