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(B) Let $f(x) = 2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$.First,we evaluate the function at integer points to locate the root.$f(-2) = 2(-32) + 5(16) + 10

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$2$

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(B) Let $f(x) = 2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$.First,we evaluate the function at integer points to locate the root.$f(-2) = 2(-32) + 5(16) + 10(-8) + 10(4) + 10(-2) + 10 = -64 + 80 - 80 + 40 - 20 + 10 = -34$.$f(-1) = 2(-1) + 5(1) + 10(-1) + 10(1) + 10(-1) + 10 = -2 + 5 - 10 + 10 - 10 + 10 = 3$.Since $f(-2)  0$,by the Intermediate Value Theorem,there exists at least one real root in the interval $(-2, -1)$.Next,we check the derivative to determine the number of real roots.$f'(x) = 10x^{4} + 20x^{3} + 30x^{2} + 20x + 10$.$f'(x) = 10(x^{4} + 2x^{3} + 3x^{2} + 2x + 1)$.Notice that $x^{4} + 2x^{3} + 3x^{2} + 2x + 1 = (x^{2} + x + 1)^{2}$.Thus,$f'(x) = 10(x^{2} + x + 1)^{2}$.Since $x^{2} + x + 1$ has a negative discriminant $(1 - 4 = -3)$,it is always positive for all real $x$.Therefore,$f'(x) > 0$ for all real $x$,which means $f(x)$ is strictly increasing and has exactly one real root.Since the root lies in $(-2, -1)$,we have $a = -2$. Thus,$|a| = |-2| = 2$.

Let $a$ be an integer such that all the real roots of the polynomial $2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$ lie in the interval $(a, a+1)$. Then,$|a|$ is equal to ...... .

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