Overview

Changing variables transforms roots and can make symmetric sums easier. Common moves include shifting and taking reciprocals.

Key Ideas

• If $r$ is a root of $P(x)$, then $r+k$ is a root of $P(x-k)$.
• If $r$ is a root of $P(x)$, then $1/r$ is a root of the reversed-coefficient polynomial when the constant term is nonzero.
• Use these to compute sums over transformed roots.

Core Skills

Shift the Variable

Replace $x$ with $x-h$ to shift roots by $h$. This is the standard way to move between a polynomial and its translated version.

Reverse Coefficients for Reciprocals

If $P(x)=a_nx^n+\cdots+a_0$ and $a_0\ne 0$, then $x^nP(1/x)$ has roots $1/r$.

Use Vieta After Transformations

Once the new polynomial is found, apply Vieta to compute sums and products.

Worked Example

If $x^2-5x+6=0$ has roots $r,s$, find the polynomial with roots $r+1, s+1$.

Replace $x$ by $x-1$: $(x-1)^2 - 5(x-1) + 6 = x^2 - 7x + 12$. So the new polynomial is $x^2 - 7x + 12$.

More Examples

Example 1: Shift Down

If $x^2-2x-3=0$ has roots $r,s$, find the polynomial with roots $r-2, s-2$.

Replace $x$ by $x+2$: $(x+2)^2-2(x+2)-3 = x^2+2x-3$.

Example 2: Reciprocal Roots

If $P(x)=x^2-3x+2$, find a polynomial with roots $1/r,1/s$.

Compute $x^2P(1/x)=1-3x+2x^2$, so the polynomial is $2x^2-3x+1$.

Example 3: Sum of Shifted Roots

If $r+s=4$, find $(r+1)+(s+1)$.

The sum increases by $2$, so it is $6$.

Strategy Checklist

• Decide whether you need a shift or a reciprocal.
• Apply the substitution carefully and simplify.
• Use Vieta to extract root sums/products.

Common Pitfalls

• Shifting in the wrong direction.
• Forgetting that reciprocal roots require a nonzero constant term.
• Dropping factors when forming $x^nP(1/x)$.

Practice Problems