Hunting for fraction roots
Tonight's idea
Hunting for the roots of a cubic by trial looks endless. With whole-number coefficients it is not: only a short list of fractions can possibly work.
Rational root theorem. Suppose \(a_n x^n + \dots + a_1 x + a_0 = 0\) has whole-number coefficients and a root \(\tfrac{p}{q}\), a fraction in lowest terms. Then \(p\) divides \(a_0\) and \(q\) divides \(a_n\).
Proof, for a cubic \(ax^3 + bx^2 + cx + d = 0\). Put \(x = \tfrac{p}{q}\) and multiply through by \(q^3\):
\[ap^3 + bp^2q + cpq^2 + dq^3 = 0.\]
Every term except the last contains \(p\), so \(p\) divides \(dq^3\). But \(p\) and \(q\) share no factor, so \(p\) divides \(d\). In the same way every term except the first contains \(q\), so \(q\) divides \(ap^3\), and so \(q\) divides \(a\). Other degrees work the same way. \(\square\)
Two things follow. The list is finite: for \(2x^3 + \dots + 3\) the only candidates are \(\pm 1\), \(\pm 3\), \(\pm\tfrac12\), \(\pm\tfrac32\). And if the leading coefficient is 1, every rational root is a whole number, so any other real root is irrational. That proves irrationality without the contradiction argument from grade 10.
Watch one: \(\sqrt2 + \sqrt3\) is irrational
First find a polynomial with this number as a root. Let \(x = \sqrt2 + \sqrt3\). Squaring, \(x^2 = 5 + 2\sqrt6\), so \(x^2 - 5 = 2\sqrt6\). Square again:
\[x^4 - 10x^2 + 25 = 24, \quad\text{so}\quad x^4 - 10x^2 + 1 = 0.\]
The leading coefficient is 1 and the constant is 1, so the only candidates are \(1\) and \(-1\). Both give \(1 - 10 + 1 = -8\), not 0. So this polynomial has no rational root at all. But \(\sqrt2 + \sqrt3\) is a root. So it is not rational.
Your turn
- List every candidate for a rational root of \(2x^3 + 3x^2 - 8x + 3 = 0\). Test them and find all three roots.
- \(3x^3 - 2x^2 - 6x + 4 = 0\) has exactly one rational root. Find it, factorise, and find the other two roots.
- Use the theorem to show that \(\sqrt[3]{2}\) is irrational. Then do the same for the golden ratio \(\tfrac{1 + \sqrt5}{2}\), which is a root of \(x^2 - x - 1 = 0\).
- The formula \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\) with \(\theta = 20°\) shows that \(x = \cos 20°\) is a root of \(8x^3 - 6x - 1 = 0\). Explain why, then prove that \(\cos 20°\) is irrational. (Substituting \(y = 2x\) first makes the list of candidates much shorter.)
Stretch: working backwards
For which whole numbers \(c\), positive or negative, does \(x^3 + cx + 2 = 0\) have a rational root? There are exactly three such values. For every other \(c\), all the real roots are irrational.
Answers
- \(p\) divides 3 and \(q\) divides 2: candidates \(\pm1\), \(\pm3\), \(\pm\tfrac12\), \(\pm\tfrac32\). \(x = 1\) gives \(2 + 3 - 8 + 3 = 0\). Dividing, \((x - 1)(2x^2 + 5x - 3) = (x - 1)(2x - 1)(x + 3)\). Roots \(1\), \(\tfrac12\), \(-3\).
- Candidates: \(\pm1\), \(\pm2\), \(\pm4\), \(\pm\tfrac13\), \(\pm\tfrac23\), \(\pm\tfrac43\). \(x = \tfrac23\) gives \(\tfrac89 - \tfrac89 - 4 + 4 = 0\). Then \(3x^3 - 2x^2 - 6x + 4 = (3x - 2)(x^2 - 2)\), so the other roots are \(\pm\sqrt2\).
- \(\sqrt[3]{2}\) is a root of \(x^3 - 2 = 0\). Leading coefficient 1, so a rational root would be a whole number dividing 2: \(\pm1\), \(\pm2\). These give \(-1\), \(-3\), \(6\), \(-10\), never 0. So \(\sqrt[3]{2}\) is irrational. For \(x^2 - x - 1\) the candidates are \(\pm1\), giving \(-1\) and \(1\). No rational roots, so the golden ratio is irrational.
- \(\cos 60° = \tfrac12\), so \(\tfrac12 = 4x^3 - 3x\); doubling gives \(8x^3 - 6x - 1 = 0\). Put \(y = 2x\): \(y^3 - 3y - 1 = 0\). Leading coefficient 1 and constant \(-1\), so the candidates are \(y = \pm1\), giving \(-3\) and \(1\). No rational \(y\), so \(x = \tfrac{y}{2}\) is not rational either. (Directly: the eight candidates \(\pm1\), \(\pm\tfrac12\), \(\pm\tfrac14\), \(\pm\tfrac18\) all fail.)
Stretch. A rational root must be \(\pm1\) or \(\pm2\). \(x = 1\): \(3 + c = 0\), \(c = -3\). \(x = -1\): \(1 - c = 0\), \(c = 1\). \(x = 2\): \(10 + 2c = 0\), \(c = -5\). \(x = -2\): \(-6 - 2c = 0\), \(c = -3\) again. So \(c = 1\), \(-3\) or \(-5\).