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Question: How do you find all real and complex roots of \(f\left( x \right)={{x}^{3}}-4{{x}^{2}}-16x+64\)?...

How do you find all real and complex roots of f(x)=x3−4x2−16x+64f\left( x \right)={{x}^{3}}-4{{x}^{2}}-16x+64?

Explanation

Solution

In this problem we have to calculate the roots of the given equation. We can observe that the given equation is the cubic equation. So, we will solve this problem by using the factors of the cubic equation. Now we will consider the first two terms individually and take x2{{x}^{2}} as common. After that we will consider the last two terms and take −16-16 as common. Now we will observe the obtained equation and take appropriate terms as common to get the factors of the given equation. After getting the factors of the cubic equation, we will equate each factor to zero and calculate the values of xx which are our required roots.

Complete step by step solution:
Given equation, f(x)=x3−4x2−16x+64f\left( x \right)={{x}^{3}}-4{{x}^{2}}-16x+64.
Considering the first two terms. We have the first term x3{{x}^{3}} and the second term −4x2-4{{x}^{2}}. By observing the above two terms we can take x2{{x}^{2}} as common. So, taking x2{{x}^{2}} as common from the first two terms of the given equation, then we will get
⇒x3−4x2−16x+64=x2(x−4)−16x+64\Rightarrow {{x}^{3}}-4{{x}^{2}}-16x+64={{x}^{2}}\left( x-4 \right)-16x+64
Considering the last two terms of the given equation. We have last term 6464 and the third term −16x-16x. By observing the above two terms we can take −16-16 as common. So, taking −16-16 as common from the last two terms of the given equation, then we will get
⇒x3−4x2−16x+64=x2(x−4)−16(x−4)\Rightarrow {{x}^{3}}-4{{x}^{2}}-16x+64={{x}^{2}}\left( x-4 \right)-16\left( x-4 \right)
In the above equation we can observe that we can take x−4x-4 as common. So, taking x−4x-4 as common from the above equation, then we will get
⇒x3−4x2−16x+64=(x−4)(x2−16)\Rightarrow {{x}^{3}}-4{{x}^{2}}-16x+64=\left( x-4 \right)\left( {{x}^{2}}-16 \right)
Hence the factors of the given equation f(x)=x3−4x2−16x+64f\left( x \right)={{x}^{3}}-4{{x}^{2}}-16x+64 are x2−16{{x}^{2}}-16, x−4x-4.
Considering the first factor which is x2−16{{x}^{2}}-16. Equating this factor to zero, then we will get
⇒x2−16=0\Rightarrow {{x}^{2}}-16=0
We can write 16=4216={{4}^{2}}, then we will have
⇒x2−42=0\Rightarrow {{x}^{2}}-{{4}^{2}}=0
Applying the formula a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) in the above equation, then we will have
⇒(x+4)(x−4)=0\Rightarrow \left( x+4 \right)\left( x-4 \right)=0
Equating each term individually to zero, then we will get
⇒x+4=0 or x−4=0 ⇒x=−4 or x=4 \begin{aligned} & \Rightarrow x+4=0\text{ or }x-4=0 \\\ & \Rightarrow x=-4\text{ or }x=4 \\\ \end{aligned}
Considering the second factor x−4x-4. Equating this factor to zero, then we will have
⇒x−4=0 ⇒x=4 \begin{aligned} & \Rightarrow x-4=0 \\\ & \Rightarrow x=4 \\\ \end{aligned}

Hence the roots of the given equation f(x)=x3−4x2−16x+64f\left( x \right)={{x}^{3}}-4{{x}^{2}}-16x+64 are ±4\pm 4.

Note: For this problem we have the factor x2−16{{x}^{2}}-16 which is in form of a2−b2{{a}^{2}}-{{b}^{2}}, so we have used the algebraic formula and equated it to zero. In many cases we may get a quadratic equation which is in the form of ax2+bx+ca{{x}^{2}}+bx+c. Now we will use the formula x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} and calculate the roots of the equation.