Question

How do you find all the zeros of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\]?

Answer 1

Hint: Zero of the function is also known as the root of the function. The zero of a function is any replacement for the variable that will produce an answer of zero. Graphically, the real zero of a function is where the graph of the function crosses the x‐axis; that is, the real zero of a function is the x‐intercept(s) of the graph of the function.

Complete step-by-step solution:
From the question it is clear we have to find all the zeros of of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\].
zeros of the function is also known as the root of the function.
Since the highest degree is \[3\], we have to get \[3\] roots.
Now consider the given equation
\[\Rightarrow f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\]……………..(1)
Now let us re-group the equation in the simpler form.
\[\Rightarrow f\left( x \right)=\left( {{x}^{3}}-4{{x}^{2}} \right)+\left( 16x-64 \right)\]……………..(2)
we can write \[64\]as the product of \[16\] and \[4\].
\[\Rightarrow f\left( x \right)=\left( {{x}^{3}}-4{{x}^{2}} \right)+\left( 16x-16\times 4 \right)\]………………(3)
Now take \[{{x}^{2}}\]common from \[\left( {{x}^{3}}-4{{x}^{2}} \right)\] and \[16\] common from \[\left( 16x-16\times 4 \right)\]
Now equation (3) becomes
\[\Rightarrow f\left( x \right)={{x}^{2}}\left( x-4 \right)+16\left( x-4 \right)\]………………(4)
Now take \[\left( x-4 \right)\] common from \[{{x}^{2}}\left( x-4 \right)\] and \[16\left( x-4 \right)\]
 So, equation (4) becomes
\[\Rightarrow f\left( x \right)=\left( x-4 \right)\left( {{x}^{2}}+16 \right)\]
To the values of roots, we have to equate the equation to zero. So \[f\left( x \right)=0\]
\[\Rightarrow \left( x-4 \right)\left( {{x}^{2}}+16 \right)=0\]
CASE 1:
\[\Rightarrow \left( x-4 \right)=0\]
Now add 4 on both sides
\[\Rightarrow \left( x-4+4 \right)=0+4\]
After simplification we get
\[\Rightarrow x=4\]
\[x=4\] is one of the roots of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\]
CASE 2:
\[\left( {{x}^{2}}+16 \right)=0\]
Now subtract 16 from both sides
\[\Rightarrow \left( {{x}^{2}}+16-16 \right)=0-16\]
After simplification we get
\[\Rightarrow {{x}^{2}}=-16\]
\[\Rightarrow x=\sqrt{-16}\]
\[\Rightarrow x=\sqrt{-1\times 16}\]
We know that \[{{i}^{2}}=-1\] and \[16={{4}^{2}}\]
Now we can write\[{{i}^{2}}=-1\] and \[16={{4}^{2}}\]
\[\Rightarrow x=\sqrt{{{i}^{2}}\times {{4}^{2}}}\]
\[\Rightarrow x=\pm 4i\]
\[x=+4i\] and \[x=-4i\] are the other two roots of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\].
\[x=+4i\] and \[x=-4i\] are imaginary roots.
\[4,+4i,-4i\] are the three roots of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\].
Now let us try to graph \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\]

Here \[x=4\] is the only root that can be represented on the graph because it is the real root.

Note: Students may have misconception that \[{{x}^{2}}+16=0\] is not possible so, \[x=4\] is the only root of \[f\left( x \right)={{x}^{3}}-4{{x}^{2}}+16x-64\]. But actually, it is possible by taking \[{{i}^{2}}=-1\] which gives us imaginary and complex roots.