The polynomial equation x^3-3ax^2+(27a^2+9)x+ | vedclass

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(A) Let $f(x) = x^3-3ax^2+(27a^2+9)x+2016$.To determine the number of real roots,we find the derivative $f'(x)$:$f'(x) = 3x^2 - 6ax + (27a^2 + 9)$.We

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exactly one real root for any real $a$

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(A) Let $f(x) = x^3-3ax^2+(27a^2+9)x+2016$.To determine the number of real roots,we find the derivative $f'(x)$:$f'(x) = 3x^2 - 6ax + (27a^2 + 9)$.We can rewrite the derivative by completing the square:$f'(x) = 3(x^2 - 2ax + 9a^2 + 3)$$f'(x) = 3((x-a)^2 + 8a^2 + 3)$.Since $(x-a)^2 \geq 0$,$8a^2 \geq 0$,and $3 > 0$,it follows that $f'(x) > 0$ for all real $x$ and all real $a$.Because the derivative $f'(x)$ is always positive,the function $f(x)$ is strictly increasing for all real $a$.$A$ strictly increasing cubic polynomial crosses the $x$-axis exactly once.Therefore,the equation has exactly one real root for any real $a$.

The polynomial equation $x^3-3ax^2+(27a^2+9)x+2016=0$ has

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