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-Dominant- [34]
2 years ago
14

Ayuda necesito resolver este problema con procedimiento ;)

Mathematics
1 answer:
Paraphin [41]2 years ago
4 0

x^3-2x^2+x-1 is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

x^8 - 3x^6 + x^4 - 2x^3 - 1

Expand the polynomial by adding 0's in the form of +a - a

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

Rearrange the terms

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

Factorize the expression

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)

Factor out x^6-2x^5+2x^3-3x^2+2x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)

Express x^2 + 2x + 1 as a perfect square

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)

Expand the polynomial by adding 0's in the form of +a - a

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)

Rearrange the terms

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)

Factorize the expression

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))

Factor out x^3-2x^2+x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)

One of the factors of the above polynomial is x^3-2x^2+x-1.

This is the same as the option (c)

Hence, x^3-2x^2+x-1 is one of the prime factors of the polynomial

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

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