Question

x^4 + x^3 + x^2 + x + 1 ​

Answer 1

⎝⎝⫷⫸⎠⎠ANSWER⎝⎝⫷⫸⎠⎠

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Given:

→$x^4+x^3+x^2+x+1$.

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Solution:

→($x^2$+($1/2+ \sqrt5/2$)$x+1$)($x^2+$($1/2-\sqrt5/2$)$x+1$).

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Step by step-Explanation:

→This quartic has four zeros, which are the non-Real Complex  

5  t h roots of  1 , as we can see from:

→$(x-1)(x^4+x^3+x^2+1)=x^5-1$,

→So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:

→$x^4+x^3+x^2+x+1$,

→($x-$($cos$($2\pi/5$)$+ isin$($2\pi/5$))),

→($x-$($cos$($4\pi/5$)$+isin$($4\pi/5$))),

→($x-$($cos$($6\pi/5$)$+isin$($6\pi/5$))),

→($x-$($cos$($8\pi/5$)$+isin$($8\pi/5$))),

→A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if  x = r is a zero of $x^4+x^3+x^2+x+1$,then $x=1/r$ is also zero.

→Hence there is a factorisation in the form:

→$x^4+x^3+x^2+x+1$,

       →($x-r1$)($x-1/r1$)($x-r2$)($x-1/r2$),

       →($x^2-$($r1+1/r1$)$x+1$)($x^2$$-$($r2+1/r2$)$x+1$),

→So let's look for a factorisation:

→$x^4+x^3+x^2+x+1$,

       →($x^2+ax+1$)($x^2+bx+1$),

       →$x^4+$($a+b$)$x^3+$($2+ab$)$x^2+$($a+b$)$x+1$,

→Equating coefficients we find:

→$a+b=1$,

→$2+ab=1,so\:ab=-1\:and\:b=-1/a$,

→substituting $b=-1/a\:in\:a+b=1\:we\:get:$

→$a-1/a=1$,

→Hence:

→$a^2-a-1=0$,

→Using the quadratic formula, we can deduce:

→$a=1/2$±$\sqrt5/2$,

→Since our derivation was symmetric in  a and  b , one of these roots can be used for  a and the other for  b , to find:

→$x^4+x^3+x^2+x+1$

→($x^2+$($1/2+\sqrt5/2$)$x+1$)($x^2+$($1/2-\sqrt5/2$)$x+1$),

If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.

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