 Algotssleeddynf

2021-12-29

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime.
${x}^{6}-2{x}^{3}+1$ Mary Goodson

Step 1Factorizing a polynomial is breaking a polynomial into small polynomials. If we multiply again those smaller polynomial with each other we will get the same polynomial.
There are many ways of factorizing, one such is taking GCF (greatest common factor) of each term of the polynomial and taking GCF common and represent it in terms of GCF and another factor.
Another important property is distributive property which is given by a(b+c)=ab+ac. It is taking common value from function and representing it in terms of common factor and another function.
Step 2
For the equation ${x}^{6}-2{x}^{3}+1$, let's take ${x}^{3}=t$ and put it in given equation to get equation in terms of variable t
${x}^{6}-2{x}^{3}+1={t}^{2}-2t+1$...(1)
From equation(1), factor the function of t by substituting −2t as (−t−t) and taking common values out.
${t}^{2}-2t+1={t}^{2}-t-t+1$
=t(t-1)-1(t-1)
=(t-1)(t-1) (using distributive property)
$={\left(t-1\right)}^{2}$...(2)
Step 3
Resubstitute value of $t={x}^{3}$ in equation (2) to get factor form equation ${x}^{6}-2{x}^{3}+1$
${t}^{2}-2t+1={\left({x}^{3}-1\right)}^{2}$
${x}^{6}-2{x}^{3}+1={\left({x}^{3}-1\right)}^{2}$ (from equation(1))
Therefore factored form of ${x}^{6}-2{x}^{3}+1$ is ${\left({x}^{3}-1\right)}^{2}$ Durst37

${\left({x}^{3}-1\right)}^{2}$
$\left({x}^{3}-2\right){x}^{3}+1$
${\left(x-1\right)}^{2}{\left({x}^{2}+x+1\right)}^{2}$
${x}^{6}-2{x}^{3}+1$ Vasquez

Given
$P\left(x\right)={x}^{6}-2{x}^{3}+1$
$=\left({x}^{3}-1{\right)}^{2}$
$=\left({x}^{3}-1\right)\left({x}^{3}-1\right)$
$P\left(x\right)=\left(x-1\right)\left({x}^{2}+x+1\right)\left(x-1\right)\left({x}^{2}+x+1\right)$[since ${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+b\right)$]
$P\left(x\right)=\left(x-1{\right)}^{2}\left({x}^{2}+x+1{\right)}^{2}$
To find zeros consider P(x)=0

$\left(x-1{\right)}^{2}\left({x}^{2}+x+1{\right)}^{2}=0$
$\left(x-1{\right)}^{2}=0,\left({x}^{2}+x+1{\right)}^{2}=0$
$x-1=0,{x}^{2}+x+1=0$
$x=1,x=\frac{-1±\sqrt{3}i}{2}$
Therefore, x=1 is the real root.

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