A polynomial deserve to have actually fewer x-intercepts than its variety of roots once a pair of facility conjugate roots exist.Example:Consider the 4-th level polynomialf(x) = x⁴ - x³ - x² - x - 2According to the Remainder theoremf(-1) = 1 + 1 - 1 + 1 - 2 = 0Therefore (x + 1) is a aspect.f(2) = 16 - 8 - 4 - 2 - 2 = 0As such (x - 2) is a factor.(x+1)*(x-2) = x² - 2x + x - 2 = x² - x - 2To discover the remaining variable, perform lengthy division. x² + 1 -------------------------------x²-x-2 | x⁴ - x³ - x² - x - 2 x⁴ - x³ - 2x² ----------------------------- x² - x - 2 x² - x - 2Thereforef(x) = (x+1)(x-2)(x²+1)Notice that (x² + 1) has actually no genuine factors.However before,x² + 1 = (x + i)(x - i),so it has actually a pair of conjugate zeros +i and -i.A graph of the function confirms that tright here are just two real zeros (presented in red color). You are watching: Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.

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