[tex]x^3-2x^2+x-1[/tex] is one of the prime factors of the polynomial
How to factor the expression?
The question implies that we determine one of the prime factors of the polynomial.
The polynomial is given as:
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1[/tex]
Expand the polynomial by adding 0's in the form of +a - a
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1[/tex]
Rearrange the terms
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1[/tex]
Factorize the expression
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]
Factor out x^6-2x^5+2x^3-3x^2+2x-1
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]
Express x^2 + 2x + 1 as a perfect square
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]
Expand the polynomial by adding 0's in the form of +a - a
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)[/tex]
Rearrange the terms
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)[/tex]
Factorize the expression
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))[/tex]
Factor out x^3-2x^2+x-1
[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)[/tex]
One of the factors of the above polynomial is [tex]x^3-2x^2+x-1[/tex].
This is the same as the option (c)
Hence, [tex]x^3-2x^2+x-1[/tex] is one of the prime factors of the polynomial
Read more about polynomials at:
https://brainly.com/question/4142886
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