# Find all zeros of the polynomial. P(x)=x^{3}-3x^{2}+3x-2

Find all zeros of the polynomial.
$P\left(x\right)={x}^{3}-3{x}^{2}+3x-2$
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Step 1
Given polynomial is $P\left(x\right)={x}^{3}-3{x}^{2}+3x-2$
Step 2
To find zeros of the polynomial.
First find the rational zero of polynomial by using rational root test.
The rational root test is if the polynomial has a rational zero then it must be a fraction $\frac{p}{q}$, where p is factor
of constant term and q is factor of leading coefficient.
Here p=-2 and q=1.
The factors of -2 are $±1,±2$.
Now substitute possible values to find zeroes of polynomial.
For x=1, $⇒P\left(1\right)={\left(1\right)}^{3}-3{\left(1\right)}^{2}+3\left(1\right)-2=1-3=3-2=-1$
For x=2, $⇒P\left(2\right)={\left(2\right)}^{3}-3{\left(2\right)}^{2}+3\left(1\right)-2=8-12+4=0$
Thus the one zero of the polynomial is x=2.
Step 3
Now divide the polynomial $P\left(x\right)={x}^{3}-3{x}^{2}+3x-2$ with x-2 get,
$\frac{{x}^{3}-3{x}^{2}+3x-2}{x-2}={x}^{2}-x+1$
$⇒{x}^{3}-3{x}^{2}+3x-2=\left(x-2\right)\left({x}^{2}-x+1\right)$
Now for ${x}^{2}-x+1=0$
$⇒x=\frac{-\left(-1\right)±\sqrt{{\left(-1\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}$
$⇒x=\frac{1±\sqrt{1-4}}{2}$
$⇒x=\frac{1±\sqrt{3}i}{2}$
Therefore the zeroes of the polynomial $P\left(x\right)={x}^{3}-3{x}^{2}+3x-2$ are,
$x=2,x=\frac{1+\sqrt{3}i}{2},x=\frac{1-\sqrt{3}i}{2}$

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