Question 1175000
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No, Ikleyn, this student wrote me this in his "thank-you" letter.
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It said to factor it. Sorry!
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This student apparently is studying very complicated algebraic factoring
methods.

The quartic is factorable algebraically using radicals (square roots):

{{{f(x)=x^4-8x^3+22x^2-24x+7}}}

We try to reduce it to a quadratic in y², by substituting a linear form for x,
say x=ay+b

{{{f(x)=f(ay+b)=(ay+b)^4-8(ay+b)^3+22(ay+b)^2-24(ay+b)+7}}}

We see if we can choose "a" and "b" so that the terms in y³ and y become 0.

It isn't difficult to see that the only two terms in y³ are

4a³by³ - 8a³y³ = 4a(b-2)y³, which would be 0 for all y if b=2

There are four terms in y, and they are

4ab^3y - 24 ab^2y + 44aby - 24ay

Substituting b=2

32ay - 96ay + 88ay - 24ay which, luckily, equals 0

Therefore, we can simply take a = 1, and b = 2

Then x = ay+b = y+2

{{{f((ay+b))=(ay+b)^4-8(ay+b)^3+22(ay+b)^2-24(ay+b)+7}}}

becomes

{{{f(x)=f(y+2)=(y+2)^4-8(y+2)^3+22(y+2)^2-24(y+2)+7}}}

which simplifies to

{{{y^4 - 2y^2 - 1}}}

and by the quadratic formula,

{{{y^2 =  1 +- sqrt(2) }}}

So there are two real zeros for y

{{{y=sqrt(1 + sqrt(2)) }}}, {{{y = -sqrt( 1 + sqrt(2)) }}}

And there are two imaginary zeros for y

{{{y=sqrt(1 - sqrt(2)) }}}, {{{y = -sqrt( 1 - sqrt(2)) }}}

Since √2 is greater than 1,

{{{y=i*sqrt(sqrt(2)-1) }}}, {{{y = -i*sqrt(sqrt(2)-1) }}}

So the factorization in terms of y is:

{{{ (y-sqrt(1 + sqrt(2)))(y+sqrt(1 + sqrt(2)))(y-i*sqrt(sqrt(2)-1)) (y+i*sqrt(sqrt(2)-1)   )}}}

Since x = y+2, we substitute x-2 for y

{{{ (x-2-sqrt(1 + sqrt(2)))(x-2+sqrt(1 + sqrt(2)))(x-2-i*sqrt(sqrt(2)-1)) (x-2+i*sqrt(sqrt(2)-1)   )}}}

Edwin</pre>