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Polynomial (1 Viewer)

tywebb

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Just looking at the coefficients - they are symmetrical and so instantly you can see that it is reducible quadratics. Just divide by z2 and regroup

Like z2(5(z2+1/z2)-11(z+1/z)+16) and let y=z+1/z

Then z2(5y2-11y+6), etc.
 
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Luukas.2

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Since all coefficients are real and rational, any complex roots will come in conjugate pairs.

It must be factorisable as a product of two quadratics, and with a leading term of 5x4, you could start from

(5x2 + Ax + B) (x2 + Cx + D)

expand, and equate coefficients to get equations in A, B, C, and D.

The answer is (5x2 - 6x + 5)(x2 - x + 1)
 

member 6003

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Since all coefficients are real and rational, any complex roots will come in conjugate pairs.

It must be factorisable as a product of two quadratics, and with a leading term of 5x4, you could start from

(5x2 + Ax + B) (x2 + Cx + D)

expand, and equate coefficients to get equations in A, B, C, and D.

The answer is (5x2 - 6x + 5)(x2 - x + 1)
that would be so slow, just do tywebb's method first to find complex roots then use that to find two quadratics
 

scaryshark09

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Just looking at the coefficients - they are symmetrical and so instantly you can see that it is reducible quadratics. Just divide by z2 and regroup

Like z2(5(z2+1/z2)-11(z+1/z)+16) and let y=z+1/z

Then z2(5y2-11y+6), etc.
whats after that
 

scaryshark09

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yeh so maybe im dumb, but i cant do the algebra after that part so....
 

tywebb

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A common question is how does the 16 become a 6?

It is because (z+1/z)2=z2+2+1/z2

and 5(2)+6=16
 

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this is how I did it. Maybe there's a faster way to prove the roots have magnitude 1 before finding them. I don't think there is but if that's given you could say z+1/z =2costheta and not have to solve the second quadratic.
 

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Luukas.2

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(5x2 + Ax + B) (x2 + Cx + D)

.

The values A, B, C, and D are likely all integers.

From (4), B and D have the same sign, as (likely) do A and C... and equation (1) prevents A and C both being positive.

So, (4) suggests two options: (a) B = 1 and D = 5 or (b) B = 5 and D = 1
  • (a) requires both A + 5C and 5A + C to be -11 and A = C = -11 / 6, making AC = 121 / 36 in (2) ===> unlikely.
  • (b) makes (3) and (1) the same, and requires AC = 16 - 5D - B = 6
With AC = 6 > 0, A and C both negative is confirmed, yielding either -2 x -3 or -1 x -6

Only A = -6 and C = -1 fits with A + 5C = -11. Thus, A = -6, B = 5, C = -1, and D = 1



The four solutions of are

 

Luukas.2

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The working for tywebb's factorisation:

 
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