D2 - Transforming polynomial roots | AQA A-Level Further Mathematics

Move every root of a polynomial by the same scaling and translation, and you can find the new equation without solving the old one. Learn to reverse the root transformation, substitute safely and check the result using symmetric sums.

Linear changes of roots

If $\alpha,\beta,\gamma$ solve $x^3-4x^2+x+6=0$ , what equation has roots $2\alpha+1,2\beta+1,2\gamma+1$ ? We will transform the equation itself, rather than first finding the three roots.

Start with one root: if $\alpha=-1$ , its new value is $2(-1)+1=-1$ . If $\beta=2$ , its new value is $5$ . The same rule acts on each root occurrence.

Affine root transformation

A transformation $u=px+q$ , with constants $p\ne0$ and $q$ , scales each root by $p$ , then adds $q$ . Its inverse is $x=(u-q)/p$ . Distinct old roots give distinct new roots, and multiplicities are preserved.

Why must the substitution be the inverse? The original polynomial expects an old value $x$ . Given a proposed new root $u$ , undo the transformation to recover that old value.

Worked example 1: mapping roots and reversing the map

Question. The roots of a cubic are $-1,2,3$ . Find the transformed roots under $u=2x+1$ , and express $x$ in terms of $u$ .

Plan. Apply the same rule to each root, then solve the transformation for the old variable.

\begin{aligned} -1&\mapsto -1,&2&\mapsto5,&3&\mapsto7\\ u=2x+1&\Longrightarrow u-1=2x \Longrightarrow x=\frac{u-1}{2} \end{aligned}

Answer. The roots become $-1,5,7$ ; the inverse substitution is $x=(u-1)/2$ .

Check. Undo the map on $7$ : $(7-1)/2=3$ , the root from which it came.

Change only the sign of the scale factor. A negative scale reverses the order of real roots, but it remains invertible.

Worked example 2: a negative scale

Question. Use the same old root list $-1,2,3$ , now with $u=-2x+1$ . Find the new roots and the inverse substitution.

Plan. Multiply by $-2$ before adding $1$ , then isolate $x$ .

\begin{aligned} -1&\mapsto3,&2&\mapsto-3,&3&\mapsto-5\\ u-1&=-2x \Longrightarrow x=\frac{1-u}{2} \end{aligned}

Answer. The roots are $3,-3,-5$ , and $x=(1-u)/2$ .

Check. The new root $-5$ maps back to $(1+5)/2=3$ ; applying the forward map again gives $-5$ .

If $p=0$ , every value maps to $q$ , so no inverse exists. One could construct $(u-q)^n=0$ to list $n$ transformed root occurrences, but the inverse-substitution method below requires $p\ne0$ .

Watch out

Wrong: to obtain roots $2\alpha+1$ , substitute $x=2u+1$ . That is the forward map in the wrong direction. Solve $u=2x+1$ first: substitute $x=(u-1)/2$ .

Practise reversing a new transformation.

Forming the transformed polynomial

If $f(x)=0$ has degree $n$ and leading coefficient $a\ne0$ , replace $x$ by $(u-q)/p$ . Multiplying by $p^n$ clears the powers of the scale factor from the denominators.

Inverse substitution

For $p\ne0$ , an equation whose roots are $p\alpha_i+q$ is

p^n f\!\left(\frac{u-q}{p}\right)=0

Its degree is still $n$ , and its leading coefficient is $a$ . A further nonzero constant multiple gives an equivalent equation. This method is not supplied in the formula booklet: learn the inverse substitution.

Completeness matters. If $f(x)=a\prod_{i=1}^{n}(x-\alpha_i)$ , then

p^n f\!\left(\frac{u-q}{p}\right) =a\prod_{i=1}^{n}\bigl(u-(p\alpha_i+q)\bigr)

These are exactly the required $n$ root occurrences, with no extra roots and no lost repeated roots. The proof works over the complex numbers too.

Worked example 3: scale and translate a cubic

Question. Form an equation with integer coefficients whose roots are $2\alpha+1,2\beta+1,2\gamma+1$ , where the old roots solve $x^3-4x^2+x+6=0$ .

Plan. Substitute $x=(u-1)/2$ and multiply the whole equation by $8$ .

\begin{aligned} \left(\frac{u-1}{2}\right)^3 -4\left(\frac{u-1}{2}\right)^2+\frac{u-1}{2}+6&=0\\ (u-1)^3-8(u-1)^2+4(u-1)+48&=0\\ u^3-3u^2+3u-1-8u^2+16u-8+4u-4+48&=0 \end{aligned}

The constant term comes from four contributions: $-1-8-4+48=35$ .

Answer. $u^3-11u^2+23u+35=0$ .

Check. The old polynomial factors as $(x+1)(x-2)(x-3)$ . Transforming those roots gives $-1,5,7$ , and $(u+1)(u-5)(u-7)$ expands to the answer.

Change only the translation, from $+1$ to $-1$ . The substitution now contains $u+1$ , which changes the collected coefficients.

Worked example 4: changing the translation

Question. Form an equation with integer coefficients whose roots are $2\alpha-1,2\beta-1,2\gamma-1$ , for the same cubic.

Plan. Use $x=(u+1)/2$ , then clear denominators before expanding.

\begin{aligned} (u+1)^3-8(u+1)^2+4(u+1)+48&=0\\ u^3+3u^2+3u+1-8u^2-16u-8+4u+4+48&=0\\ u^3-5u^2-9u+45&=0 \end{aligned}

Answer. $u^3-5u^2-9u+45=0$ .

Check. The new roots are $-3,3,5$ . Their factor product $(u+3)(u-3)(u-5)=(u^2-9)(u-5)$ gives exactly this cubic.

For a quartic, multiply by $p^4$ ; do not use $p^3$ just because the worked examples were cubic. If integer coefficients are requested and fractions remain, multiply the entire equation by a suitable nonzero integer.

Watch out

Wrong: multiplying only the leading term by $8$ . Every term of the equation must be multiplied, including the old constant. Finish with $=0$ : an expression alone is not the requested equation.

Apply the cubic method with a fresh polynomial.

Checking with roots and coefficients

Symmetric sums provide an alternative route and a quick independent check. For three roots, a common translation appears three times in their sum, but its effect on the pairwise sum and product needs careful expansion.

Let the old cubic have $e_1=\alpha+\beta+\gamma$ , $e_2=\alpha\beta+\alpha\gamma+\beta\gamma$ , $e_3=\alpha\beta\gamma$ . For new roots $p\alpha+q,p\beta+q,p\gamma+q$ , expansion gives

\begin{aligned} E_1&=pe_1+3q\\ E_2&=p^2e_2+2pqe_1+3q^2\\ E_3&=p^3e_3+p^2qe_2+pq^2e_1+q^3 \end{aligned}

For example, each old root appears in two of the three pair products, explaining $2pqe_1$ . These are derived expressions, not extra formula-booklet entries. The monic new cubic is $u^3-E_1u^2+E_2u-E_3=0$ .

Worked example 5: a pure scaling by symmetric sums

Question. The roots of $x^3-2x^2-5x+6=0$ are scaled by $2$ . Find their equation using coefficient relations.

Plan. Read the old sums, scale each product according to its number of factors, then rebuild the cubic.

\begin{aligned} (e_1,e_2,e_3)&=(2,-5,-6)\\ (E_1,E_2,E_3)&=(2e_1,4e_2,8e_3)=(4,-20,-48) \end{aligned}

Answer. $u^3-4u^2-20u+48=0$ .

Check. Inverse substitution gives $8f(u/2)=u^3-4u^2-20u+48$ , independently confirming all coefficients.

Keep the scaling and add a translation of $1$ . The mixed terms now matter; this is why substitution is often shorter for cubic and quartic questions.

Worked example 6: adding a translation by symmetric sums

Question. For the same old cubic, find an equation whose roots are $2\alpha+1,2\beta+1,2\gamma+1$ , using coefficient relations.

Plan. Use $p=2,q=1$ in the three derived sums.

\begin{aligned} E_1&=2(2)+3=7\\ E_2&=4(-5)+4(2)+3=-9\\ E_3&=8(-6)+4(-5)+2(2)+1=-63 \end{aligned}

Answer. $u^3-7u^2-9u+63=0$ .

Check. The inverse-substitution expression is $(u-1)^3-4(u-1)^2-20(u-1)+48$ . Expanding gives the same cubic, including constant $-1-4+20+48=63$ .

For a quartic, the sum check is $E_1=pe_1+4q$ ; the $4$ counts four root occurrences. Full expansion of all quartic symmetric sums is possible, but inverse substitution avoids managing many mixed terms. Choose the shorter route, then use a sum or product as a check.

Watch out

Wrong for a translation: $E_3=p^3e_3+q^3$ . The product also contains mixed terms. Expand it completely, or use inverse substitution. When using symmetric sums, group expressions as $p(\alpha+\beta+\gamma)+3q$ so the coefficient relation can actually be used.

Check a proposed equation by two symmetric sums.

Quick Recap

Reverse $u=px+q$ to $x=(u-q)/p$ , with $p\ne0$ .
Substitute the inverse into the old degree- $n$ equation and multiply by $p^n$ .
The factor identity proves that degree, completeness and multiplicities are preserved.
Symmetric sums offer a second route and useful coefficient checks; translations create mixed products.

The opening equation is $u^3-11u^2+23u+35=0$ , whose roots are exactly $2\alpha+1,2\beta+1,2\gamma+1$ .

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D2 - Transforming polynomial roots

Linear changes of roots

Forming the transformed polynomial

Checking with roots and coefficients

Quick Recap

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