Not as complicated as it look
x 9 − x 7 − x 6 − x 5 + x 4 + x 3 + x 2 − 1
Given that the above expression completely factors to ( x + a ) b ( x + c ) d ( x d + c ) ( x d + x + c ) for integers a , b , c , d , find a + b + c + d .
The answer is 5.
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2 solutions
Excellent!
( x 9 − 1 ) + ( − x 7 + x 4 ) + ( − x 6 + x 3 ) + ( − x 5 + x 2 ) = [ ( x 3 ) 3 − 1 ] − x 4 ( x 3 − 1 ) − x 3 ( x 3 − 1 ) − x 2 ( x 3 − 1 ) = ( x 3 − 1 ) ( x 6 + x 3 + 1 ) − ( x 3 − 1 ) ( x 4 + x 3 + x 2 ) = ( x 3 − 1 ) [ x 6 + x 3 + 1 − x 4 − x 3 − x 2 ] = ( x 3 − 1 ) ( x 6 − x 4 − x 2 + 1 ) = ( x − 1 ) ( x 2 + x + 1 ) [ x 4 ( x 2 − 1 ) − 1 ( x 2 − 1 ) ] = ( x − 1 ) ( x 2 + x + 1 ) ( x 4 − 1 ) ( x 2 − 1 ) = ( x − 1 ) ( x 2 + x + 1 ) ( x 2 + 1 ) ( x 2 − 1 ) ( x + 1 ) ( x − 1 ) = ( x − 1 ) 3 ( x + 1 ) 2 ( x 2 + 1 ) ( x 2 + x + 1 )
Isto posto, temos que ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ a = − 1 b = 3 c = 1 d = 2 .
Com efeito,
a + b + c + d = − 1 + 3 + 1 + 2 a + b + c + d = 5
x 9 − x 7 − x 6 − x 5 + x 4 + x 3 + x 2 − 1 = ( x 9 − x 7 ) − ( x 6 − x 4 ) − ( x 5 − x 3 ) + ( x 2 − 1 ) = x 7 ( x 2 − 1 ) − x 4 ( x 2 − 1 ) − x 3 ( x 2 − 1 ) + ( x 2 − 1 ) = ( x 7 − x 4 − x 3 + 1 ) ( x 2 − 1 ) = [ ( x 4 ) ( x 3 − 1 ) − ( x 3 − 1 ) ] [ x 2 − 1 ] = ( x 4 − 1 ) ( x 3 − 1 ) ( x 2 − 1 ) = ( x − 1 ) 3 ( x + 1 ) 2 ( x 2 + 1 ) ( x 2 + x + 1 ) ⇒ a = − 1 , b = 3 , c = 1 , d = 2 . . . . . . . . . . . . . ∴ a + b + c + d = 5 .