Expressions & Transformations of Roots
Build any symmetric expression in the roots from the coefficients alone, then construct a brand-new equation whose roots are a shifted, scaled or inverted version of the original.
Expressions relating to the roots of a polynomial
Any expression in $\alpha,\beta,\gamma,\ldots$ that is unchanged by swapping the roots around — a symmetric function of the roots — can always be rewritten purely in terms of the elementary symmetric functions $\sum\alpha$, $\sum\alpha\beta$, $\alpha\beta\gamma$, etc. from the previous lesson. The two most useful identities for doing this are:
the second obtained simply by putting the three fractions over the common denominator $\alpha\beta\gamma$.
$\sum\alpha=4$, $\sum\alpha\beta=5$, $\alpha\beta\gamma=2$.
Linear transformations of roots
Given a polynomial equation in $x$ with known roots, an equation with roots that are a fixed linear transformation of the originals — such as $2\alpha,2\beta,2\gamma$, or $\alpha+2,\beta+2,\gamma+2$ — can be found by substitution. If the new root is $y=mx+c$ (so $x=\dfrac{y-c}{m}$), substituting $x=\dfrac{y-c}{m}$ into the original equation and clearing fractions produces the new equation, now expressed in $y$.
To find an equation with roots $k\alpha,k\beta,k\gamma,\ldots$ (a scaling), replace every $x$ in the original equation with $\dfrac{x}{k}$. To find an equation with roots $\alpha+c,\beta+c,\gamma+c,\ldots$ (a shift), replace every $x$ with $x-c$. Multiply through to clear any fractions, and the result is an equation in $x$ whose roots are the transformed originals.
Replace $x$ with $\dfrac{x}{2}$ in the original equation:
Multiplying through by $8$: $x^3-8x^2+20x-16=0$.
Replace $x$ with $x-2$ in the original equation and expand:
$(x-2)^3=x^3-6x^2+12x-8$; $-4(x-2)^2=-4x^2+16x-16$; $5(x-2)=5x-10$. Collecting all terms:
Always double-check which direction the substitution goes: to shift the roots up by $c$ (i.e. new root $=$ old root $+c$), you substitute $x\to x-c$ into the original equation — not $x+c$. Testing the substitution on a single known root (if one is easy to spot) is a fast way to confirm the direction is correct before committing to the full expansion.
Exam-style questions
Question 1 6 marks
The roots of $2x^2-3x-1=0$ are $\alpha,\beta$. Find, with integer coefficients, the quadratic equation with roots $\alpha^2,\beta^2$.
$\alpha+\beta=\dfrac32$, $\alpha\beta=-\dfrac12$ (B1)
New sum: $\alpha^2+\beta^2=\left(\dfrac32\right)^2-2\left(-\dfrac12\right)=\dfrac94+1=\dfrac{13}{4}$ (M1, A1)
New product: $\alpha^2\beta^2=(\alpha\beta)^2=\dfrac14$ (M1)
Equation: $x^2-\dfrac{13}{4}x+\dfrac14=0 \Rightarrow 4x^2-13x+1=0$ (A1)
Question 2 7 marks
The roots of $x^3+x^2-4x-4=0$ are $\alpha,\beta,\gamma$. Given the identity $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma=(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$, find the exact value of $\alpha^3+\beta^3+\gamma^3$.
$\sum\alpha=-1$, $\sum\alpha\beta=-4$, $\alpha\beta\gamma=4$ (B1)
$\alpha^2+\beta^2+\gamma^2=(-1)^2-2(-4)=1+8=9$ (M1, A1)
$\alpha^2+\beta^2+\gamma^2-\sum\alpha\beta = 9-(-4)=13$ (M1)
$\alpha^3+\beta^3+\gamma^3 = 3\alpha\beta\gamma+(\sum\alpha)(13) = 3(4)+(-1)(13)=12-13=-1$ (M1, A1)