Roots of polynomialsEdexcel A-Level Further Maths: Revision notes

If a polynomial equation has roots $\alpha,\beta,\ldots$, then its coefficients are fixed by symmetric sums of the roots, with alternating signs.

• Cubic $ax^3+bx^2+cx+d=0$: $\sum\alpha=-\frac ba$, $\sum\alpha\beta=\alpha\beta+\beta\gamma+\gamma\alpha=\frac ca$, $\alpha\beta\gamma=-\frac da$.
• Quartic $ax^4+bx^3+cx^2+dx+e=0$: $\sum\alpha=-\frac ba$, $\sum\alpha\beta=\frac ca$, $\sum\alpha\beta\gamma=-\frac da$, $\alpha\beta\gamma\delta=\frac ea$. The pattern is: sum of roots $=-\frac{\text{next coefficient}}{a}$, then the sign alternates. For $2x^3-6x^2+3x+5=0$: $\sum\alpha=3$, $\sum\alpha\beta=\frac32$, $\alpha\beta\gamma=-\frac52$. The roots may be real or complex; the relationships hold either way.

Rewrite the expression using only the symmetric sums, then substitute.

• $\alpha^2+\beta^2+\gamma^2=(\sum\alpha)^2-2\sum\alpha\beta$.
• $\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\sum\alpha\beta}{\alpha\beta\gamma}$.
• $(k+\alpha)(k+\beta)(k+\gamma)=k^3+k^2\sum\alpha+k\sum\alpha\beta+\alpha\beta\gamma$ (or $-f(-k)$ for a monic cubic $f$).
• $\alpha^3+\beta^3+\gamma^3$: either use the identity $\alpha^3+\beta^3+\gamma^3=(\sum\alpha)^3-3\sum\alpha\sum\alpha\beta+3\alpha\beta\gamma$, or use the fact that each root satisfies the equation, so $\alpha^3=-\frac ba\alpha^2-\frac ca\alpha-\frac da$, and sum over the three roots. Worked example: $x^3+3x^2-5x+2=0$ has $\sum\alpha=-3$, $\sum\alpha\beta=-5$, $\alpha\beta\gamma=-2$. Then $\sum\alpha^2=9+10=19$ and $\sum\alpha^3=-3(19)+5(-3)-3(2)=-78$.

The same method works for four roots. For $x^4+2x^3-7x^2+4x-3=0$: $\sum\alpha=-2$, $\sum\alpha\beta=-7$, $\sum\alpha\beta\gamma=-4$, $\alpha\beta\gamma\delta=-3$.

• $\sum\alpha^2=(\sum\alpha)^2-2\sum\alpha\beta=4+14=18$.
• $\sum\frac1\alpha=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}=\frac{-4}{-3}=\frac43$. A useful check: for real roots, $\sum\alpha^2$ cannot be negative. If your formula gives a negative value, the equation must have at least one complex root. For $x^4+3x^2+6x+10=0$, $\sum\alpha=0$ and $\sum\alpha\beta=3$, so $\sum\alpha^2=-6<0$.

To find an equation whose roots are $y=px+q$ where $x$ are the roots of a given equation, there are two methods. Substitution. Rearrange to $x=\frac{y-q}{p}$, substitute into the original equation, then clear fractions. For $x^3-4x^2+2x+5=0$ and $y=2x+1$: $x=\frac{y-1}{2}$ gives $(y-1)^3-8(y-1)^2+8(y-1)+40=0$, so $y^3-11y^2+27y+23=0$. Sums of roots. Find the new sum, sum of pairs and product of roots, then write $y^3-(\sum y)y^2+(\sum yy')y-yy'y''=0$. Here $\sum y=2(4)+3=11$, $\sum yy'=27$ and $yy'y''=-23$, which gives the same equation. For roots $\frac1\alpha$, substitute $x=\frac1y$ and multiply through by the highest power of $y$: the coefficients appear in reverse order. For roots $k\alpha$, replace $x$ by $\frac yk$.

• Identify the polynomial's degree and write the symmetric sums with signs first.
• Convert the expression to sums; never solve for the roots themselves.
• Give exact fractions rather than decimals.
• When transforming, check one coefficient with the sums method.
• A negative value of $\sum\alpha^2$ means a complex root exists.