sunmoonlight
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How to do this question?
that would be so slow, just do tywebb's method first to find complex roots then use that to find two quadraticsSince 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)
wait can someone do the full working for this question please
How to do this question?
whats after thatJust 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.
Well z can’t be 0 so you can get rid of the z2 and 5y2-11y+6=0, etcwhats after that
Not *that* slow, but I do agree that tywebb's approach is preferable.that would be so slow, just do tywebb's method first to find complex roots then use that to find two quadratics