Chapter 4.5, Problem 67AYU

In Problems $57-68$, solve each equation in the real number system.

$2x^4-19x^3+57x^2-64x+20=0$

To determine

The solution of the equation $2x^4-19x^3+57x^2-64x+20=0$ in the real number system.

Answer to Problem 67AYU

Solution:

The real solutions of the equation $2x^4-19x^3+57x^2-64x+20=0$ are $x=\frac{1}{2},2,5$.

Explanation of Solution

Given Information:

The equation $2x^4-19x^3+57x^2-64x+20=0$

Step-1: The degree of the polynomial $f\left(x\right)=2x^4-19x^3+57x^2-64x+20$ is 4; therefore, there are at the most 4 real zeros.

Step-2: Since there are 4 variations in the sign of the non-zero coefficients of $f\left(x\right)$ by using the Descartes’ Rule of Signs, there are either 4 positive real zeros or 2 or 0 positive real roots.

$f\left(-x\right)=2{\left(-x\right)}^4-19{\left(-x\right)}^3+57{\left(-x\right)}^2-64\left(-x\right)+20$

$\Rightarrow f\left(-x\right)=2x^4-19x^3+57x^2+64x+20$.

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 the possible rational zeroes are of the form $x=\frac{p}{q}$, where $p$ means all the factors of the constant term in the polynomial and $q$ means all the factors of the leading term of the polynomial.

In the polynomial $f\left(x\right)=2x^4-19x^3+57x^2-64x+20$, the constant term is $20$ and the leading term is 2.

Factors of the constant term $20=p=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

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

Therefore, all the possible rational zeros are:

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

Simplifies and reduces to

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

Thus, all the potential rational zeros of the polynomial function $f\left(x\right)=2x^4-19x^3+57x^2-64x+20$ are $x=1,-1,\frac{1}{2},-\frac{1}{2},2,-2,4,-4,5,-5,\frac{5}{2},-\frac{5}{2},10,-10,20,-20$. $$

Now, test $x=1$ using synthetic division

$\begin{array}{l}1\overline{)2-1957-6420}\\ \underset{\_}{2-1740-24}\\ 2-1740-24-4\end{array}$

Here, since the remainder is $-4$, $x=1$ is not a zero of $f$

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

$\begin{array}{l}\frac{1}{2}\overline{)2-1957-6420}\\ \underset{\_}{1-924-20}\\ 2-1848-400\end{array}$

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 $f$

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

$f\left(x\right)=2x^4-19x^3+57x^2-64x+20=\left(x-\frac{1}{2}\right)\left(2x^3-18x^2+48x-40\right)$.

$\Rightarrow f\left(x\right)=2x^4-19x^3+57x^2-64x+20=2\left(x-\frac{1}{2}\right)\left(x^3-9x^2+24x-20\right)$

Here, $f_1\left(x\right)=x^3-9x^2+24x-20=0$ is called a depressed equation

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

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

In the polynomial $f_1\left(x\right)=x^3-9x^2+24x-20$, the constant term is $-20$ and the leading term is 1.

Factors of the constant term $-20=p=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

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

Therefore, all the possible rational zeros are:

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

Simplifies and reduces to

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

Thus, all the potential rational zeros of the depressed equation $f_1\left(x\right)=x^3-9x^2+24x-20$ are $x=1,-1,2,-2,4,-4,5,-5,10,-10,20,-20$. $$

Now, test $x=2$ using synthetic division

$\begin{array}{l}2\overline{)1-924-20}\\ \underset{\_}{2-1420}\\ 1-7100\end{array}$

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

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

$f\left(x\right)=2x^4-19x^3+57x^2-64x+20=2\left(x-\frac{1}{2}\right)\left(x-2\right)\left(x^2-7x+10\right)$.

Here, $f_2\left(x\right)=x^2-7x+10=0$ is called a depressed equation and it can be factored as follows:

$x^2-7x+10=\left(x-5\right)\left(x-2\right)=0$.

$\Rightarrow \left(x-5\right)\left(x-2\right)=0$

$\Rightarrow \left(x-5\right)=0$ or $\left(x-2\right)=0$

$\Rightarrow x=2$ or $x=5$

Thus, the real zeros of the equation $2x^4-19x^3+57x^2-64x+20=0$ are $x=\frac{1}{2},2,5$ with 2 being

a repeated zero of multiplicity 2