Subjective Type

Show that $$x^{5} + 5x^{4} - 20x^{2} - 19x - 2 = 0$$ has a root between $$2$$ and $$3$$, and a root between $$-4$$ and $$-5$$.
Solution
Given equation, $$x^5+5x^4-20x^2-19x-2=0$$
Consider $$f(x) = x^5+5x^4-20x^2-19x-2$$
Then, $$f(2) = -8$$ and $$f(3) = 409$$
Since the signs of $$f(2)$$ and $$f(3)$$ are opposite, $$f(x)$$ must cross x-axis atleast once in the interval $$(2,3)$$.
$$\therefore\,\, f(x) = 0$$ must have a root between $$2$$ and $$3$$
Similarly, $$f(-4) = 10$$ and $$f(-5) = -407$$.
Since the signs of $$f(-4)$$ and $$f(-5)$$ are different, $$f(x) = 0$$ must have one root between $$-4$$ and $$-5$$.
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(Note: This question was asked in Maths subject in JEE Mains $$2015$$ online exam held on $$11$$ April $$2015$$)
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