2 polynomials questions

**differentiate** · April 1st 2010, 07:40 PM

1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers
(i) show that P(x) has exactly one real root which is greater than zero
(ii) Hence show that P(x) has four complex roots with at least two of them with negative real parts

2. If the polynomial $x^3 + 3mx^2 + 3nx + r = 0$ has a double root. Show that:
(i) The double root is x = $\frac {mn-r}{2(n-m^2)}$
(ii) $(mn-r)^2 = 4(m^2 -n)(n^2 - mr)$

Thanks

**Drexel28** · April 1st 2010, 08:57 PM

Originally Posted by differentiate

1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers
(i) show that P(x) has exactly one real root which is greater than zero

$P(0)=0$ and $\lim_{x\to\infty}P(x)=\infty$ . That guarantees the root. To see that it only has one note that $P'(x)=5x^4+3ax^2+b>0,\text{ }x>0$ so assuming there was another would violate Rolle's theorem.

**Drexel28** · April 1st 2010, 09:04 PM

Originally Posted by differentiate

1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers

(ii) Hence show that P(x) has four complex roots with at least two of them with negative real parts

Note that the same works for the negative numbers since $P'(x)$ is symmetric about the y-axis. Namely, $P'(x)=5x^4+3ax^2+b>0,\text{ }x<0$ . Appealing to Rolle's theorem again guarantees that $P(x)\ne 0,\text{ }x<0$ . It follows that $P^{-1}(\{0\})=\{0\}$ . But, by the FTA we must have that $P:\mathbb{C}\to\mathbb{C}$ has five zeros counting multiplicity. But, by a very easy to prove theorem, we have that since the coefficients of $P$ are real that the zeros come in conjugate pairs, so that $P(x)=x(x-z_1)(x-\overline{z_1})(x-z_2)(x-\overline{z_2})$ . The negative part buisness follows since $\text{Im}(z_k),\text{Im}(\overline{z_k})\ne 0,\text{ }k=1,2$ since they aren't real. Thus, either $\text{Im}(z_1)<0$ or $\text{Im}(\overline{z_1})<0$ . A similar analysis shows that either $z_1,z_2$ has negative imaginary parts.

**HallsofIvy** · April 2nd 2010, 01:44 AM

Originally Posted by Drexel28

$P(0)=0$

You mean P(0)= -c

and $\lim_{x\to\infty}P(x)=\infty$ . That guarantees the root. To see that it only has one note that $P'(x)=5x^4+3ax^2+b>0,\text{ }x>0$ so assuming there was another would violate Rolle's theorem.

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