Chapter 8, Problem 10CR

To determine

The solution of the equation $3x^5-10x^4+21x^3-42x^2+36x-8=0$ in complex number system.

Answer to Problem 10CR

Solution:

The solution of the equation is $x=1,2,\frac{1}{3},2i,-2i$

Explanation of Solution

Given information:

The equation $3x^5-10x^4+21x^3-42x^2+36x-8=0$

Consider the equation $3x^5-10x^4+21x^3-42x^2+36x-8=0$

Let $f\left(x\right)=3x^5-10x^4+21x^3-42x^2+36x-8$

To solve this equation, use the rational root theorem to find its possible roots

Rational root theorem states that for a polynomial equation with integer coefficients

  $a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{.......}+a_1{x}^1+a_0=0$, if there exists a rational solution, then the possible solution can be found by checking all the numbers $\pm \frac{\text{Divisors}\text{of}{a}_0}{\text{Divisors}\text{of}{a}_n}$

On comparing with above equation $a_0=-8$ and $a_n=3$

  $\text{Divisors}\text{of}{a}_0$ are $\pm 1,\pm 2,\pm 4,\pm 8$

  $\text{Divisors}\text{of}{a}_n$ are $\pm 1,\pm 3$

Thus, possible roots are $\pm \frac{1,2,4,8}{1,3}$

That is, $\pm \left(1,\frac{1}{3},2,\frac{2}{3},4,\frac{4}{3},8,\frac{8}{3}\right)$

Consider $x=1$, check whether it is root or not

Substitute $x=1$ in $3x^5-10x^4+21x^3-42x^2+36x-8$

  $\Rightarrow 3{\left(1\right)}^5-10{\left(1\right)}^4+21{\left(1\right)}^3-42{\left(1\right)}^2+36\left(1\right)-8$

  $\Rightarrow 3-10+21-42+36-8=0$

Therefore, $x=1$ is a root of $3x^5-10x^4+21x^3-42x^2+36x-8=0$

Therefore, $\left(x-1\right)$ is factor of $f\left(x\right)$

Factor out $f\left(x\right)$ by synthetic division with $x=1$ as root

Therefore,

  $\begin{array}{ccccccc}1& 3& -10& 21& -42& 36& -8\\ & & 3& -7& 14& -28& 8\\ & 3& -7& 14& -28& 8& 0\end{array}$ .

Thus, the polynomial $f\left(x\right)$ is $\left(x-1\right)\left(3x^4-7x^3+14x^2-28x+8\right)$

Consider $x=2$, check whether $x=2$ is root or not of $f\left(x\right)$

Substitute $x=2$ in $3x^4-7x^3+14x^2-28x+8$

  $\Rightarrow 3{\left(2\right)}^4-7{\left(2\right)}^3+14{\left(2\right)}^2-28\left(2\right)+8$

  $\Rightarrow 3\left(16\right)-7\left(8\right)+14\left(4\right)-56+8$

  $\Rightarrow 48-56+56-56+8=0$

Therefore, $x=2$ is root of above polynomial

Therefore, $\left(x-2\right)$ is factor of $f\left(x\right)$

That is, $\left(x-2\right)$ is a factor of $3x^4-7x^3+14x^2-28x+8$

By synthetic division with $x=2$ as root factor out $3x^4-7x^3+14x^2-28x+8$

Therefore, $\begin{array}{cccccc}2& 3& -7& 14& -28& 8\\ & & 6& -2& 24& -8\\ & 3& -1& 12& -4& 0\end{array}$

Thus, the term $3x^4-7x^3+14x^2-28x+8$ is $\left(x-2\right)\left(3x^3-x^2+12x-4\right)$

To find the factors of $3x^3-x^2+12x-4$

Consider $x=\frac{1}{3}$, check whether it is root or not of $f\left(x\right)$

It could be checked by substituting $x=\frac{1}{3}$ in $3x^3-x^2+12x-4$

  $\Rightarrow 3{\left(\frac{1}{3}\right)}^3-{\left(\frac{1}{3}\right)}^2+12\left(\frac{1}{3}\right)-4$

  $\Rightarrow \frac{1}{9}-\frac{1}{9}+4-4=0$

Thus, $x=\frac{1}{3}$ is a root of $f\left(x\right)$

By using synthetic division with $x=\frac{1}{3}$ as a root gives

  $\begin{array}{ccccc}\frac{1}{3}& 3& -1& 12& -4\\ & & 1& 0& 4\\ & 3& 0& 12& 0\end{array}$

Thus, the term $3x^3-x^2+12x-4$ is written as $\left(x-\frac{1}{3}\right)\left(3x^2+12\right)$

Now, to factor the polynomial $3x^2+12$

Factor out $3$ in $3x^2+12$ it gives $3\left(x^2+4\right)$

Consider $x^2+4=0$

  $\Rightarrow x^2=-4$

  $\Rightarrow x=2i,-2i$

Therefore, the equation is $\left(x-1\right)\left(x-2\right)\left(3x-1\right)\left(x-2i\right)\left(x+2i\right)=0$

Using the Zero Factor Principle: If $ab=0$, then $a=0$ or $b=0$ or both $a=0,b=0$ .

Thus, solution of equation is $x=1,2,\frac{1}{3},2i,-2i$