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

Last updated date: 20th Jun 2024
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Hint: 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 ${{x}^{2}}$ as common. After that we will consider the last two terms and take $-16$ 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 $x$ which are our required roots.

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

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

Note: For this problem we have the factor ${{x}^{2}}-16$ which is in form of ${{a}^{2}}-{{b}^{2}}$, 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 $a{{x}^{2}}+bx+c$. Now we will use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and calculate the roots of the equation.