Fun Finding Factors!
Algebra
Level
2
If x^{3}+px+q and 3x^{2}+p have a common factor then 4p^{3}+27q^{2} is equal to-
The answer is 0.
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1 solution
If both polynomials have a common factor ( x + A ) , then each can be written in the following factored forms:
x 3 + p x + q = ( x + A ) ( x 2 + B x + C ) = x 3 + ( A + B ) x 2 + ( A B + C ) x + A C (i)
3 x 2 + p = ( x + A ) ( 3 x 2 + D ) = 3 x 2 + ( 3 A + D ) x + A D (ii)
From (i) above we find that: B = − A , A B + C = − A 2 + C = p ; A C = q . From (ii) above we get: D = − 3 A , A D = − 3 A 2 = p . Substituting this very last value into (i)'s findings, we can obtain:
− A 2 + C = p = − 3 A 2 ⇒ C = − 2 A 2 and A C = q = − 2 A 3 .
Finally, we calculate 4 p 3 + 2 7 q 2 = 4 ( − 3 A 2 ) 3 + 2 7 ( − 2 A 3 ) 2 = − 1 0 8 A 6 + 1 0 8 A 6 = 0 .