Chapter 4.5, Problem 47E

To determine

To solve: the given equation using two methods. Explain the easier one.

Answer to Problem 47E

The real solutions are $-3,4$ , and $3$ .

Factorization method is easier.

Explanation of Solution

Given information:

Given equation,

  $x^3-4x^2-9x+36=0$

Calculation:

Consider the equation.

  $x^3-4x^2-9x+36=0$

Method 1:

Solve the above equation by factorizing the left hand side.

  $\begin{array}{l}{x}^3-4x^2-9x+36=0\\ x^2\left(x-4\right)-9\left(x-4\right)=0\\ \left(x-4\right)\left(x^2-9\right)=0\\ \left(x-4\right)\left(x+3\right)\left(x-3\right)=0\left[a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\end{array}$

Equating each factor with 0, we get

  $\begin{array}{c}\left(x-4\right)=0\\ x=4\\ x+3=0\\ x=-3\\ x-3=0\\ x=3\end{array}$

The solutions are $-3,4$ , and $3$ .

Method 2:

The above equation is of degree 3, so it has 3 solutions.

Here, the leading coefficient is $1$ and constant term is $36$ .

The possible rational solutions are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 9,\pm 12,\pm 18\text{and}\pm 36$ .

Test possible solution using synthetic division until a solution is found.

For $x=3$

  $\begin{array}{ccccc}3& 1& -4& -9& 36\\ & & 3& -3& -36\\ & 1& -1& -12& 0\end{array}$

Therefore, $x-3$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}{x}^3-4x^2-9x+36=0\\ \left(x-3\right)\left(x^2-x-12\right)=0\\ \left(x-3\right)\left(x-4\right)\left(x+3\right)=0\end{array}$

Solve further,

  $\begin{array}{c}\left(x-3\right)\left(x-4\right)\left(x+3\right)=0\\ x=-3,4,3\end{array}$

Therefore, all real solutions are $-3,4$ , and $3$ .

Method 1 is easier as it takes fewer steps than method 2.