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1 Answer

f(x+1) = 2(x+1)^3 - 5(x+1) + 6

= 2(x^3 + 3x^2 + 3x + 1) - 5x - 5 + 6

= 2x^3 + 6x^2 + 6x + 2 - 5x + 1

= 2x^3 + 6x^2 + x + 3

f(x) = 2x^3 - 5x + 6

f(x+1) - f(x) = 2x^3 + 6x^2 + x + 3 - (2x^3 - 5x + 6)

= 2x^3 + 6x^2 + x + 3 - 2x^3 + 5x - 6

= 6x^2 + 6x - 3

6x^2 + 6x - 3 = 0

2x^2 + 2x - 1 = 0

(2x   2)(x   1) or (2x   1)(x   2). . .no, doesn't look like it factors nicely.

Quadratic

x = (-2 +- sqrt(2^2 - 4(2)(-1))) / 2(2)

x = (-2 +- sqrt(4 + 8)) / 4

x = (-2 +- sqrt(12)) / 4

x = (-2 +- 2sqrt(3)) / 4

x = (-1 +- sqrt(3)) / 2

Answer:  x = (-1 + sqrt(3)) / 2, or x = (-1 - sqrt(3)) / 2
by Level 13 User (103k points)

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