Roots of Polynomials (Edexcel A-Level Further Mathematics): Revision Notes

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3.1.2 Linear Transformations of Roots

Introduction

A linear transformation of the roots of a polynomial involves creating a new polynomial whose roots are related to the roots of the original polynomial by a specific linear equation.

For example, if the roots of a polynomial are α,β,γ\alpha, \beta, \gamma, a linear transformation might be new roots=pα+q\text{new roots} = p\alpha + q, where pp and qq are constants.

This topic is about understanding how to determine the equation of a new polynomial when the roots of the original polynomial are subjected to such transformations.

Original Polynomial

Suppose the given polynomial is:

f(x)=anxn+an−1xn−1+⋯+a1x+a0=0f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 = 0

with roots α,β,γ,…\alpha, \beta, \gamma, \dots

Transformation of Roots

If the roots are transformed by y=pα+qy = p\alpha + q, the new polynomial can be found by:

  1. Expressing xx in terms of yy: From the transformation y=pα+qy = p\alpha + q, rearrange to α=y−qp\alpha = \frac{y - q}{p}
  2. Substituting α=y−qp\alpha = \frac{y - q}{p} into the original polynomial: Replace xx in the original polynomial with y−qp\frac{y - q}{p}
  3. Clearing the fraction: Multiply through by pnp^n to eliminate the denominator and simplify.

Worked Examples

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Example 1:

Linear Transformation y=2α+3y = 2\alpha + 3

Given Polynomial

f(x)=x3−6x2+11x−6=0f(x) = x^3 - 6x^2 + 11x - 6 = 0

with roots α,β,γ\alpha, \beta, \gamma

Step 1**: Express** xx in terms of yy

The transformation is y=2x+3y = 2x + 3

Rearrange to find xx

x=y−32x = \frac{y - 3}{2}

Step 2**: Substitute** x=y−32x = \frac{y - 3}{2} into the original polynomial

Replace xx in f(x)f(x) with y−32\frac{y - 3}{2}

f(y−32)=(y−32)3−6(y−32)2+11(y−32)−6f\left(\frac{y - 3}{2}\right) = \left(\frac{y - 3}{2}\right)^3 - 6\left(\frac{y - 3}{2}\right)^2 + 11\left(\frac{y - 3}{2}\right) - 6

Step 3**: Expand each term**

Expand :

(y−32)3\left(\frac{y - 3}{2}\right)^3(y−32)3=(y−3)38\left(\frac{y - 3}{2}\right)^3 = \frac{(y - 3)^3}{8}

Expand :

(y−32)2\left(\frac{y - 3}{2}\right)^2(y−32)2=(y−3)24\left(\frac{y - 3}{2}\right)^2 = \frac{(y - 3)^2}{4}

Simplify all terms.

f(y−32)=(y−3)38−6(y−3)24+11(y−3)2−6f\left(\frac{y - 3}{2}\right) = \frac{(y - 3)^3}{8} - \frac{6(y - 3)^2}{4} + \frac{11(y - 3)}{2} - 6

Step 4**: Clear fractions**

Multiply through by 8 (the denominator of 18\frac{1}{8}) to eliminate fractions:

(y−3)3−12(y−3)2+44(y−3)−48=0(y - 3)^3 - 12(y - 3)^2 + 44(y - 3) - 48 = 0

Step 5**: Simplify**

Expand (y−3)3(y - 3)^3 and simplify:

y3−9y2+27y−27−12(y2−6y+9)+44(y−3)−48=0y^3 - 9y^2 + 27y - 27 - 12(y^2 - 6y + 9) + 44(y - 3) - 48 = 0y3−9y2+27y−27−12y2+72y−108+44y−132−48=0y^3 - 9y^2 + 27y - 27 - 12y^2 + 72y - 108 + 44y - 132 - 48 = 0

Combine like terms:

y3−21y2+143y−315=0y^3 - 21y^2 + 143y - 315 = 0

The new polynomial is:

y3−21y2+143y−315=0y^3 - 21y^2 + 143y - 315 = 0
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Example 2:

Linear Transformation y=3α−4y = 3\alpha - 4

Given Polynomial

f(x)=x3+2x2−5x+6=0f(x) = x^3 + 2x^2 - 5x + 6 = 0

with roots α,β,γ\alpha, \beta, \gamma

Step 1**: Express** xx in terms of yy

The transformation is y=3x−4y = 3x - 4.

Rearrange to find xx

x=y+43.x = \frac{y + 4}{3}.

Step 2**: Substitute** x=y+43x = \frac{y + 4}{3} into the original polynomial

Replace xx in f(x)f(x):

f(y+43)=(y+43)3+2(y+43)2−5(y+43)+6f\left(\frac{y + 4}{3}\right) = \left(\frac{y + 4}{3}\right)^3 + 2\left(\frac{y + 4}{3}\right)^2 - 5\left(\frac{y + 4}{3}\right) + 6

Step 3**: Clear fractions**

Multiply through by 27 (since 3^3$$ = 27):

(y+4)3+6(y+4)2−45(y+4)+162=0(y + 4)^3 + 6(y + 4)^2 - 45(y + 4) + 162 = 0

Step 4**: Simplify**

Expand and combine terms:

y3+12y2+48y+64+6(y2+8y+16)−45y−180+162=0y^3 + 12y^2 + 48y + 64 + 6(y^2 + 8y + 16) - 45y - 180 + 162 = 0y3+12y2+48y+64+6y2+48y+96−45y−180+162=0y^3 + 12y^2 + 48y + 64 + 6y^2 + 48y + 96 - 45y - 180 + 162 = 0y3+18y2+51y+142=0y^3 + 18y^2 + 51y + 142 = 0

The new polynomial is:

y3+18y2+51y+142=0y^3 + 18y^2 + 51y + 142 = 0

Note Summary

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Common Mistakes

  1. Forgetting to rearrange the transformation properly: Ensure xx is correctly expressed in terms of yy.
  2. Errors in substitution: Be careful when substituting y−qp\frac{y - q}{p} into the original polynomial.
  3. Fraction mishandling: Ensure all terms are multiplied through by the appropriate power of pp to clear fractions.
  4. Expanding incorrectly: Mistakes often occur when expanding (y−q)n(y - q)^n
  5. Combining terms inaccurately: Check carefully for errors when simplifying after substitution.
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Key Formulas

  1. Linear transformation of roots: y=pα+qy = p\alpha + q

  2. Express xx in terms of: x=y−qpx = \frac{y - q}{p}

  3. New polynomial: Substitute x=y−qpx = \frac{y - q}{p} into f(x)=0f(x) = 0

  4. Clear fractions: Multiply by pnp^n, where nn is the degree of the original polynomial.

  5. Simplify to find the new polynomial equation.

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