All the real zeros of the given polynomial are integers. Find the zeros, and write...

Step 1
Given:
${\displaystyle P\left(x\right)=x^3-6x^2+12x-8}$
Step 2
We can write the given polynomial as:
${\displaystyle P\left(x\right)={\left(x\right)}^3-3{\left(x\right)}^2\left(2\right)+3\left(x\right){\left(2\right)}^2-{\left(2\right)}^3}$
We have the formula,
${\displaystyle {(a-b)}^3=a^3-3a^{2}b+3a{b}^2-b^3}$
Hence,
${\displaystyle P\left(x\right)={(x-2)}^3}$
=(x-2)(x-2)(x-2)
This is the polynomial P(x) in factored form.
Step 3
To find the zeros of polynomial P(x), equate it to 0.
P(x)=0
(x-2)(x-2)(x-2)=0
x-2=0 or x-2=0 or x-2=0
x=2,2,2
Hence, the polynomial P(x) has all integer real zeros and they are equal to 2 with multiplicity 3.

The given polynomial is ${\displaystyle P\left(x\right)=x^3-6x^2+12x-8}$
Since the leading coefficient is 1, any rational zero must be a divisior of the constant term -8.
So the possible rational zeros are ${\displaystyle \pm 1,\pm 2,\pm 4,\pm 8}$
We test each of these possibilities
${\displaystyle P\left(1\right)={\left(1\right)}^3-6{\left(1\right)}^2+12\left(1\right)-8}$
=1-6+12-8
=-1
${\displaystyle P(-1)={(-1)}^3-6{(-1)}^2+12(-1)-8}$
=-1-6-12-8
=-27
${\displaystyle P\left(2\right)={\left(2\right)}^3-\left(6\right){\left(2\right)}^2+12\left(2\right)-8}$
=8-24+24-8
=0
${\displaystyle P(-2)={(-2)}^3-6{(-2)}^2+12(-2)-8}$
=-8-24-24-8
=-64
${\displaystyle P\left(4\right)={\left(4\right)}^3-6{\left(4\right)}^2+12\left(4\right)-8}$
=64-96+48-8
=8
${\displaystyle P(-4)={(-4)}^3-6{(-4)}^2+12(-4)-8}$
=-64-96-48-8
=-216
${\displaystyle P\left(8\right)={\left(8\right)}^3-6{\left(8\right)}^2+12\left(8\right)-8}$
=512-384+96-8
=216
${\displaystyle P(-8)={(-8)}^3-6{(-8)}^2+12(-8)-8}$
=-512-384-96-8
=-1000
The rational zero of P is 2