interdicoxd

2021-12-20

All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{3}-6{x}^{2}+12x-8$

jgardner33v4

Expert

Step 1
Given:
$P\left(x\right)={x}^{3}-6{x}^{2}+12x-8$
Step 2
We can write the given polynomial as:
$P\left(x\right)={\left(x\right)}^{3}-3{\left(x\right)}^{2}\left(2\right)+3\left(x\right){\left(2\right)}^{2}-{\left(2\right)}^{3}$
We have the formula,
${\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}$
Hence,
$P\left(x\right)={\left(x-2\right)}^{3}$
=(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.

Marcus Herman

Expert

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

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