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Sagot :
2(x²+1)(x+1)(x-1)
1) Let's examine that polynomial to check whether that polynomial can be factored or not.
[tex]2x^4-2[/tex]2) Examining we can see that the coefficient of the x-term (2) and the constant (2), are the same. So we can find the GCD, between 2,2 and write it as a factor.
GCD (2, 2) = 2
Now we'll divide every term by the GCD, and write it inside the parentheses just like this
[tex]\begin{gathered} 2x^4-2 \\ \frac{2x^4}{2}=x^4 \\ \frac{-2}{2}=-1 \\ \\ 2(x^4-1)^{} \end{gathered}[/tex]2.2) Now let's proceed with the factorization of the binomial x^4-1
Remembering that (a+n)(a-b)=a²-b²
Notice that we are dealing with a 4th-degree polynomial, so we can write it as:
So we can rewrite it as
[tex]2(x^4-1)=2(x^2+1)(x^2-1^2)\text{ =}2(x^2+1)(x+1)(x-1)[/tex]3) So our polynomial is not prime, for it can be reduced. And it can be written as 2(x²+1)(x+1)(x-1) in its simplest form.
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