Find all the roots of (x2+3x+2)(x2−7x+12)(x2−2x−1)+24=0.
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Solution
We re-factor as (x+1)(x−3)(x+2)(x−4)(x2−2x−1)+24, or (x2−2x−3)(x2−2x−8)(x2−2x−1)+24, and this becomes (y−4)(y−9)(y−2)+24 where y=(x−1)2. Now, (y−4)(y−9)(y−2)+24=(y−8)(y−6)(y−1), so y is 1, 6, or 8. Thus the roots of the original polynomial are 0,2,1±6,1±22.
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