116. Differentiation of polynomials

1. Show that if $\varphi (x)$ is a polynomial then ${\varphi }^{\prime }(x)$ is the coefficient of $h$ in the expansion of $\varphi (x+h)$ in powers of $h$.

2. If $\varphi (x)$ is divisible by $(x–\alpha {)}^2$, then ${\varphi }^{\prime }(x)$ is divisible by $x–\alpha$: and generally, if $\varphi (x)$ is divisible by $(x–\alpha {)}^m$, then ${\varphi }^{\prime }(x)$ is divisible by $(x–\alpha {)}^{m-1}$.

3. Conversely, if $\varphi (x)$ and ${\varphi }^{\prime }(x)$ are both divisible by $x–\alpha$, then $\varphi (x)$ is divisible by $(x–\alpha {)}^2$; and if $\varphi (x)$ is divisible by $x–\alpha$ and ${\varphi }^{\prime }(x)$ by $(x–\alpha {)}^{m-1}$, then $\varphi (x)$ is divisible by $(x–\alpha {)}^m$.

4. Show how to determine as completely as possible the multiple roots of $P(x)=0$, where $P(x)$ is a polynomial, with their degrees of multiplicity, by means of the elementary algebraical operations.

[If $H_1$ is the highest common factor of $P$ and $P^{\prime }$, $H_2$ the highest common factor of $H_1$ and $P”$, $H_3$ that of $H_2$ and $P{”}^{\prime }$, and so on, then the roots of $H_1{H}_3/H_2^2=0$ are the double roots of $P=0$, the roots of $H_2{H}_4/H_3^2=0$ the treble roots, and so on. But it may not be possible to complete the solution of $H_1{H}_3/H_2^2=0$, $H_2{H}_4/H_3^2=0$, …. Thus if $P(x)=(x–1{)}^3(x^5–x–7{)}^2$ then $H_1{H}_3/H_2^2=x^5–x–7$ and $H_2{H}_4/H_3^2=x–1$; and we cannot solve the first equation.]

5. Find all the roots, with their degrees of multiplicity, of $$x^4+3x^3–3x^2–11x–6=0,x^6+2x^5–8x^4–14x^3+11x^2+28x+12=0.$$

6. If $a{x}^2+2bx+c$ has a double root, is of the form $a(x–\alpha {)}^2$, then $2(ax+b)$ must be divisible by $x–\alpha$, so that $\alpha =-b/a$. This value of $x$ must satisfy $a{x}^2+2bx+c=0$. Verify that the condition thus arrived at is $ac–b^2=0$.

7. The equation $1/(x–a)+1/(x–b)+1/(x–c)=0$ can have a pair of equal roots only if $a=b=c$.

8. Show that $$a{x}^3+3b{x}^2+3cx+d=0$$ has a double root if $G^2+4H^3=0$, where $H=ac–b^2$, $G=a^{2}d–3abc+2b^3$.

[Put $ax+b=y$, when the equation reduces to $y^3+3Hy+G=0$. This must have a root in common with $y^2+H=0$.]

9. The reader may verify that if $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $$a{x}^4+4b{x}^3+6c{x}^2+4dx+e=0,$$ then the equation whose roots are $$\frac{1}{12}a\{(\alpha –\beta )(\gamma –\delta )–(\gamma –\alpha )(\beta –\delta )\},$$ and two similar expressions formed by permuting $\alpha$, $\beta$, $\gamma$ cyclically, is $$4{\theta }^3–g_2\theta –g_3=0,$$ where $$g_2=ae–4bd+3c^2,g_3=ace+2bcd–a{d}^2–e{b}^2–c^3.$$ It is clear that if two of $\alpha$, $\beta$, $\gamma$, $\delta$ are equal then two of the roots of this cubic will be equal. Using the result of Ex. 8 we deduce that $g_2^3–27g_3^2=0$.

10. Rolle’s Theorem for polynomials. If $\varphi (x)$ is any polynomial, then between any pair of roots of $\varphi (x)=0$ lies a root of ${\varphi }^{\prime }(x)=0$.

A general proof of this theorem, applying not only to polynomials but to other classes of functions, will be given later. The following is an algebraical proof valid for polynomials only. We suppose that $\alpha$, $\beta$ are two successive roots, repeated respectively $m$ and $n$ times, so that $$\varphi (x)=(x–\alpha {)}^m(x–\beta {)}^n\theta (x),$$ where $\theta (x)$ is a polynomial which has the same sign, say the positive sign, for $\alpha \le x\le \beta$. Then $$\begin{array}{rl}{\varphi }^{\prime }(x)& =(x–\alpha {)}^m(x–\beta {)}^n{\theta }^{\prime }(x)+\{m(x–\alpha {)}^{m-1}(x–\beta {)}^n+n(x–\alpha {)}^m(x–\beta {)}^{n-1}\}\theta (x)\\ & =(x–\alpha {)}^{m-1}(x–\beta {)}^{n-1}[(x–\alpha )(x–\beta ){\theta }^{\prime }(x)+\{m(x–\beta )+n(x–\alpha )\}\theta (x)]\\ & =(x–\alpha {)}^{m-1}(x–\beta {)}^{n-1}F(x),\end{array}$$ say. Now $F(\alpha )=m(\alpha –\beta )\theta (\alpha )$ and $F(\beta )=n(\beta –\alpha )\theta (\beta )$, which have opposite signs. Hence $F(x)$, and so ${\varphi }^{\prime }(x)$, vanishes for some value of $x$ between $\alpha$ and $\beta$.