Question

How do you find all the zeros of ${x^3} - 3{x^2} + 4x - 2$?

Answer 1

Hint: In this question, they have given an equation and we are asked to find all of its zeros. We will first find whether 1 is a zero or not by adding all the coefficients and checking it whether it equals zero or not. Then we have taken out the factor and find the remaining zeros of the equation given.

Complete step-by-step solution:
In this question, they have given an equation and we are asked to find all of its zeros.
First we will check whether the sum of the coefficients is zero. If it is zero, then $1$ is one of the zeros of the equation.
That is:\[1 - 3 + 4 - 2 = 0\]
Hence \[x = 1\] is one of the zero of the given equation ${x^3} - 3{x^2} + 4x - 2$
That implies that \[(x - 1)\;\]is a factor of the given equation,
If \[(x - 1)\;\] is one of the factors of the given equation${x^3} - 3{x^2} + 4x - 2$, then
$ \Rightarrow \dfrac{{{x^3} - 3{x^2} + 4x - 2}}{{x - 1}}$
We need to find the other factor,
$ \Rightarrow \dfrac{{{x^3} - {x^2} - 2{x^2} + 2x + 2x - 2}}{{x - 1}}$
$ \Rightarrow \dfrac{{{x^2}\left( {x - 1} \right) - 2x\left( {x - 1} \right) + 2\left( {x - 1} \right)}}{{x - 1}}$
Taking the common out,
\[ \Rightarrow \dfrac{{{x^2} - 2x + 2\left( {x - 1} \right)}}{{x - 1}}\]
Cancelling $x - 1$ in both numerator and denominator,
We get \[{x^2} - 2x + 2\] , so this the other factor
Therefore,
$ \Rightarrow {x^3} - 3{x^2} + 4x - 2 = (x - 1)({x^2} - 2x + 2)$
The remaining quadratic factor only has Complex zeros, which we can find by completing the square and using the difference of squares identity:
${a^2} - {b^2} = (a - b)(a + b)$, with\[\;a = (x - 1)\;\] and\[\;b = i\] as follows:
We know that, ${x^2} - 2x + 1$ can be written as ${(x - 1)^2}$
Then,\[{x^2} - 2x + 2\]
On rewriting we get
\[ \Rightarrow {x^2} - 2x + 1 + 1\]
Then, we get
$ \Rightarrow {(x - 1)^2} + 1$
$1$ can be written as ${i^2}$ then,
$ \Rightarrow {(x - 1)^2} - {i^2}$
On rewriting we get
\[ \Rightarrow ((x - 1) - i)((x - 1) + i)\]
Then, we get
\[ \Rightarrow (x - 1 - i)(x - 1 + i)\]
Hence zeros \[x = 1 \pm i\]

Therefore the zeros of ${x^3} - 3{x^2} + 4x - 2$ are $1,1 + i,1 - i$

Note: The zero of a function or any equation is any replacement for the variable that will produce an answer of zero to the equation.
In mathematics, a zero of a real-, complex-, or generally vector-valued function f, is a member x of the domain of f such that f(x) attains the value of 0 at x, or equivalently, x is the solution to the equation \[f\left( x \right) = 0\].
Note that, the number of zeros of an equation can be determined as the highest value of degree. If the highest degree in the equation is three, then there are three zeros or roots available for the equation.