Home / A-Level Further Maths 9231 / Roots of polynomial equations / Transforming polynomial equations
Cambridge International AS & A Level · Topic 1.1
A-Level Further Maths 9231 / Topic 1.1 / AS & A Level

New roots, no solving.

You can build the equation whose roots are \(\frac{1}{\alpha}\), \(\alpha^{2}\) or \(2\alpha + 1\) without ever finding \(\alpha\). One substitution turns the old equation into the new one, and its coefficients then hand you awkward symmetric functions almost for free.

Topic 1.1All roots of polynomial equations lessons →
The key idea

To find the equation whose roots are \(y = f(\alpha)\) for each root \(\alpha\) of \(p(x) = 0\): write \(y\) in terms of \(x\), rearrange to get \(x\) in terms of \(y\), substitute into \(p(x) = 0\) and clear fractions or surds. Reciprocals \(y = \frac{1}{x}\) reverse the coefficients; squares need \(x = \sqrt{y}\), then isolate the \(\sqrt{y}\) terms and square.

1. New root in terms of old y = 2x + 1 2. Invert: old in terms of new x = (y − 1) ÷ 2 3. Substitute into p(x) = 0 p((y − 1) ÷ 2) = 0 4. Clear fractions or surds tidy to integer coefficients Squares: x = √y, put the √y terms on one side, then square both sides
Fig. 1: the four steps of a substitution. Step 2 is the one students skip: you substitute for x, so you need x in terms of y
Section 01

The method: invert, then substitute.

Let \(\alpha\), \(\beta\), \(\gamma\) be the roots of \(p(x) = 0\) and suppose you want the equation whose roots are \(f(\alpha)\), \(f(\beta)\), \(f(\gamma)\). Put \(y = f(x)\). Since \(x\) is a root of \(p(x) = 0\) exactly when \(y\) is a root of the new equation, rearrange to \(x = g(y)\) and substitute: \(p(g(y)) = 0\) is the equation you want. The exam will not give you the substitution for reciprocals, squares or simple linear changes, so choose it yourself.

New rootsPutSubstitute
\(\dfrac{1}{\alpha}\)\(y = \dfrac{1}{x}\)\(x = \dfrac{1}{y}\), then multiply by \(y^{n}\): the coefficients reverse
\(k\alpha\)\(y = kx\)\(x = \dfrac{y}{k}\)
\(\alpha + h\)\(y = x + h\)\(x = y - h\)
\(2\alpha + 1\)\(y = 2x + 1\)\(x = \dfrac{y - 1}{2}\)
Worked example Reciprocal roots
The roots of \(x^{3} - 2x^{2} + 5x - 3 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\), \(\dfrac{1}{\gamma}\).
  1. Put \(y = \dfrac{1}{x}\), so \(x = \dfrac{1}{y}\).
  2. \(\dfrac{1}{y^{3}} - \dfrac{2}{y^{2}} + \dfrac{5}{y} - 3 = 0\).
  3. Multiply by \(y^{3}\): \(1 - 2y + 5y^{2} - 3y^{3} = 0\).
  4. Multiply by \(-1\) and order the terms. The coefficients \(1, -2, 5, -3\) have simply reversed (up to sign).
Answer: \(3y^{3} - 5y^{2} + 2y - 1 = 0\)
Worked example A linear change of root
The roots of \(2x^{3} + 3x^{2} - x + 4 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(2\alpha + 1\), \(2\beta + 1\), \(2\gamma + 1\).
  1. Put \(y = 2x + 1\), so \(x = \dfrac{y - 1}{2}\).
  2. \(2\left(\dfrac{y - 1}{2}\right)^{3} + 3\left(\dfrac{y - 1}{2}\right)^{2} - \dfrac{y - 1}{2} + 4 = 0\).
  3. Multiply by 4: \((y - 1)^{3} + 3(y - 1)^{2} - 2(y - 1) + 16 = 0\).
  4. Expand: \(y^{3} - 3y^{2} + 3y - 1 + 3y^{2} - 6y + 3 - 2y + 2 + 16 = 0\).
  5. Check: the new sum of roots is \(2(\alpha + \beta + \gamma) + 3 = 2\left(-\dfrac{3}{2}\right) + 3 = 0\), matching the missing \(y^{2}\) term.
Answer: \(y^{3} - 5y + 20 = 0\)
Your turn
The roots of \(x^{3} + 4x^{2} - 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(\alpha - 3\), \(\beta - 3\), \(\gamma - 3\).
Examiner trap

Substitute the INVERSE. For roots \(\alpha + 2\) you put \(x = y - 2\), not \(x = y + 2\); the second gives roots \(\alpha - 2\). A quick check on the sum of the new roots catches this every time.

Section 02

Squares and given substitutions.

For roots \(\alpha^{2}\) put \(y = x^{2}\), so \(x = \sqrt{y}\). The odd powers of \(x\) leave \(\sqrt{y}\) behind. Collect every term containing \(\sqrt{y}\) on one side, everything else on the other, and square both sides once. Harder substitutions, such as \(y = \dfrac{x}{x - 1}\) or \(y = x^{2} + 1\), are usually given, but the method is the same: make \(x\) the subject.

Worked example Square roots of a cubic
The roots of \(x^{3} + 2x^{2} - 3x + 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(\alpha^{2}\), \(\beta^{2}\), \(\gamma^{2}\).
  1. Put \(x = \sqrt{y}\): \(y\sqrt{y} + 2y - 3\sqrt{y} + 1 = 0\).
  2. Isolate the \(\sqrt{y}\) terms: \(\sqrt{y}\,(y - 3) = -(2y + 1)\).
  3. Square both sides: \(y(y - 3)^{2} = (2y + 1)^{2}\).
  4. \(y^{3} - 6y^{2} + 9y = 4y^{2} + 4y + 1\).
  5. Check: \(\sum \alpha^{2} = (\sum \alpha)^{2} - 2\sum \alpha\beta = 4 - 2(-3) = 10\), the new sum of roots.
Answer: \(y^{3} - 10y^{2} + 5y - 1 = 0\)
Worked example A given substitution
The roots of \(x^{3} - 3x + 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Use the substitution \(y = \dfrac{x}{x - 1}\) to find a cubic equation in \(y\), and hence find \(\dfrac{\alpha}{\alpha - 1} + \dfrac{\beta}{\beta - 1} + \dfrac{\gamma}{\gamma - 1}\).
  1. Make \(x\) the subject: \(yx - y = x\), so \(x(y - 1) = y\) and \(x = \dfrac{y}{y - 1}\).
  2. \(\dfrac{y^{3}}{(y - 1)^{3}} - \dfrac{3y}{y - 1} + 1 = 0\).
  3. Multiply by \((y - 1)^{3}\): \(y^{3} - 3y(y - 1)^{2} + (y - 1)^{3} = 0\).
  4. Expand: \(y^{3} - 3y^{3} + 6y^{2} - 3y + y^{3} - 3y^{2} + 3y - 1 = -y^{3} + 3y^{2} - 1 = 0\).
  5. So \(y^{3} - 3y^{2} + 1 = 0\), whose roots are the three fractions; their sum is \(3\).
Answer: \(y^{3} - 3y^{2} + 1 = 0\), and the sum is \(3\)
Your turn
The roots of \(x^{3} + 3x - 2 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(\alpha^{2}\), \(\beta^{2}\), \(\gamma^{2}\).
Examiner trap

Do not square term by term. \((y\sqrt{y} - 3\sqrt{y})^{2}\) is not \(y^{3} + 9y\): isolate first, then square whole sides, and expand the brackets fully.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

The roots of \(2x^{3} - 5x^{2} + x + 7 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Which equation has roots \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\), \(\dfrac{1}{\gamma}\)?

Quick check+10 XP

To find the equation with roots \(\alpha - 3\), \(\beta - 3\), \(\gamma - 3\) from \(p(x) = 0\), substitute

Quick check+10 XP

The roots of \(x^{3} + 4x^{2} - x + 6 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Which equation has roots \(2\alpha\), \(2\beta\), \(2\gamma\)?

Quick check+10 XP

After putting \(x = \sqrt{y}\) into a cubic to get roots \(\alpha^{2}\), what is the next step?

Section 03

Symmetric functions the quick way.

Once you have the transformed equation, its coefficients give the sums and products of the NEW roots directly. A sum such as \(\sum \dfrac{1}{\alpha^{2}}\) is just the sum of the roots of the equation with roots \(\dfrac{1}{\alpha^{2}}\). Compare this with the direct method, where you rewrite the expression in terms of \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\).

Worked example Sum of reciprocal squares, two ways
The roots of \(x^{3} + 2x^{2} - 3x + 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find \(\dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} + \dfrac{1}{\gamma^{2}}\).
  1. From the previous section, \(y^{3} - 10y^{2} + 5y - 1 = 0\) has roots \(\alpha^{2}\), \(\beta^{2}\), \(\gamma^{2}\).
  2. Reverse the coefficients for the reciprocals: \(-y^{3} + 5y^{2} - 10y + 1 = 0\), that is \(y^{3} - 5y^{2} + 10y - 1 = 0\), with roots \(\dfrac{1}{\alpha^{2}}\), \(\dfrac{1}{\beta^{2}}\), \(\dfrac{1}{\gamma^{2}}\).
  3. Their sum is \(5\).
  4. Direct check: \(\sum \dfrac{1}{\alpha^{2}} = \dfrac{(\sum \alpha\beta)^{2} - 2\alpha\beta\gamma \sum \alpha}{(\alpha\beta\gamma)^{2}} = \dfrac{9 - 2(-1)(-2)}{1} = 5\).
Answer: \(5\)
Worked example A product with y = x squared plus 1
The roots of \(x^{3} - 2x + 4 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Use \(y = x^{2} + 1\) to find a cubic in \(y\), and hence find \((\alpha^{2} + 1)(\beta^{2} + 1)(\gamma^{2} + 1)\).
  1. \(x = \sqrt{y - 1}\), and \(x^{3} = (y - 1)\sqrt{y - 1}\).
  2. \((y - 1)\sqrt{y - 1} - 2\sqrt{y - 1} + 4 = 0\), so \(\sqrt{y - 1}\,(y - 3) = -4\).
  3. Square: \((y - 1)(y - 3)^{2} = 16\), so \(y^{3} - 7y^{2} + 15y - 9 = 16\).
  4. \(y^{3} - 7y^{2} + 15y - 25 = 0\). The product of its roots is \(-\dfrac{-25}{1} = 25\).
  5. Check on the sum: \(\sum \alpha^{2} + 3 = (0 - 2(-2)) + 3 = 7\). Correct.
Answer: \(y^{3} - 7y^{2} + 15y - 25 = 0\); the product is \(25\)
Your turn
Using the equation \(y^{3} - 7y^{2} + 15y - 25 = 0\) above, find \(\dfrac{1}{\alpha^{2} + 1} + \dfrac{1}{\beta^{2} + 1} + \dfrac{1}{\gamma^{2} + 1}\).
Examiner trap

Read the constant term's sign with care. For a cubic \(ay^{3} + by^{2} + cy + d = 0\) the product of roots is \(-\dfrac{d}{a}\), so \(-25\) gives a product of \(+25\), not \(-25\).

Stage 2 · Exam

Exam-style questions

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Exam style+20 XP

The roots of \(x^{3} - 4x^{2} + 2x - 5 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find the equation with roots \(\alpha + 2\), \(\beta + 2\), \(\gamma + 2\).

Exam style+20 XP

The roots of \(x^{3} + 3x - 2 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Which equation has roots \(\alpha^{2}\), \(\beta^{2}\), \(\gamma^{2}\)?

Exam style+20 XP

The roots of \(x^{3} - x^{2} + 2x + 4 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Find \(\dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} + \dfrac{1}{\gamma^{2}}\).

Exam style+20 XP

The roots of \(x^{3} + 2x - 1 = 0\) are \(\alpha\), \(\beta\), \(\gamma\). Using \(y = x^{2} + 1\), which equation has roots \(\alpha^{2} + 1\), \(\beta^{2} + 1\), \(\gamma^{2} + 1\)?

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Six original Cambridge-style questions
Reciprocal and shifted roots, squared roots and \(\sum \alpha^{4}\), the given substitutions \(y = \frac{x}{x + 1}\) and \(y = x^{2} + 1\), and a proof that a cubic has non-real roots.