Linear Transformations of Roots
Substituting for x in terms of a new variable y builds a brand-new polynomial whose roots are a scaled, shifted, or inverted version of the original — without ever solving either equation.
Building a new equation from transformed roots
If $\alpha,\beta,\gamma$ are the roots of a polynomial in $x$, and a new polynomial is required whose roots are a linear transformation of these (e.g. $2\alpha,2\beta,2\gamma$ or $\alpha+3,\beta+3,\gamma+3$), one reliable method is substitution: if the new root is $y=f(x)$, express $x$ in terms of $y$ and substitute into the original equation. The resulting equation in $y$ has exactly the transformed roots.
An alternative to substitution is to work directly with the sums and products of the new roots, built from the sums and products of the original roots via Vieta's formulae. For roots $\alpha,\beta,\gamma$ transformed to $f(\alpha),f(\beta),f(\gamma)$:
- Write down $\sum\alpha$, $\sum\alpha\beta$, $\alpha\beta\gamma$ (or the quartic equivalents) for the original equation using Vieta's formulae.
- Expand the new sum $\sum f(\alpha)$, new pair-sum $\sum f(\alpha)f(\beta)$, and new product $f(\alpha)f(\beta)f(\gamma)$ in terms of these original symmetric sums.
- Use the results to write down the new equation directly, without ever introducing a second variable.
Both methods always agree — substitution is usually faster for a single clean transformation like $y=2x$ or $y=x-3$, while the symmetric-functions method is often clearer when the transformation is more awkward to invert, or when you want an independent check on a substitution answer.
Scaling the roots
Let $y=2x$, so $x=\dfrac{y}{2}$. Substitute into the original equation:
$\dfrac{y^3}{8}-\dfrac{4y^2}{4}+\dfrac{5y}{2}-2=0 \Rightarrow \dfrac{y^3}{8}-y^2+\dfrac{5y}{2}-2=0$.
Multiplying through by 8: $y^3-8y^2+20y-16=0$.
From the original equation: $\alpha+\beta+\gamma=4$, $\alpha\beta+\beta\gamma+\gamma\alpha=5$, $\alpha\beta\gamma=2$.
New sum: $2\alpha+2\beta+2\gamma=2(\alpha+\beta+\gamma)=2(4)=8$.
New pair-sum: $(2\alpha)(2\beta)+(2\beta)(2\gamma)+(2\gamma)(2\alpha)=4(\alpha\beta+\beta\gamma+\gamma\alpha)=4(5)=20$.
New product: $(2\alpha)(2\beta)(2\gamma)=8\alpha\beta\gamma=8(2)=16$.
New equation: $y^3-8y^2+20y-16=0$ — matching Example 1 exactly.
The scaling method can always be verified: if roots are scaled by $k$, then $\sum(k\alpha)=k\sum\alpha$, $\sum(k\alpha)(k\beta)=k^2\sum\alpha\beta$, and $(k\alpha)(k\beta)(k\gamma)=k^3\alpha\beta\gamma$ — each symmetric sum picks up a factor of $k$ to the power matching how many roots are multiplied. This gives a quick sanity check without re-deriving the whole equation.
Shifting the roots
Let $y=x-1$, so $x=y+1$. Substitute:
Expanding: $(y^3+3y^2+3y+1)-4(y^2+2y+1)+5(y+1)-2=0$
$y^3+3y^2+3y+1-4y^2-8y-4+5y+5-2=0$
$y^3-y^2+0y+0=0 \Rightarrow y^3-y^2=0$.
New sum: $(\alpha-1)+(\beta-1)+(\gamma-1)=(\alpha+\beta+\gamma)-3=4-3=1$.
New pair-sum: $(\alpha-1)(\beta-1)+(\beta-1)(\gamma-1)+(\gamma-1)(\alpha-1)$, which expands to $(\alpha\beta+\beta\gamma+\gamma\alpha)-2(\alpha+\beta+\gamma)+3=5-2(4)+3=0$.
New product: $(\alpha-1)(\beta-1)(\gamma-1)=\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha+\beta+\gamma)-1=2-5+4-1=0$.
New equation: $y^3-y^2+0y-0=0$, i.e. $y^3-y^2=0$ — matching Example 3.
For $(\alpha+k)$-style shifts, the symmetric-functions method reuses exactly the same expansions as "shifted product" questions, so if you've already found $\sum\alpha$, $\sum\alpha\beta$ and $\alpha\beta\gamma$ for another part of a question, it's often less work than restarting with a substitution. For scalings, reciprocals or anything that doesn't expand into a short polynomial in $k$, substitution is usually quicker.
Reciprocal and other transformations
Let $y=\dfrac1x$, so $x=\dfrac1y$. Substitute:
Multiply through by $y^3$: $2-3y+4y^2-y^3=0$, i.e. $y^3-4y^2+3y-2=0$ (multiplying by $-1$ to make the leading coefficient positive).
Let $y=2x+1$, so $x=\dfrac{y-1}{2}$. Substitute:
Multiply through by 4: $(y-1)^2-10(y-1)+12=0$.
$y^2-2y+1-10y+10+12=0 \Rightarrow y^2-12y+23=0$.
Always isolate $x$ in terms of $y$ first, then substitute — trying to substitute $y=f(x)$ directly without rearranging almost always leads to an equation still tangled in both variables. Multiply through by whatever denominator clears all fractions as the final algebraic step.
Extending to a quartic
From the original equation: $\sum\alpha=2$, $\sum\alpha\beta=3$, $\sum\alpha\beta\gamma=1$, $\alpha\beta\gamma\delta=6$.
New sum: each of the four roots contributes a $+1$, so $\sum(2\alpha+1)=2\sum\alpha+4(1)=2(2)+4=8$.
New product: $(2\alpha+1)(2\beta+1)(2\gamma+1)(2\delta+1)$ expands to $16\alpha\beta\gamma\delta+8\sum\alpha\beta\gamma+4\sum\alpha\beta+2\sum\alpha+1$ (the standard binomial-style expansion of four linear factors).
$=16(6)+8(1)+4(3)+2(2)+1=96+8+12+4+1=121$.
So the transformed equation has sum of roots $8$ and product of roots $121$ — the same result a full substitution $y=2x+1$ would give, but reached without ever writing an equation in two variables.
Exam-style questions
Question 1 6 marks
The roots of $x^3-2x^2-5x+6=0$ are $\alpha,\beta,\gamma$. Find the equation with roots $3\alpha,3\beta,3\gamma$.
Let $y=3x$, $x=y/3$. Substitute: $(y/3)^3-2(y/3)^2-5(y/3)+6=0$ (M1, A1)
$\dfrac{y^3}{27}-\dfrac{2y^2}{9}-\dfrac{5y}{3}+6=0$. Multiply by 27: $y^3-6y^2-45y+162=0$ (M1, A1, A1, A1)
Question 2 7 marks
The roots of $x^3+x^2-4x-4=0$ are $\alpha,\beta,\gamma$. Find the equation with roots $\alpha+2,\beta+2,\gamma+2$.
Let $y=x+2$, $x=y-2$. Substitute: $(y-2)^3+(y-2)^2-4(y-2)-4=0$ (M1, A1)
$(y^3-6y^2+12y-8)+(y^2-4y+4)-4y+8-4=0$ (M1, A1)
$y^3-5y^2+4y+0=0 \Rightarrow y^3-5y^2+4y=0$ (A1, A1, A1)
Question 3 7 marks
The roots of $x^3-3x^2+x-7=0$ are $\alpha,\beta,\gamma$. Find the equation with roots $\dfrac{1}{\alpha},\dfrac1\beta,\dfrac1\gamma$.
Let $y=1/x$, $x=1/y$. Substitute: $(1/y)^3-3(1/y)^2+(1/y)-7=0$ (M1, A1)
Multiply by $y^3$: $1-3y+y^2-7y^3=0$ (M1, A1)
Rearranged with positive leading coefficient: $7y^3-y^2+3y-1=0$ (A1, A1, A1)
Question 4 8 marks
The roots of $x^3-5x^2+2x-1=0$ are $\alpha,\beta,\gamma$. Using the symmetric-functions method (i.e. without substitution), find the equation with integer coefficients that has roots $\alpha+3,\beta+3,\gamma+3$.
$\alpha+\beta+\gamma=5$, $\alpha\beta+\beta\gamma+\gamma\alpha=2$, $\alpha\beta\gamma=1$ (B1)
New sum: $(\alpha+3)+(\beta+3)+(\gamma+3)=5+9=14$ (M1, A1)
New pair-sum: $(\alpha+3)(\beta+3)+(\beta+3)(\gamma+3)+(\gamma+3)(\alpha+3)=(\alpha\beta+\beta\gamma+\gamma\alpha)+6(\alpha+\beta+\gamma)+27=2+30+27=59$ (M1, A1)
New product: $(\alpha+3)(\beta+3)(\gamma+3)=\alpha\beta\gamma+3(\alpha\beta+\beta\gamma+\gamma\alpha)+9(\alpha+\beta+\gamma)+27=1+6+45+27=79$ (M1, A1)
Equation: $y^3-14y^2+59y-79=0$ (A1)