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3 years ago

6. Determine the Distance between (8, -2) and (9, 3) ?

Mathematics

2 answers:

cluponka [151]3 years ago

4 0

D none of the above is the answer

Greeley [361]3 years ago

4 0

Answer : D
Step by step: D

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$x^3-2x^2+x-1$ 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