Finding zeroes by factorization
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1 Answer

x^5+2x^4-3x^3-8x^2-4x = 0

The first factor we can take out is x.

x(x^4 + 2x^3 - 3x^2 - 8x - 4) = 0

By inspection x = -1 is a root, hence (x+1) is a factor. Taking out this factor,

x(x+1)(x^3 + x^2 - 4x - 4) = 0

By inspection x = -1 is again a root, hence (x+1) is again a factor, Taking out this factor, 

x(x+1)(x+1)(x^2 - 4) = 0

The end term is a difference of two squares and factorises as,

x(x+1)^2(x-2)(x+2) = 0

The roots of the equation are: x = 0, x = -1 (twice), x = 2, x = -2

by Level 11 User (81.5k points)

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