4.05b - Transforming polynomial equations | OCR A-Level Further Mathematics A

Find an equation with prescribed changes to its roots without solving the original equation. You will translate and scale roots, take their reciprocals, and replace them by their squares, using a substitution that connects the old unknown to the new one.

Translating every root

How can you replace every root of $x^3 - 3x^2 + 2x + 6 = 0$ by its square without finding the roots? This lesson builds towards that question. First, make a simpler change: move every root the same distance.

You already know

If $f(\alpha) = 0$ , then $\alpha$ is a root of $f(x) = 0$ . Multiplying the whole equation by a non-zero constant leaves its roots unchanged. The coefficients give sums and products of roots, so they can check a transformed equation.

Suppose $f(x) = 0$ has roots $1$ , $2$ , $3$ , and you want roots $-1$ , $0$ , $1$ . Each new root is the corresponding old root minus $2$ . Introduce $w = x - 2$ . To put this change into the old equation you need $x = w + 2$ , so the new equation is $f(w + 2) = 0$ .

[DIAGRAM: asset_name: Translating the roots of a cubic; asset_slug: h245_4_05b_root_translation; caption: The rule w = x - 2 moves each old root two units left, but the substitution into f is x = w + 2; description: Two horizontal number lines with identical scales. The upper line is labelled x and has blue roots at 1, 2 and 3. The lower line is labelled w and has clay roots at -1, 0 and 1. Parallel arrows join corresponding roots, each moving two units left. An annotation gives w = x - 2, x = w + 2 and f(w + 2) = 0.]
The rule w = x - 2 moves each old root two units left, but the substitution into f is x = w + 2

This reversal is the whole decision. The root rule describes where roots go; the inverse rule describes what replaces $x$ in the given polynomial.

Translating roots by a constant

If $f(x) = 0$ has roots $\alpha_1, \ldots, \alpha_n$ , an equation with roots $\alpha_1 + h, \ldots, \alpha_n + h$ is

f(w - h) = 0

because $w = x + h$ gives $x = w - h$ . This holds for real or complex roots and any constant $h$ .
Not in the formula booklet: learn it.

Write both directions of the relationship before expanding. That small step prevents a long calculation with the wrong roots.

Worked example 1: adding one to each root

Question. The equation $x^3 - 6x^2 + 11x - 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Determine an equation whose roots are $\alpha + 1$ , $\beta + 1$ , $\gamma + 1$ .

Plan. Set $w = x + 1$ , substitute $x = w - 1$ , then collect powers of $w$ .

\begin{aligned} 0 &= (w - 1)^3 - 6(w - 1)^2 + 11(w - 1) - 6 && x = w - 1 \\ &= w^3 - 3w^2 + 3w - 1 \\ &\quad - 6w^2 + 12w - 6 + 11w - 11 - 6 && \text{expand each bracket} \\ &= w^3 - 9w^2 + 26w - 24 && \text{collect like powers} \end{aligned}

Why this step? Substituting $x = w + 1$ would produce roots one smaller than the old roots, which is the opposite change.

Answer. $w^3 - 9w^2 + 26w - 24 = 0$ .

Check. The old root sum is $6$ ; adding $1$ to each of three roots gives $9$ , agreeing with the new quadratic coefficient. In this case the roots $1$ , $2$ , $3$ become $2$ , $3$ , $4$ , whose product is $24$ .

Keep the translation $w = x + 1$ , but change the degree from cubic to quartic. The substitution does exactly the same job; only the expansion has an extra power.

Worked example 2: removing a quartic's cubic term

Question. The equation $x^4 + 4x^3 + 11x^2 + 14x + 10 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ , $\delta$ . Determine an equation with roots $\alpha + 1$ , $\beta + 1$ , $\gamma + 1$ , $\delta + 1$ .

Plan. Set $w = x + 1$ , so $x = w - 1$ , then expand every bracket.

\begin{aligned} 0 &= (w - 1)^4 + 4(w - 1)^3 + 11(w - 1)^2 + 14(w - 1) + 10 \\ &= w^4 - 4w^3 + 6w^2 - 4w + 1 && \text{fourth power} \\ &\quad + 4w^3 - 12w^2 + 12w - 4 && \text{cubic term} \\ &\quad + 11w^2 - 22w + 11 + 14w - 14 + 10 && \text{remaining terms} \\ &= w^4 + 5w^2 + 4 && \text{collect powers} \end{aligned}

Why this step? The old root sum is $-4$ , so their mean is $-1$ . Adding $1$ makes the new sum zero, which removes the cubic coefficient.

Answer. $w^4 + 5w^2 + 4 = 0$ .

Check. $(w^2 + 1)(w^2 + 4) = 0$ gives $w = \pm i, \pm 2i$ . Hence the old roots are $-1 \pm i$ , $-1 \pm 2i$ , which make $\{(x + 1)^2 + 1\}\{(x + 1)^2 + 4\}$ zero. Expanding those factors gives the original quartic.

Removing the mean works for complex roots too, because the coefficient relationship is algebraic. You do not need a graph with real intercepts.

Watch out

Wrong: “New roots $\alpha + 3$ , so replace $x$ by $w + 3$ .” Right: write $w = x + 3$ , then rearrange to $x = w - 3$ . The sign in the substitution is opposite to the translation of the roots.

In a “show that” question, expand the brackets and show the collection of terms. Stating the target equation does not demonstrate the change.

Scaling and translating together

If each new root is $2$ times the old root, write $w = 2x$ , not $x = 2w$ . Adding a translation changes the rearrangement, but no new principle is needed.

A linear change of roots

If the new roots are $k\alpha_1 + h, \ldots, k\alpha_n + h$ , where $k \ne 0$ , set

w = kx + h \qquad x = \frac{w - h}{k}

The transformed equation is

f\left(\frac{w - h}{k}\right) = 0

Multiplication by a suitable non-zero constant can clear the coefficient denominators.
Not in the formula booklet: learn it.

The condition $k \ne 0$ makes the change reversible. If $k = 0$ , every proposed root would collapse to $h$ ; this substitution cannot be inverted.

Worked example 3: doubling the roots

Question. The equation $2x^3 - 3x^2 + 4x - 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Determine a cubic equation with integer coefficients and roots $2\alpha$ , $2\beta$ , $2\gamma$ .

Plan. Substitute $x = w/2$ , then multiply the entire equation by $4$ .

\begin{aligned} 0 &= 2\left(\frac w2\right)^3 - 3\left(\frac w2\right)^2 + 4\left(\frac w2\right) - 6 && x = w/2 \\ &= \frac{w^3}{4} - \frac{3w^2}{4} + 2w - 6 && \text{simplify powers} \\ 0 &= w^3 - 3w^2 + 8w - 24 && \text{multiply every term by } 4 \end{aligned}

Why this step? Each power of $x$ receives a different power of the scale factor. Scaling the coefficients directly would miss that.

Answer. $w^3 - 3w^2 + 8w - 24 = 0$ .

Check. The old product is $3$ . Doubling three roots multiplies their product by $2^3$ , giving $24$ , as the new constant requires.

Change only the root rule by adding $1$ after doubling. Use the same denominator-clearing step, then expand the shifted powers.

Worked example 4: twice each root plus one

Question. For the same cubic, determine an equation with integer coefficients whose roots are $2\alpha + 1$ , $2\beta + 1$ , $2\gamma + 1$ .

Plan. Set $w = 2x + 1$ , giving $x = (w - 1)/2$ , and multiply by $4$ .

\begin{aligned} 0 &= (w - 1)^3 - 3(w - 1)^2 + 8(w - 1) - 24 && \text{substitute and clear fractions} \\ &= w^3 - 3w^2 + 3w - 1 \\ &\quad - 3w^2 + 6w - 3 + 8w - 8 - 24 && \text{expand} \\ &= w^3 - 6w^2 + 17w - 36 && \text{collect powers} \end{aligned}

Why this step? Solve the complete rule for $x$ before substituting; subtracting $1$ and dividing by $2$ must occur in that order.

Answer. $w^3 - 6w^2 + 17w - 36 = 0$ .

Check. The old root sum is $3/2$ , so the new sum is $2(3/2) + 3 = 6$ . Also $x = 3/2$ is an old root, so $w = 4$ must work: $64 - 96 + 68 - 36 = 0$ .

Sometimes the scale is disguised. For roots of this cubic, $\alpha\beta\gamma = 3$ , so $\alpha^2\beta\gamma = 3\alpha$ , $\alpha\beta^2\gamma = 3\beta$ and $\alpha\beta\gamma^2 = 3\gamma$ . Such a question is a scaling problem after using the product relationship.

Exam tip

If asked for an equation with integer coefficients, clear every coefficient denominator and finish with an equation, including $= 0$ . An expression alone does not state the required equation.

You can test a proposed change before expanding: put a known old root into the rule and check the corresponding new root afterwards.

Taking reciprocals of roots

To replace each root by its reciprocal, write $w = 1/x$ , so $x = 1/w$ . This needs every old root to be non-zero: check the constant coefficient before you begin.

For a cubic,

\begin{aligned} a\left(\frac1w\right)^3 + b\left(\frac1w\right)^2 + c\left(\frac1w\right) + d &= 0 \\ a + bw + cw^2 + dw^3 &= 0 && \text{multiply by } w^3 \end{aligned}

The coefficients reverse their order. This also works for quartics and other degrees, but the substitution explains the reversal and its conditions.

Reciprocal root equation

For $f(x) = ax^n + \cdots + d$ , where $a \ne 0$ and $d \ne 0$ , an equation whose roots are the reciprocals of the roots of $f(x) = 0$ is

w^n f\left(\frac1w\right) = 0

The old roots are non-zero because $d \ne 0$ . The resulting polynomial has non-zero constant $a$ , so $w = 0$ has not been added as a root.
Not in the formula booklet: learn it.

Keep the non-zero restriction while clearing the variable denominator; afterwards check the new polynomial at zero.

Worked example 5: reciprocal roots

Question. Determine a cubic equation with integer coefficients whose roots are the reciprocals of the roots of $2x^3 - 3x^2 + 4x - 6 = 0$ .

Plan. Put $x = 1/w$ , clear $w^3$ , then arrange descending powers.

\begin{aligned} 0 &= \frac{2}{w^3} - \frac{3}{w^2} + \frac4w - 6 && w \ne 0 \\ 0 &= 2 - 3w + 4w^2 - 6w^3 && \text{multiply by } w^3 \\ 0 &= 6w^3 - 4w^2 + 3w - 2 && \text{multiply by } -1 \end{aligned}

Why this step? The constant $-6$ in the original polynomial rules out a zero old root, so every required reciprocal exists.

Answer. $6w^3 - 4w^2 + 3w - 2 = 0$ .

Check. The old product is $3$ , so the new product should be $1/3$ , agreeing with $-(-2)/6$ . The new constant is non-zero.

Change only the scale of the reciprocal: use $2/x$ instead of $1/x$ . Reversing the coefficients alone is no longer enough.

Worked example 6: scaled reciprocals

Question. For the same cubic, determine an equation with roots $2/\alpha$ , $2/\beta$ , $2/\gamma$ .

Plan. Set $w = 2/x$ , so $x = 2/w$ , then clear the denominator.

\begin{aligned} 0 &= 2\left(\frac2w\right)^3 - 3\left(\frac2w\right)^2 + 4\left(\frac2w\right) - 6 && \text{substitute} \\ 0 &= 16 - 12w + 8w^2 - 6w^3 && \text{multiply by } w^3 \\ 0 &= 3w^3 - 4w^2 + 6w - 8 && \text{divide by } -2 \end{aligned}

Why this step? A factor of $2$ in $x$ contributes $2^3$ to the cubic term and $2^2$ to the square term, so the full substitution is safer than a coefficient shortcut.

Answer. $3w^3 - 4w^2 + 6w - 8 = 0$ .

Check. Its root product is $8/3$ , agreeing with $(2/\alpha)(2/\beta)(2/\gamma) = 8/3$ .

If the original polynomial has a zero root, factor out that root first. An equation for reciprocals of the remaining non-zero roots has lower degree; there is no reciprocal corresponding to zero.

Watch out

Wrong: $w = 2/x$ gives $x = 1/(2w)$ . Right: multiply by $x$ to get $wx = 2$ , so $x = 2/w$ . Check the rearrangement before taking powers.

The product relationship gives a particularly quick independent check on a reciprocal transformation.

Replacing each root by its square

The substitution $w = x^2$ is not reversible in the same way as a translation: opposite roots have the same square. You can nevertheless eliminate $x$ without calculating any roots.

For $ax^3 + bx^2 + cx + d = 0$ , group the even and odd powers after putting $w = x^2$ :

\begin{aligned} x(aw + c) &= -(bw + d) && x^3 = xw \\ w(aw + c)^2 &= (bw + d)^2 && \text{square and use } x^2 = w \end{aligned}

The result is cubic in $w$ . Each old root supplies one new squared root, counted with multiplicity. If two opposite roots occur, the new equation has a repeated root. For example, $i$ and $-i$ both square to $-1$ : squaring can turn distinct roots into a repeated one.

A cubic for squared roots

If $ax^3 + bx^2 + cx + d = 0$ , with $a \ne 0$ , has roots $\alpha$ , $\beta$ , $\gamma$ , an equation with roots $\alpha^2$ , $\beta^2$ , $\gamma^2$ is

w(aw + c)^2 - (bw + d)^2 = 0

This follows by the substitution $w = x^2$ and elimination shown above. It is a method to derive, rather than a separate rule to memorise.
Not in the formula booklet: learn it.

There is no need to divide by $aw + c$ . Avoiding that division retains cases in which it is zero and prevents the loss of a root.

Worked example 7: two roots with the same square

Question. The equation $x^3 - 3x^2 + 2x - 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Determine an equation with roots $\alpha^2$ , $\beta^2$ , $\gamma^2$ .

Plan. Set $w = x^2$ , isolate the terms containing $x$ , then square to eliminate $x$ .

\begin{aligned} x(w + 2) &= 3w + 6 && \text{group odd and even powers} \\ w(w + 2)^2 &= (3w + 6)^2 && \text{square and use } x^2 = w \\ w^3 + 4w^2 + 4w &= 9w^2 + 36w + 36 && \text{expand} \\ w^3 - 5w^2 - 32w - 36 &= 0 && \text{collect powers} \end{aligned}

Why this step? Dividing by $w + 2$ would discard $w = -2$ . That is a required squared root, not an unwanted case.

Answer. $w^3 - 5w^2 - 32w - 36 = 0$ .

Check. The old polynomial is $(x - 3)(x^2 + 2)$ , with roots $3$ , $\sqrt{2}\,i$ , $-\sqrt{2}\,i$ . Their squares are $9$ , $-2$ , $-2$ . The new polynomial is $(w - 9)(w + 2)^2$ .

Change only the constant from $-6$ to $6$ . The same elimination now gives the answer to the opening question, even though the original cubic no longer has that convenient factorisation.

Worked example 8: the opening cubic

Question. The equation $x^3 - 3x^2 + 2x + 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Determine an equation whose roots are their squares.

Plan. Use $w = x^2$ , keeping all terms until the final expansion.

\begin{aligned} x(w + 2) &= 3w - 6 && \text{rearrange the original equation} \\ w(w + 2)^2 &= (3w - 6)^2 && \text{square} \\ w^3 + 4w^2 + 4w &= 9w^2 - 36w + 36 && \text{expand both sides} \\ w^3 - 5w^2 + 40w - 36 &= 0 && \text{collect powers} \end{aligned}

Why this step? Squaring the isolated odd part leaves only $x^2$ , which the substitution replaces by $w$ . No individual root is needed.

Answer. $w^3 - 5w^2 + 40w - 36 = 0$ .

Check. The old coefficient relationships give $S_1 = 3$ , $S_2 = 2$ , $S_3 = -6$ . The new sum is $3^2 - 2(2) = 5$ , the new pair sum is $2^2 - 2(3)(-6) = 40$ , and the new product is $(-6)^2 = 36$ , agreeing with all three coefficients.

Squaring eliminated the sign of $x$ , but it did not ask you to choose the positive square root. In particular, complex roots do not have an ordering into positive and negative values.

Watch out

Wrong: from $w = x^2$ , substituting $x = \sqrt{w}$ and keeping only positive real values. Right: eliminate $x$ algebraically and keep multiplicities. Two opposite old roots may produce the same new root twice.

The last calculation answers the opening question with a cubic in squared roots. The coefficients provide a check; the substitution provides the requested construction.

Quick Recap

Define the new root $w$ in terms of the old root $x$ , then rearrange before substituting into the given equation.
Translate and scale using $x = (w - h)/k$ ; multiply the whole equation to clear coefficient denominators.
Reciprocal changes require non-zero old roots; clear variable denominators and check that zero has not become a root.
For squared roots, put $w = x^2$ , group odd and even powers, then eliminate $x$ by squaring.

The squares of the roots of $x^3 - 3x^2 + 2x + 6 = 0$ satisfy $w^3 - 5w^2 + 40w - 36 = 0$ , obtained without solving the original cubic.

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4.05b - Transforming polynomial equations

Translating every root

Scaling and translating together

Taking reciprocals of roots

Replacing each root by its square

Quick Recap

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