CP 4.2 - Equations with transformed roots

CP 4.2 - Equations with transformed roots

You can move every root of a polynomial without finding the roots individually. You will form equations for shifted and scaled roots by reversing the transformation, substituting into the original equation and clearing numerical denominators.

Reversing the root transformation

Suppose a cubic has roots α,β,γ\alpha,\beta,\gamma. How could you write an equation with roots 2α+1,2β+1,2γ+12\alpha+1,2\beta+1,2\gamma+1 while leaving the original cubic unsolved?

Start with one root: an old root x=3x=3 would become a new root y=7y=7. To use the old equation, you need to turn that new value back into the old one: x=(y−1)/2x=(y-1)/2. The substitution in the polynomial therefore runs in the reverse direction from the stated root transformation.

You already know

For a polynomial with roots α1,…,αn\alpha_1,\ldots,\alpha_n, factorisation gives f(x)=a∏j=1n(x−αj)f(x)=a\prod_{j=1}^{n}(x-\alpha_j). Roots are counted with multiplicity, and multiplying an equation by a non-zero constant leaves its roots unchanged. The sum of the roots is the negative of the next-to-leading coefficient divided by the leading coefficient.

Here “linear transformation of the roots” allows a scale followed by a shift, y=px+qy=px+q, with constants p,qp,q. The scale must be non-zero if the transformed equation is to retain the same degree and root multiplicities.

Equation for linearly transformed roots

If f(x)=0f(x)=0 has degree nn and roots αj\alpha_j, then an equation with roots pαj+qp\alpha_j+q, where p≠0p\ne0, is

f(y−qp)=0f\left(\frac{y-q}{p}\right)=0

Multiplying by pnp^n clears the substitution's numerical denominators. Any further non-zero constant multiple is an equivalent equation.

Not in the formula booklet: learn it.

Why does this capture every root? The equation y=pαj+qy=p\alpha_j+q is equivalent to (y−q)/p=αj(y-q)/p=\alpha_j, so the substituted polynomial vanishes exactly at the required new roots. This works for real and non-real roots alike; repeated roots keep their multiplicities. If p=0p=0, all the new values collapse to qq, so the inverse substitution is unavailable.

Key idea

Name the new variable, write its relationship to the old variable, then solve for the old variable before substituting. That one line prevents the common reversal of a shift.

The following check asks for the substitution, before any expansion gets in the way.

Building the transformed cubic

For a cubic, substitute the reversed transformation into all four terms and clear a denominator cubed before collecting powers of the new variable.

Do not transform each coefficient as though it were a root. A shift changes several coefficients because every power expands. A missing power in the final equation is allowed: it simply has coefficient zero.

Worked example 1: shifting every root

Question. The equation x3−2x2−x+2=0x^3-2x^2-x+2=0 has roots α,β,γ\alpha,\beta,\gamma. Find a monic cubic equation with roots α+1,β+1,γ+1\alpha+1,\beta+1,\gamma+1.

Plan. Set y=x+1y=x+1, substitute x=y−1x=y-1 and expand.

0=(y−1)3−2(y−1)2−(y−1)+2substitute the old variable=y3−3y2+3y−1−2y2+4y−2−y+1+2expand each term=y3−5y2+6ycollect powers\begin{aligned} 0&=(y-1)^3-2(y-1)^2-(y-1)+2 &&\text{substitute the old variable}\\ &=y^3-3y^2+3y-1-2y^2+4y-2-y+1+2 &&\text{expand each term}\\ &=y^3-5y^2+6y &&\text{collect powers} \end{aligned}

Why this step? The root α+1\alpha+1 must become α\alpha inside the original polynomial, so you subtract 11, rather than add it again.

Answer. y3−5y2+6y=0y^3-5y^2+6y=0.

Check. The old root sum is 22. Shifting three roots by 11 gives new sum 2+3=52+3=5, agreeing with the coefficient −5-5. Also f(1)=0f(1)=0, so the new root 22 should work: 8−20+12=08-20+12=0.

Worked example 2 changes only the scale in y=px+1y=px+1, from 11 to −2-2. This includes a reflection and introduces a denominator in the inverse substitution.

Worked example 2: a negative scale and a shift

Question. For the same original cubic, find a monic equation with roots 1−2α,1−2β,1−2γ1-2\alpha,1-2\beta,1-2\gamma.

Plan. Set y=1−2xy=1-2x, substitute x=(1−y)/2x=(1-y)/2 and multiply by 88.

0=(1−y2)3−2(1−y2)2−1−y2+2inverse substitution0=(1−y)3−4(1−y)2−4(1−y)+16multiply every term by 8=−y3−y2+9y+9expand and collect\begin{aligned} 0&=\left(\frac{1-y}{2}\right)^3 -2\left(\frac{1-y}{2}\right)^2-\frac{1-y}{2}+2 &&\text{inverse substitution}\\ 0&=(1-y)^3-4(1-y)^2-4(1-y)+16 &&\text{multiply every term by }8\\ &=-y^3-y^2+9y+9 &&\text{expand and collect} \end{aligned}

Why this step? A cubic denominator requires 23=82^3=8. The lower-degree terms must also be multiplied by 88, leaving factors of 44 rather than 22.

Answer. Multiplying by −1-1 gives the monic equation y3+y2−9y−9=0y^3+y^2-9y-9=0.

Check. The original roots are −1,1,2-1,1,2, as factorisation (x+1)(x−1)(x−2)(x+1)(x-1)(x-2) confirms. Their transformed values are 3,−1,−33,-1,-3, and (y−3)(y+1)(y+3)=y3+y2−9y−9(y-3)(y+1)(y+3)=y^3+y^2-9y-9.

A useful surprise follows from the root sum: shifting every root by its mean removes the next-to-leading term. If nn roots have sum S1S_1, the shifted roots αj−S1/n\alpha_j-S_1/n sum to S1−n(S1/n)=0S_1-n(S_1/n)=0. This is why a suitable shift can remove the squared term of any cubic. This consequence is not in the formula booklet; derive it from the root sum when needed.

Watch out

Wrong: roots α+1\alpha+1 require f(y+1)=0f(y+1)=0. That equation has roots α−1\alpha-1. Right: state y=x+1y=x+1, rearrange to x=y−1x=y-1, and use f(y−1)=0f(y-1)=0.

The next transformation changes the scale to a new positive value. Use the same inverse-substitution line before expanding.

Quartics and preserved multiplicities

The inverse-substitution method works for quartics as well: use a fourth power when clearing the substitution denominator, and preserve every repeated root.

A change of variable can make a complicated-looking quartic much simpler. The transformation does not discard roots; a repeated old root becomes the corresponding repeated new root.

Worked example 3: a shift that simplifies a quartic

Question. The roots of x4−4x3+2x2+4x−3=0x^4-4x^3+2x^2+4x-3=0 are α,β,γ,δ\alpha,\beta,\gamma,\delta. Find a monic equation with roots α−1,β−1,γ−1,δ−1\alpha-1,\beta-1,\gamma-1,\delta-1.

Plan. Set y=x−1y=x-1, substitute x=y+1x=y+1 and collect powers.

0=(y+1)4−4(y+1)3+2(y+1)2+4(y+1)−3=y4+(4−4)y3+(6−12+2)y2+(4−12+4+4)y+(1−4+2+4−3)collect by degree=y4−4y2simplify coefficients\begin{aligned} 0&=(y+1)^4-4(y+1)^3+2(y+1)^2+4(y+1)-3\\ &=y^4+(4-4)y^3+(6-12+2)y^2\\ &\quad +(4-12+4+4)y+(1-4+2+4-3) &&\text{collect by degree}\\ &=y^4-4y^2 &&\text{simplify coefficients} \end{aligned}

Why this step? The old root sum is 44, so shifting all four roots by −1-1 makes the new sum zero. The missing cubic term is expected, not a lost term.

Answer. y4−4y2=0y^4-4y^2=0, or y2(y−2)(y+2)=0y^2(y-2)(y+2)=0.

Check. The transformed roots are −2,0,0,2-2,0,0,2. Shifting back gives −1,1,1,3-1,1,1,3, and (x+1)(x−1)2(x−3)(x+1)(x-1)^2(x-3) expands to the original quartic. The double root has survived.

Worked example 4 changes only the scale of those shifted roots, from 11 to 33. The centred form just found saves a long second expansion.

Worked example 4: scaling a centred quartic

Question. For the same quartic, find a monic equation with roots 3α−3,3β−3,3γ−3,3δ−33\alpha-3,3\beta-3,3\gamma-3,3\delta-3.

Plan. Set y=3(x−1)y=3(x-1). Use x−1=y/3x-1=y/3 in the equivalent centred equation (x−1)4−4(x−1)2=0(x-1)^4-4(x-1)^2=0.

0=(y3)4−4(y3)2substitute the centred variable=y481−4y29evaluate the powers0=y4−36y2multiply by 81\begin{aligned} 0&=\left(\frac y3\right)^4-4\left(\frac y3\right)^2 &&\text{substitute the centred variable}\\ &=\frac{y^4}{81}-\frac{4y^2}{9} &&\text{evaluate the powers}\\ 0&=y^4-36y^2 &&\text{multiply by }81 \end{aligned}

Why this step? The previous expansion established the whole centred polynomial. Reusing that identity is shorter than expanding the original quartic again, and still applies the correct inverse transformation.

Answer. y4−36y2=0y^4-36y^2=0.

Check. The roots are −6,0,0,6-6,0,0,6, three times the shifted roots −2,0,0,2-2,0,0,2. Their sum is zero and their pair sum is −36-36, matching both coefficients.

These particular transformed equations are not in the formula booklet: derive them by substitution. The opening puzzle has the same solution for a cubic or quartic: reverse the desired scale and shift inside the original polynomial.

Watch out

Wrong: a repeated root is listed only once when forming the new polynomial. Right: preserve its multiplicity. In Worked example 3, y=0y=0 appears twice, so the factor is y2y^2, and the equation still has degree four.

A new quartic and a different shift give another chance to check denominator powers and multiplicities together.

Quick Recap

  • Write the new root variable as y=px+qy=px+q, then substitute x=(y−q)/px=(y-q)/p into the original equation.
  • Multiply through by pnp^n for a degree-nn polynomial before collecting coefficients.
  • A non-zero scale preserves the degree and every root's multiplicity, including non-real or repeated roots.
  • Check the transformed root sum, or map one convenient old root to its new value.

The opening transformation 2α+12\alpha+1 needs f((y−1)/2)=0f((y-1)/2)=0: the original cubic can stay unsolved.