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Question:

Suppose the equation $x^3-hx^2+kx-9=0$ has only one real root which has a value of $1$. Find the range of values of $k$.

I really have no idea when it comes to a cubic equation. Any advice to solving?

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    $\begingroup$ use strictly increasing or strictly decreasing condition for polynomials. $\endgroup$
    – Nosrati
    Jan 9, 2017 at 12:11

2 Answers 2

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You know that one factor of this cubic equation is $(x-1)$ from the root. Therefore, you can equate the above polynomial to $(x-1)(x^2 + ax +b)$ and use that to get the relationship between $h$ and $k$ on the one side and $a$ and $b$ on the other.

You also know that $(x^2 +ax +b)$ has no real roots. Using the discriminant should give you enough to finish the question.

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I would start by factoring the polynomial based on the known route. By doing polynomial division $(x^3-hx^2+kx-9) : (x-1)$.

By doing so, you will get a residue, that must be zero (since you factoring by a root), so you will get one restriction on $k$ and $h$, to ensure $1$ being a root.

In addition, you will have a quadratic polynomial, that has to have complex roots. By inspecting the root of the $pq$-formula to be negative, there is an additional restriction on $k$ and $h$.

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