Roots and coefficients of polynomialsAQA A-Level Further Maths: Revision notes

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$, then $ax^2+bx+c=a(x-\alpha)(x-\beta)$, and comparing coefficients gives $\alpha+\beta=-\frac ba,\qquad \alpha\beta=\frac ca.$ For a cubic $ax^3+bx^2+cx+d=0$ with roots $\alpha,\beta,\gamma$, expanding $a(x-\alpha)(x-\beta)(x-\gamma)$ gives $\sum\alpha=-\frac ba,\qquad \sum\alpha\beta=\frac ca,\qquad \alpha\beta\gamma=-\frac da,$ where $\sum\alpha\beta=\alpha\beta+\beta\gamma+\gamma\alpha$ is the sum of products taken two at a time. Example: $x^3-4x^2+5x-7=0$ has $\sum\alpha=4$, $\sum\alpha\beta=5$ and $\alpha\beta\gamma=7$.

For $ax^4+bx^3+cx^2+dx+e=0$ with roots $\alpha,\beta,\gamma,\delta$: $\sum\alpha=-\frac ba,\quad \sum\alpha\beta=\frac ca,\quad \sum\alpha\beta\gamma=-\frac da,\quad \alpha\beta\gamma\delta=\frac ea.$ The pattern is that the signs alternate, starting with a minus: the sum of the roots is $-\frac ba$, the pairs $+\frac ca$, the triples $-\frac da$ and the product of all four $+\frac ea$. For a cubic the product of all three roots is $-\frac da$ because the last sign follows the same alternating pattern. Example: $x^4-6x^3+3x^2+2x-5=0$ has $\sum\alpha=6$, $\sum\alpha\beta=3$, $\sum\alpha\beta\gamma=-2$ and $\alpha\beta\gamma\delta=-5$.

Many expressions can be written using $\sum\alpha$, $\sum\alpha\beta$ and $\alpha\beta\gamma$ without finding the roots.

• $\alpha^2+\beta^2+\gamma^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta$.
• $\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\sum\alpha\beta}{\alpha\beta\gamma}$.
• $\frac1{\alpha\beta}+\frac1{\beta\gamma}+\frac1{\gamma\alpha}=\frac{\sum\alpha}{\alpha\beta\gamma}$.
• $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$ is found by writing each bracket using $\sum\alpha$, for example $\alpha+\beta=\sum\alpha-\gamma$. For the cubic $x^3+3x^2-5x+1=0$: $\sum\alpha^2=(-3)^2-2(-5)=19$.

To form an equation whose roots are $y=px+q$, where $x$ ranges over the roots of the original, substitute $x=\frac{y-q}{p}$ into the original equation and tidy up (multiply to clear fractions). Example: the roots of $2x^3-x^2+4x-5=0$ are $\alpha,\beta,\gamma$. For roots $2\alpha,2\beta,2\gamma$ put $x=\frac y2$: $2\left(\frac y2\right)^3-\left(\frac y2\right)^2+4\left(\frac y2\right)-5=0\ \Rightarrow\ y^3-y^2+8y-20=0.$ Check with the relationships: the new sum is $2\sum\alpha=1$, the new sum of pairs is $4\sum\alpha\beta=8$ and the new product is $8\alpha\beta\gamma=20$. A cubic with sum $S$, pairs $P$ and product $Q$ is $y^3-Sy^2+Py-Q=0$, which matches.

Questions often give a condition on the roots, such as roots in arithmetic progression ($a-d,\ a,\ a+d$) or geometric progression ($\frac ar,\ a,\ ar$). Write the roots using the condition, then use the relationships. Example: $x^3-14x^2+kx-64=0$ has roots in GP. With roots $\frac ar,a,ar$, the product is $a^3=64$, so $a=4$. The sum gives $\frac4r+4+4r=14$, so $2r^2-5r+2=0$ and $r=2$ or $\frac12$. The roots are $2,4,8$ and $k=2\times4+2\times8+4\times8=56$. A good first step is the relationship that gives a single unknown, such as the product $a^3=64$.