A very Difficult Problem

Algebra Level 2

f ( x ) = x ( x + 3 ) ( x − 9 ) ( x + 10 ) ( x 2 − 1 ) ( x + 21 ) ( x − 4 ) \large f(x)=x(x+3)(x-9)(x+10)(x^2-1)(x+21)(x-4)

Evaluate the sum of all zeroes of f ( x ) f(x) .


The answer is -21.

Bloons Qoth by Bloons Qoth , 21, USA

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2 solutions

Viki Zeta
Jul 16, 2016

x ( x + 3 ) ( x − 9 ) ( x + 10 ) ( x 2 − 1 ) ( x + 21 ) ( x − 4 ) = 0 ⟹ x = 0 , x + 3 = 0 , x − 9 = 0 , x + 10 = 0 , x 2 − 1 = 0 , x + 21 = 0 , x − 4 = 0 ⟹ x = 0 , x = − 3 , x = 9 , x = − 10 , x = 1 , x = − 1 , x = − 21 , x = 4 ⟹ Sum of all values = 0 + ( − 3 ) + 9 + ( − 10 ) + 1 + ( − 1 ) + ( − 21 ) + 4 = -21 x(x+3)(x-\color{#20A900}{9})(x+\color{#20A900}{10})(x^2-1)(x+\color{#20A900}{21})(x-4) = 0\\ \implies x = 0, x + 3 = 0, x-9 = 0, x+10 = 0, x^2 - 1 = 0, x+21 = 0, x-4 = 0\\ \implies x = 0, x = -3, x=9, x=-10, x=1, x = -1, x=-21, x=4\\ \implies \text{Sum of all values = } 0 + (-3) + 9 + (-10) + 1 + (-1) + (-21) + 4 \\ = \fbox{-21}

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Theodore Lietz
Jul 16, 2016

0-3+9-10+1-1-21+4=-21

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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