Polynomials of Real Coefficients and Multiplicity

The equation $4x^3-27x+k=0$ has a double root. Find the possible values of $k$.

When we differentiate a polynomial function, we get its derivative, which provides information about the slope and behaviour of the original function. In particular, for a polynomial function, the multiplicity of a root can be determined by looking at the behaviour of the derivative at that root.

1. If $P(\alpha) \ne 0$ and $P^{\prime}(\alpha) \ne 0$: In this case, the root at $x = \alpha$ is a single root or has a multiplicity of $1$. The graph of the function crosses the $x$-axis at $x = \alpha$, and the derivative is nonzero at that point. The function behaves like a straight line passing through the $x$-axis at that point.
2. If $P(\alpha)= 0$ and $P^{\prime}(\alpha) \ne 0$: Here, the root at $x = \alpha$ has a multiplicity greater than $1$. The graph of the function touches the $x$-axis at $x = \alpha$ but doesn’t cross it. The derivative is still nonzero at that point, indicating that the function has a flat tangent line at that root.
3. If $P(\alpha) = P^{\prime}(\alpha) = 0$: In this case, the root at $x = \alpha$ has a multiplicity greater than $1$, and the graph of the function also touches the $x$-axis at $x = \alpha$. However, unlike the previous case, the derivative is also zero at that point, indicating that the function has a “bouncing” behaviour near the root.

If $P(x)$ is a polynomial with real coefficients, has a root $\alpha$ of multiplicity $r$, then $P^{\prime} (x)$ has the root $\alpha$ with multiplicity $r-1$.

\displaystyle \begin{align} P(x) &= 4x^3-27x+k \\ P^{\prime} (x) &= 12x^2-27 \end{align}

If $P(x)$ has a double root $\alpha$, then $\alpha$ is a single root of $P^{\prime}(x)$.

\displaystyle \begin{align} P^{\prime}(\alpha) &= 12 \alpha^2-27 \\ 12 \alpha^2-27 &= 0 \\ \alpha^2 &= \frac{9}{4} \\ \alpha &= \pm \frac{3}{2} \end{align}

Thus $\displaystyle \alpha = \frac{3}{2}$ or $\displaystyle \alpha = -\frac{3}{2}$ is a double root of $P(x) = 0$.

\displaystyle \begin{align} 4 \left( \frac{3}{2} \right)^3-27 \left( \frac{3}{2} \right) + k &= 0 \\ -27 + k &= 0 \\ \therefore k &= 27 \\ 4 \left( -\frac{3}{2} \right)^3-27 \left( -\frac{3}{2} \right) + k &= 0 \\ 27 + k &= 0 \\ \therefore k &= -27 \end{align}

Thus, the possible values of $k=\pm 27$.

Mastering Integration by Parts: The Ultimate Guide

Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s…

High School Math for Life: Making Sense of Earnings

Salary Salary refers to the fixed amount of money that an employer pays an employee at regular intervals, typically on a monthly or biweekly basis,…

Induction Made Simple: The Ultimate Guide

“Induction Made Simple: The Ultimate Guide” is your gateway to mastering the art of mathematical induction, demystifying a powerful tool in mathematics. This ultimate guide…

Math Made Easy: Volumes by Cross Sections Simplified

Welcome to the world of mathematics, where complex calculations often seem daunting. You’re not alone if you’ve ever scratched your head over finding the volume…

The Best Practices for Using Two-Way Tables in Probability

Welcome to a comprehensive guide on mastering probability through the lens of two-way tables. If you’ve ever found probability challenging, fear not. We’ll break it…