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 $\alpha, \beta, \gamma$ , a linear transformation might be $\text{new roots} = p\alpha + q$ , where $p$ and $q$ 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) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 = 0

with roots $\alpha, \beta, \gamma, \dots$

Transformation of Roots

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

Expressing $x$ in terms of $y$ : From the transformation $y = p\alpha + q$ , rearrange to $\alpha = \frac{y - q}{p}$
Substituting $\alpha = \frac{y - q}{p}$ into the original polynomial: Replace $x$ in the original polynomial with $\frac{y - q}{p}$
Clearing the fraction: Multiply through by $p^n$ to eliminate the denominator and simplify.

Worked Examples

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Example 1:

Linear Transformation $y = 2\alpha + 3$

Given Polynomial

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

with roots $\alpha, \beta, \gamma$

Step 1**: Express** $x$ in terms of $y$

The transformation is $y = 2x + 3$

Rearrange to find $x$

x = \frac{y - 3}{2}

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

Replace $x$ in $f(x)$ with $\frac{y - 3}{2}$

f\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 :

\left(\frac{y - 3}{2}\right)^3

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

Expand :

\left(\frac{y - 3}{2}\right)^2

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

Simplify all terms.

f\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 $\frac{1}{8}$ ) to eliminate fractions:

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

Step 5**: Simplify**

Expand $(y - 3)^3$ and simplify:

y^3 - 9y^2 + 27y - 27 - 12(y^2 - 6y + 9) + 44(y - 3) - 48 = 0

y^3 - 9y^2 + 27y - 27 - 12y^2 + 72y - 108 + 44y - 132 - 48 = 0

Combine like terms:

y^3 - 21y^2 + 143y - 315 = 0

The new polynomial is:

y^3 - 21y^2 + 143y - 315 = 0

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Example 2:

Linear Transformation $y = 3\alpha - 4$

Given Polynomial

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

with roots $\alpha, \beta, \gamma$

Step 1**: Express** $x$ in terms of $y$

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

Rearrange to find $x$

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

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

Replace $x$ in $f(x)$ :

f\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

Step 4**: Simplify**

Expand and combine terms:

y^3 + 12y^2 + 48y + 64 + 6(y^2 + 8y + 16) - 45y - 180 + 162 = 0

y^3 + 12y^2 + 48y + 64 + 6y^2 + 48y + 96 - 45y - 180 + 162 = 0

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

The new polynomial is:

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

Note Summary

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Common Mistakes

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

infoNote

Key Formulas

Linear transformation of roots: $y = p\alpha + q$
Express $x$ in terms of: $x = \frac{y - q}{p}$
New polynomial: Substitute $x = \frac{y - q}{p}$ into $f(x) = 0$
Clear fractions: Multiply by $p^n$ , where $n$ is the degree of the original polynomial.
Simplify to find the new polynomial equation.

Linear Transformations of Roots Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

3.1.2 Linear Transformations of Roots

Introduction

Original Polynomial

Transformation of Roots

Worked Examples

Example 1:

Example 2:

Note Summary

Common Mistakes

Key Formulas

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