Figure 1.5: Cubic Curve Roots & Conjugate Complex Pairs in the Argand Plane
1. Fundamental Theorem of Algebra & Conjugate Root Theorem
A general polynomial $P(z)$ of degree $n \ge 1$ with complex coefficients is defined as:
$$P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0, \quad a_n \neq 0$$- Fundamental Theorem of Algebra (FTA): Every non-constant polynomial of degree $n$ has at least one root in $\mathbb{C}$. Consequently, $P(z)$ factors completely into exactly $n$ linear complex factors: $$P(z) = a_n (z - \alpha_1)(z - \alpha_2)\dots(z - \alpha_n)$$ where roots $\alpha_i$ are counted according to their algebraic multiplicities.
- Complex Conjugate Root Theorem: If all coefficients $a_0, a_1, \dots, a_n \in \mathbb{R}$, and $z = u + iv$ ($v \neq 0$) is a root of $P(z) = 0$, then its complex conjugate $z^* = u - iv$ is also a root.
Proof Sketch: Taking the conjugate of the equation: $$\overline{P(z)} = \overline{\sum_{k=0}^n a_k z^k} = \sum_{k=0}^n \overline{a_k} (\bar{z})^k = \sum_{k=0}^n a_k (\bar{z})^k = P(\bar{z})$$ since $a_k \in \mathbb{R} \implies \overline{a_k} = a_k$. If $P(z) = 0$, then $P(\bar{z}) = \overline{0} = 0$. $\blacksquare$
Irreducible Real Factors: The product of the conjugate pair gives a strictly real quadratic factor: $$(z - (u+iv))(z - (u-iv)) = (z - u)^2 - (iv)^2 = z^2 - 2uz + (u^2 + v^2) \in \mathbb{R}[z]$$ Hence, every polynomial with real coefficients factors completely over $\mathbb{R}$ into linear factors and irreducible quadratic factors with $\Delta < 0$. - Odd Degree Implication: Because non-real roots of real polynomials occur in pairs of two, any polynomial with real coefficients of odd degree must have at least one real root.
2. Polynomial Division, Remainder Theorem & Factor Theorem
For any polynomial $P(x)$ and non-zero divisor $D(x)$, there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that:
$$P(x) = D(x) Q(x) + R(x), \quad \text{where } \deg(R) < \deg(D) \text{ or } R(x) = 0$$- Remainder Theorem: When divided by a linear factor $(x - c)$, the remainder is a constant $R = P(c)$. If divided by $(ax - b)$, the remainder is $P\left(\frac{b}{a}\right)$.
- Factor Theorem: $(x - c)$ is a factor of $P(x) \iff P(c) = 0 \iff x = c$ is a root of $P(x) = 0$.
- Repeated Roots and Derivatives: A root $\alpha$ has multiplicity $m \ge 2$ if and only if $P(\alpha) = 0$ and $P'(\alpha) = 0$. For multiplicity $m$, $P(\alpha) = P'(\alpha) = \dots = P^{(m-1)}(\alpha) = 0$ while $P^{(m)}(\alpha) \neq 0$.
3. Vieta’s Formulas (Roots and Coefficients)
By expanding $P(x) = a_n \prod_{i=1}^n (x - \alpha_i) = a_n \left( x^n - e_1 x^{n-1} + e_2 x^{n-2} - \dots + (-1)^n e_n \right)$ and matching coefficients with $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$, we obtain Vieta’s relations for elementary symmetric sums $e_k$:
| Polynomial | Sum of Roots: $\sum \alpha$ | Sum of Products in Pairs: $\sum \alpha\beta$ | Sum of Products in Triples: $\sum \alpha\beta\gamma$ | Product of All Roots |
|---|---|---|---|---|
| Quadratic $ax^2 + bx + c = 0$ |
$\alpha + \beta = -\frac{b}{a}$ | — | — | $\alpha\beta = \frac{c}{a}$ |
| Cubic $ax^3 + bx^2 + cx + d = 0$ |
$\alpha + \beta + \gamma = -\frac{b}{a}$ | $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ | — | $\alpha\beta\gamma = -\frac{d}{a}$ |
| Quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ |
$\sum \alpha = -\frac{b}{a}$ | $\sum \alpha\beta = \frac{c}{a}$ | $\sum \alpha\beta\gamma = -\frac{d}{a}$ | $\alpha\beta\gamma\delta = \frac{e}{a}$ |
4. Key Symmetric Sums & Newton's Sums Recurrence
In IB exam problems, you will evaluate power sums $S_k = \sum \alpha^k$ without solving for the roots explicitly:
- Sum of Squares: $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2 \sum \alpha\beta$
- Sum of Reciprocals: $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{c/a}{-d/a} = -\frac{c}{d}$
- Sum of Reciprocal Squares: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)^2 - 2\left(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}\right) = \left(\frac{\sum \alpha\beta}{\alpha\beta\gamma}\right)^2 - 2\left(\frac{\sum \alpha}{\alpha\beta\gamma}\right)$
- Newton’s Sums (Power Sum Recurrence): For $P(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0 = 0$, every root $\alpha_i$ satisfies $\alpha_i^k + a_{n-1}\alpha_i^{k-1} + \dots = 0$. Summing across all roots gives: $$S_k + a_{n-1} S_{k-1} + a_{n-2} S_{k-2} + \dots + a_{n-k} \cdot k = 0 \quad (\text{for } 1 \le k \le n)$$ and for $k \ge n$: $$S_k + a_{n-1} S_{k-1} + a_{n-2} S_{k-2} + \dots + a_0 S_{k-n} = 0$$
5. Transforming Polynomial Equations
Instead of calculating symmetric sums, you can find a polynomial equation whose roots are simple transformations of the original roots by substitution:
- Reciprocal roots $\frac{1}{\alpha}$: Let $y = \frac{1}{x} \implies x = \frac{1}{y}$. Substitute into $P(x) = 0 \implies P\left(\frac{1}{y}\right) = 0$.
- Shifted roots $\alpha + k$: Let $y = x + k \implies x = y - k$. Substitute $P(y - k) = 0$.
- Scaled roots $k\alpha$: Let $y = kx \implies x = \frac{y}{k}$. Substitute $P\left(\frac{y}{k}\right) = 0$.
- Squared roots $\alpha^2$: Let $y = x^2 \implies x = \sqrt{y}$. Rearrange $P(x)$ to isolate odd powers on one side, square both sides, and substitute $x^2 = y$.
Worked Examples
Worked Example 1: Factorization of Real Quartic with Complex Roots
Question: A quartic polynomial with real coefficients is given by $P(z) = z^4 - 6z^3 + 23z^2 - 34z + 26$.
Given that $z_1 = 1 + i$ is a zero of $P(z)$:
(a) Write down another zero $z_2$ of $P(z)$. [1 mark]
(b) Find the quadratic factor corresponding to these two zeros. [2 marks]
(c) Hence find all four zeros of $P(z)$ in the form $a \pm bi$. [4 marks]
Solution:
(a) Complex Conjugate Root: Because all coefficients of $P(z)$ are real, non-real roots occur in complex conjugate pairs. Thus, $z_2 = z_1^* = 1 - i$. [A1]
(b) Real Quadratic Factor: [M1]
$$(z - (1 + i))(z - (1 - i)) = ((z - 1) - i)((z - 1) + i) = (z - 1)^2 - i^2 = z^2 - 2z + 1 - (-1) = z^2 - 2z + 2 \quad [\text{A1}]$$(c) Finding Remaining Zeros: Perform polynomial long division of $P(z)$ by $(z^2 - 2z + 2)$: [M1]
$$z^4 - 6z^3 + 23z^2 - 34z + 26 = (z^2 - 2z + 2)(z^2 - 4z + 13) \quad [\text{A1}]$$Set the quadratic quotient to zero: $z^2 - 4z + 13 = 0$. [M1]
$$z = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(13)}}{2} = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i \quad [\text{A1}]$$The four zeros of $P(z)$ are: $1 + i, \; 1 - i, \; 2 + 3i, \; 2 - 3i$.
Worked Example 2: Symmetric Sums & Factored Products
Question: The roots of the cubic equation $2x^3 - 8x^2 + 6x - 5 = 0$ are $\alpha, \beta, \gamma$. Without solving the equation, determine the exact value of:
(a) $\alpha^2 + \beta^2 + \gamma^2$ [3 marks]
(b) $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$ [2 marks]
(c) $(2 - \alpha)(2 - \beta)(2 - \gamma)$ [3 marks]
Solution:
Identify Vieta's relations with $a = 2, b = -8, c = 6, d = -5$: [M1]
$$\sum \alpha = -\frac{-8}{2} = 4, \quad \sum \alpha\beta = \frac{6}{2} = 3, \quad \alpha\beta\gamma = -\frac{-5}{2} = \frac{5}{2} \quad [\text{A1}]$$(a) Sum of Squares: [M1]
$$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 4^2 - 2(3) = 16 - 6 = 10 \quad [\text{A1}]$$(b) Sum of Reciprocals: [M1]
$$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{3}{5/2} = \frac{6}{5} \quad [\text{A1}]$$(c) Factored Product:
Method 1 (Polynomial evaluation): Notice that $P(x) = 2(x - \alpha)(x - \beta)(x - \gamma)$. [M1]
Therefore, $(2 - \alpha)(2 - \beta)(2 - \gamma) = \frac{P(2)}{2}$. [A1]
Evaluate $P(2) = 2(2)^3 - 8(2)^2 + 6(2) - 5 = 16 - 32 + 12 - 5 = -9$.
Hence $(2 - \alpha)(2 - \beta)(2 - \gamma) = \frac{-9}{2} = -4.5$. [A1]
Method 2 (Expansion): $8 - 4(\alpha+\beta+\gamma) + 2(\sum \alpha\beta) - \alpha\beta\gamma = 8 - 4(4) + 2(3) - \frac{5}{2} = 8 - 16 + 6 - 2.5 = -4.5$.
Worked Example 3: Roots in Geometric Progression
Question: The cubic polynomial $x^3 - 14x^2 + 56x - 64 = 0$ has three positive roots $\alpha, \beta, \gamma$ which form a geometric progression.
(a) Find all three roots. [5 marks]
(b) Verify that the sum of pairwise products equals $56$. [2 marks]
Solution:
(a) Finding the Roots: Let the roots in GP be $\frac{k}{r}, k, kr$ with $r > 0$. [M1]
By Vieta's third formula for the product of roots: [M1]
By Vieta's first formula for the sum of roots: [M1]
$$\frac{4}{r} + 4 + 4r = 14 \implies \frac{4}{r} + 4r = 10 \implies 4 + 4r^2 = 10r \implies 2r^2 - 5r + 2 = 0$$ $$(2r - 1)(r - 2) = 0 \implies r = 2 \text{ or } r = \frac{1}{2} \quad [\text{A1}]$$For $r = 2$, roots are $\frac{4}{2} = 2, 4, 4(2) = 8$. (For $r = 1/2$, roots are $8, 4, 2$, which is identical).
The three roots are $2, 4, 8$. [A1]
(b) Verification: [M1]
$$\sum \alpha\beta = (2)(4) + (4)(8) + (2)(8) = 8 + 32 + 16 = 56 \quad [\text{AG}] [\text{A1}]$$Worked Example 4: Polynomial Transformation
Question: The roots of the cubic equation $x^3 - 3x^2 + 4x - 2 = 0$ are $\alpha, \beta, \gamma$.
Find a cubic equation with integer coefficients whose roots are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$. [4 marks]
Solution:
Let $y = \frac{1}{x} \implies x = \frac{1}{y}$. [M1]
Substitute $x = \frac{1}{y}$ into the original polynomial equation: [M1]
Multiply through by $y^3$ (since $y \neq 0$): [M1]
$$1 - 3y + 4y^2 - 2y^3 = 0$$Multiply by $-1$ to make the leading coefficient positive: [A1]
$$2y^3 - 4y^2 + 3y - 1 = 0$$Check using Vieta: For new equation, sum of roots $= \frac{4}{2} = 2$.
Original $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\sum \alpha\beta}{\alpha\beta\gamma} = \frac{4}{2} = 2$. Verified!
Worked Example 5: Trigonometric Roots & Chebyshev Product
Question:
(a) Use the identity $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$ to show that $x_1 = \cos\left(\frac{\pi}{9}\right)$, $x_2 = \cos\left(\frac{7\pi}{9}\right)$, and $x_3 = \cos\left(\frac{13\pi}{9}\right)$ are roots of $8x^3 - 6x - 1 = 0$. [4 marks]
(b) Hence evaluate the exact value of $\cos\left(\frac{\pi}{9}\right)\cos\left(\frac{7\pi}{9}\right)\cos\left(\frac{13\pi}{9}\right)$. [2 marks]
Solution:
(a) Linking to Triple Angle Identity: Let $x = \cos\theta$. [M1]
Substitute into $8x^3 - 6x - 1 = 0$: [M1]
Solve $\cos(3\theta) = \frac{1}{2}$ in the general interval: [M1]
$$3\theta = \frac{\pi}{3}, \; 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}, \; 2\pi + \frac{\pi}{3} = \frac{7\pi}{3}, \; 4\pi - \frac{\pi}{3} = \frac{11\pi}{3}, \; 4\pi + \frac{\pi}{3} = \frac{13\pi}{3}$$Dividing by 3 gives distinct angles: $\theta = \frac{\pi}{9}, \; \frac{7\pi}{9}, \; \frac{13\pi}{9}$. [A1]
Since a cubic polynomial has at most three distinct roots, these three distinct cosine values are precisely the three roots of $8x^3 - 6x - 1 = 0$. [AG]
(b) Product of Roots via Vieta: By Vieta's formula for the cubic $8x^3 + 0x^2 - 6x - 1 = 0$ with $a = 8, d = -1$: [M1]
$$\cos\left(\frac{\pi}{9}\right)\cos\left(\frac{7\pi}{9}\right)\cos\left(\frac{13\pi}{9}\right) = -\frac{d}{a} = -\frac{-1}{8} = \frac{1}{8} \quad [\text{A1}]$$Practice Exercises (10 Tiered Questions)
Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.
When $P(x) = 2x^3 + ax^2 - 5x + 6$ is divided by $(x - 2)$, the remainder is 16. Find the value of $a$, and hence determine whether $(x + 3)$ is a factor of $P(x)$.
Show Full Solution
By Remainder Theorem: $P(2) = 16$. [M1]
$2(2)^3 + a(2)^2 - 5(2) + 6 = 16 \implies 16 + 4a - 10 + 6 = 16 \implies 4a + 12 = 16 \implies 4a = 4 \implies a = 1$. [A1]
Now $P(x) = 2x^3 + x^2 - 5x + 6$.
Test $(x + 3)$: $P(-3) = 2(-27) + (9) - 5(-3) + 6 = -54 + 9 + 15 + 6 = -24 \neq 0$. [M1]
Since $P(-3) \neq 0$, $(x + 3)$ is not a factor of $P(x)$. [A1]
The quadratic $3x^2 - 12x + 7 = 0$ has roots $\alpha$ and $\beta$. Find the value of $\alpha^2 + \beta^2$.
Show Full Solution
From Vieta: $\alpha + \beta = -\frac{-12}{3} = 4$, and $\alpha\beta = \frac{7}{3}$. [M1]
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4^2 - 2\left(\frac{7}{3}\right) = 16 - \frac{14}{3} = \frac{34}{3}$. [A1]
A polynomial with real coefficients has degree 4. Two of its roots are $2 + 3i$ and $-4i$. State the remaining two roots.
Show Full Solution
By the Complex Conjugate Root Theorem, because all coefficients are real, non-real roots must occur in conjugate pairs. [R1]
• The conjugate of $2 + 3i$ is $2 - 3i$. [A1]
• The conjugate of $-4i = 0 - 4i$ is $4i = 0 + 4i$. [A1]
Remaining two roots: $2 - 3i$ and $4i$.
The roots of $x^3 - 6x^2 + 11x - 6 = 0$ are $\alpha, \beta, \gamma$. Calculate $\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}$.
Show Full Solution
Combine over a common denominator: [M1]
$\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} = \frac{\gamma + \alpha + \beta}{\alpha\beta\gamma} = \frac{\alpha + \beta + \gamma}{\alpha\beta\gamma}$. [A1]
From Vieta: $\alpha + \beta + \gamma = 6$, and $\alpha\beta\gamma = -\frac{-6}{1} = 6$. [M1]
Result: $\frac{6}{6} = 1$. [A1]
The roots of $x^3 - 9x^2 + 23x - 15 = 0$ form an arithmetic progression. Find all three roots.
Show Full Solution
Let the roots in AP be $k - d, k, k + d$. [M1]
By Vieta for sum: $(k - d) + k + (k + d) = 3k = 9 \implies k = 3$. [A1]
By Vieta for product: $(3 - d)(3)(3 + d) = 3(9 - d^2) = 15 \implies 9 - d^2 = 5 \implies d^2 = 4 \implies d = \pm 2$. [M1]
For $d = 2$, roots are $3 - 2 = 1, 3, 3 + 2 = 5$. [A1]
Roots are $1, 3, 5$.
The roots of $x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$. Find a quadratic equation with integer coefficients whose roots are $2\alpha + 1$ and $2\beta + 1$.
Show Full Solution
$\alpha + \beta = 5$, $\alpha\beta = 3$. [M1]
Sum of new roots: $(2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2(5) + 2 = 12$. [A1]
Product of new roots: $(2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 4(3) + 2(5) + 1 = 12 + 10 + 1 = 23$. [A1]
Equation: $x^2 - 12x + 23 = 0$. [A1]
The quartic polynomial $P(z) = z^4 - 2z^3 + 6z^2 - 8z + 8$ has real coefficients. Given that $z = 2i$ is a root, factorize $P(z)$ completely into four linear factors over $\mathbb{C}$.
Show Full Solution
Since coefficients are real, $z = -2i$ is also a root. [R1]
Quadratic factor: $(z - 2i)(z + 2i) = z^2 + 4$. [M1]
Polynomial division of $P(z)$ by $(z^2 + 4)$: [M1]
$z^4 - 2z^3 + 6z^2 - 8z + 8 = (z^2 + 4)(z^2 - 2z + 2)$. [A1]
Solve $z^2 - 2z + 2 = 0 \implies z = \frac{2 \pm \sqrt{4 - 8}}{2} = 1 \pm i$. [A1]
Linear factorization: $P(z) = (z - 2i)(z + 2i)(z - (1+i))(z - (1-i))$. [A1]
The roots of the quartic $x^4 - 8x^3 + 14x^2 + 8x - 15 = 0$ satisfy the relation $\alpha + \beta = \gamma + \delta$. Find all four roots.
Show Full Solution
By Vieta for sum: $(\alpha + \beta) + (\gamma + \delta) = 8$.
Since $\alpha + \beta = \gamma + \delta$, both must equal $4$: $\alpha + \beta = 4$ and $\gamma + \delta = 4$. [M1]
Let $P(x) = (x^2 - 4x + p)(x^2 - 4x + q) = 0$. [M1]
Expand: $x^4 - 8x^3 + (p + q + 16)x^2 - 4(p + q)x + pq$. [M1]
Match coefficients: $p + q + 16 = 14 \implies p + q = -2$, and $pq = -15$.
Roots of $t^2 + 2t - 15 = 0$ are $3$ and $-5$.
Factors: $(x^2 - 4x + 3)(x^2 - 4x - 5) = 0$. [A1]
• $x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0 \implies x = 1, 3$.
• $x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0 \implies x = 5, -1$.
The four roots are $-1, 1, 3, 5$. [A1]
Let $\alpha, \beta, \gamma$ be the roots of $x^3 - 3x + 1 = 0$. Using the identity $S_n = \alpha^n + \beta^n + \gamma^n$ where $S_n - 3S_{n-2} + S_{n-3} = 0$, calculate $S_4 = \alpha^4 + \beta^4 + \gamma^4$.
Show Full Solution
From the equation $x^3 = 3x - 1$, multiplying by $x^{n-3}$ gives $x^n = 3x^{n-2} - x^{n-3}$, so $S_n = 3S_{n-2} - S_{n-3}$. [M1]
• $S_0 = 1 + 1 + 1 = 3$.
• $S_1 = \alpha + \beta + \gamma = 0$ (no $x^2$ term). [A1]
• $S_2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0^2 - 2(-3) = 6$. [A1]
• $S_3 = 3S_1 - S_0 = 3(0) - 3 = -3$. [A1]
• $S_4 = 3S_2 - S_1 = 3(6) - 0 = 18$. [A1]
Show that the roots of $8x^3 - 6x - 1 = 0$ are $\cos\left(\frac{\pi}{9}\right)$, $\cos\left(\frac{7\pi}{9}\right)$, and $\cos\left(\frac{13\pi}{9}\right)$, and deduce the exact value of $\cos\left(\frac{\pi}{9}\right)\cos\left(\frac{7\pi}{9}\right)\cos\left(\frac{13\pi}{9}\right)$.
Show Full Solution
Recall the triple angle identity: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. [M1]
Let $x = \cos\theta$. Then $8x^3 - 6x = 2(4x^3 - 3x) = 2\cos(3\theta)$.
The equation becomes $2\cos(3\theta) - 1 = 0 \implies \cos(3\theta) = \frac{1}{2}$. [M1]
Solutions for $3\theta$: $3\theta = \frac{\pi}{3}, \frac{7\pi}{3}, \frac{13\pi}{3} \implies \theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}$. [A1]
Since the cubic has degree 3, these 3 distinct cosine values are all three roots of $8x^3 - 6x - 1 = 0$.
By Vieta's formula for the product of roots: [M1]
$\cos\left(\frac{\pi}{9}\right)\cos\left(\frac{7\pi}{9}\right)\cos\left(\frac{13\pi}{9}\right) = -\frac{-1}{8} = \frac{1}{8}$. [A1]
Formula Booklet References
Syllabus AHL 1.16: Polynomials & Vieta’s Relations
For polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$ with roots $\alpha_1, \alpha_2, \dots, \alpha_n$:
$$\sum \alpha_i = -\frac{a_{n-1}}{a_n}, \quad \sum_{i < j} \alpha_i \alpha_j = \frac{a_{n-2}}{a_n}, \quad \dots, \quad \prod_{i=1}^n \alpha_i = (-1)^n \frac{a_0}{a_n}$$Key Formulas & Identities
1. Complex Conjugate Root Theorem
If coefficients are real, any non-real root $z = a + bi$ must be paired with $z^* = a - bi$. Any real polynomial of odd degree must have at least one real root.
2. Useful Symmetric Root Expansions
- $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum\alpha\beta$
- $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\sum\alpha\beta}{\alpha\beta\gamma} = -\frac{c}{d}$
- $(\alpha + 1)(\beta + 1)(\gamma + 1) = 1 + \sum\alpha + \sum\alpha\beta + \alpha\beta\gamma = -P(-1) \text{ (monic)}$
3. Newton's Sums Recurrence Formula
For monic cubic $x^3 + px^2 + qx + r = 0$ with power sums $S_k = \alpha^k + \beta^k + \gamma^k$:
$$S_1 + p = 0 \implies S_1 = -p$$ $$S_2 + p S_1 + 2q = 0 \implies S_2 = -p S_1 - 2q$$ $$S_3 + p S_2 + q S_1 + 3r = 0 \implies S_3 = -p S_2 - q S_1 - 3r$$ $$S_k + p S_{k-1} + q S_{k-2} + r S_{k-3} = 0 \quad (\text{for } k \ge 4)$$Common Exam Pitfalls
Sign Flips in Vieta’s Product
Remember that the signs alternate: $\sum \alpha_i$ has a negative sign, $\sum \alpha_i \alpha_j$ is positive, and the product is $(-1)^n \frac{a_0}{a_n}$. For a cubic ($n=3$), product is $-\frac{d}{a}$, while for a quartic ($n=4$), product is $+\frac{e}{a}$.
Conjugate Theorem Requires REAL Coefficients
The conjugate root theorem does NOT apply if any coefficient is complex! For instance, $z^2 - iz = 0$ has roots $0$ and $i$, which are not conjugates because the coefficient $-i$ is not real.
Division Algorithm with Quadratic Remainder
When dividing $P(x)$ by a quadratic $(x-a)(x-b)$, the remainder must be written as a linear expression $R(x) = Ax + B$. Do NOT assume the remainder is a numerical constant.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
If $P(x) = x^3 - 5x^2 + kx - 12$ has roots that multiply to 12, what is the value of the sum of the roots?
A polynomial $Q(x)$ with real coefficients has degree 5. What is the maximum number of non-real roots it can possess?
If the roots of $2x^3 - 6x^2 + 4x - 1 = 0$ are $\alpha, \beta, \gamma$, what is $\alpha^2 + \beta^2 + \gamma^2$?
If $(x - 3)$ is a factor of $x^3 + ax^2 - 5x + 6$, what is the value of $a$?
Section B: Examination-Style Free Response Questions
The cubic polynomial $P(x) = x^3 - 7x^2 + kx - 8$ has three positive real roots $\alpha, \beta, \gamma$ that form a geometric progression.
(a) Show that the middle term of the geometric progression is 2. [2 marks]
(b) Hence determine the common ratio of the progression and find all three roots. [3 marks]
(c) Find the value of $k$. [1 mark]
Show Full Markscheme
(a) Middle root value: [2 marks]
Let the roots be $\frac{b}{r}, b, br$. [M1]
By Vieta's formula for the product of roots:
$\left(\frac{b}{r}\right)(b)(br) = b^3 = -\frac{-8}{1} = 8 \implies b = 2$ [A1].
(b) Finding the roots: [3 marks]
By Vieta's formula for the sum of roots:
$\frac{2}{r} + 2 + 2r = 7 \implies \frac{2}{r} + 2r = 5$ [M1]
Multiply by $r$: $2r^2 - 5r + 2 = 0 \implies (2r - 1)(r - 2) = 0 \implies r = 2$ or $r = \frac{1}{2}$ [A1].
Both ratios yield the same set of roots: $1, 2, 4$ [A1].
(c) Finding $k$: [1 mark]
$k = \sum \alpha\beta = (1)(2) + (2)(4) + (1)(4) = 2 + 8 + 4 = 14$ [A1].
Consider the polynomial $P(z) = z^4 - 2z^3 + 6z^2 - 8z + 8$.
(a) Given that $z = 2i$ is a root of $P(z) = 0$, write down another imaginary root. [1 mark]
(b) Find the quadratic factor of $P(z)$ corresponding to these imaginary roots. [2 marks]
(c) Factorize $P(z)$ completely over the complex field $\mathbb{C}$, and sketch the four roots on an Argand diagram. [4 marks]
Show Full Markscheme
(a) Conjugate root: [1 mark]
Since all coefficients of $P(z)$ are real, $z = -2i$ is another root [A1].
(b) Quadratic factor: [2 marks]
$(z - 2i)(z + 2i) = z^2 - (2i)^2 = z^2 + 4$ [M1][A1].
(c) Complete factorization & Argand sketch: [4 marks]
Perform polynomial division of $P(z)$ by $(z^2 + 4)$: [M1]
$z^4 - 2z^3 + 6z^2 - 8z + 8 = (z^2 + 4)(z^2 - 2z + 2)$ [A1].
Solve $z^2 - 2z + 2 = 0$:
$z = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i$ [A1].
The four roots are $2i, -2i, 1 + i, 1 - i$.
Argand diagram: Points plotted at $(0, 2), (0, -2), (1, 1), (1, -1)$ with symmetric reflection across the real axis [A1].