A-Level Further Mathematics – Chapter 7: Roots of Polynomials Further Pure
Contents
The relationship between a polynomial's roots and its coefficients is one of the most elegant results in algebra. Named after François Viète (Latinised as Vieta), these identities allow us to deduce properties of a polynomial's roots without ever solving the equation explicitly. In this chapter we develop Vieta's formulas for quadratics, cubics, and quartics, learn how to form equations with transformed roots, and explore the role of complex roots in polynomials with real coefficients.
7.1 Roots and Coefficients of Quadratics
Consider the general quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$. Since we can write the polynomial in factored form as $a(x - \alpha)(x - \beta)$, expanding and comparing coefficients immediately gives us Vieta's formulas.
Theorem 7.1 — Vieta's Formulas for a Quadratic
If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, then:
$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$
Proof. Write $ax^2+bx+c = a(x-\alpha)(x-\beta) = a\bigl(x^2 - (\alpha+\beta)x + \alpha\beta\bigr)$. Comparing coefficients of $x$ and the constant term gives the result.
Definition — Symmetric Functions of Roots
A symmetric function of $\alpha$ and $\beta$ is one whose value is unchanged when $\alpha$ and $\beta$ are swapped. All symmetric functions can be expressed in terms of $\alpha+\beta$ and $\alpha\beta$. Common examples include:
- $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
- $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$
- $\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha+\beta)$
- $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$
Example 7.1.1 — Finding symmetric functions
The roots of $2x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$. Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
Step 1 Read off Vieta's formulas: $\alpha+\beta = \tfrac{5}{2}$, $\alpha\beta = \tfrac{3}{2}$.
Step 2 $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \tfrac{25}{4} - 3 = \tfrac{13}{4}$.
Step 3 $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = \tfrac{125}{8} - 3\cdot\tfrac{3}{2}\cdot\tfrac{5}{2} = \tfrac{125}{8} - \tfrac{45}{4} = \tfrac{35}{8}$.
Example 7.1.2 — Forming a quadratic from symmetric functions
Given that $\alpha + \beta = 4$ and $\alpha^2 + \beta^2 = 10$, find a quadratic with integer coefficients whose roots are $\alpha$ and $\beta$.
Step 1 Find $\alpha\beta$: $\alpha\beta = \tfrac{(\alpha+\beta)^2 - (\alpha^2+\beta^2)}{2} = \tfrac{16-10}{2} = 3$.
Step 2 The quadratic is $x^2 - (\alpha+\beta)x + \alpha\beta = 0$, i.e. $x^2 - 4x + 3 = 0$.
Example 7.1.3 — Condition for real roots
The roots of $x^2 - px + (p-1) = 0$ are $\alpha$ and $\beta$. Show that $(\alpha - \beta)^2 = p^2 - 4p + 4 = (p-2)^2 \ge 0$, and state when the roots are equal.
$\alpha+\beta = p$, $\alpha\beta = p-1$. Then $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = p^2 - 4(p-1) = p^2 - 4p + 4 = (p-2)^2$. This is always $\ge 0$, so the roots are always real. They are equal when $p = 2$, giving the repeated root $x = 1$.
Example 7.1.4 — Finding $\alpha^4 + \beta^4$
The roots of $x^2 - 3x + 1 = 0$ are $\alpha$ and $\beta$. Find $\alpha^4 + \beta^4$.
$\alpha+\beta=3$, $\alpha\beta=1$. $\alpha^2+\beta^2=9-2=7$. $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2=49-2=47$.
Exam Tip
Always state $\alpha+\beta$ and $\alpha\beta$ at the start of any roots-of-polynomials question — examiners award method marks for this even if subsequent working contains an error. Be careful with signs: $\alpha+\beta = -b/a$, not $+b/a$.
7.2 Roots and Coefficients of Cubics
For a cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$, we can write $ax^3+bx^2+cx+d = a(x-\alpha)(x-\beta)(x-\gamma)$ and expand to obtain three symmetric relations.
Theorem 7.2 — Vieta's Formulas for a Cubic
If $\alpha, \beta, \gamma$ are the roots of $ax^3 + bx^2 + cx + d = 0$, then:
$$\alpha+\beta+\gamma = -\frac{b}{a}$$ $$\alpha\beta+\alpha\gamma+\beta\gamma = \frac{c}{a}$$ $$\alpha\beta\gamma = -\frac{d}{a}$$
Newton's Identity for Cubics
Let $p_k = \alpha^k + \beta^k + \gamma^k$ (called the power sums). Then:
$$p_2 = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\alpha\gamma+\beta\gamma)$$
$$p_3 = \alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)^3 - 3(\alpha\beta+\alpha\gamma+\beta\gamma)(\alpha+\beta+\gamma) + 3\alpha\beta\gamma$$
Example 7.2.1 — Reading off symmetric functions of a cubic
The roots of $x^3 - 6x^2 + 11x - 6 = 0$ are $\alpha, \beta, \gamma$. Find $\alpha+\beta+\gamma$, $\alpha\beta+\alpha\gamma+\beta\gamma$, $\alpha\beta\gamma$, and $\alpha^2+\beta^2+\gamma^2$.
Here $a=1, b=-6, c=11, d=-6$, so: $\alpha+\beta+\gamma = 6$, $\alpha\beta+\alpha\gamma+\beta\gamma = 11$, $\alpha\beta\gamma = 6$.
$\alpha^2+\beta^2+\gamma^2 = 36 - 2(11) = 14$.
Example 7.2.2 — Using Newton's identity for $p_3$
For the cubic $2x^3 + 3x^2 - x - 4 = 0$ with roots $\alpha, \beta, \gamma$, find $\alpha^3 + \beta^3 + \gamma^3$.
$\alpha+\beta+\gamma = -\tfrac{3}{2}$, $\quad \alpha\beta+\alpha\gamma+\beta\gamma = -\tfrac{1}{2}$, $\quad \alpha\beta\gamma = 2$.
$p_3 = \bigl(-\tfrac{3}{2}\bigr)^3 - 3\bigl(-\tfrac{1}{2}\bigr)\bigl(-\tfrac{3}{2}\bigr) + 3(2) = -\tfrac{27}{8} - \tfrac{9}{4} + 6 = -\tfrac{27}{8} - \tfrac{18}{8} + \tfrac{48}{8} = \tfrac{3}{8}$.
Example 7.2.3 — Forming a cubic from given symmetric functions
A cubic has roots $\alpha, \beta, \gamma$ such that $\alpha+\beta+\gamma = 2$, $\alpha\beta+\alpha\gamma+\beta\gamma = -5$, and $\alpha\beta\gamma = -6$. Write down the cubic equation with leading coefficient 1.
$$x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\alpha\gamma+\beta\gamma)x - \alpha\beta\gamma = 0$$ $$x^3 - 2x^2 - 5x + 6 = 0$$
Example 7.2.4 — Finding $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$
For $x^3 + 4x^2 - 2x + 8 = 0$ with roots $\alpha, \beta, \gamma$, find $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{-2}{-8} = \frac{1}{4}$.
Exam Tip
Expressions such as $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ or $\alpha^2\beta+\alpha^2\gamma+\beta^2\alpha+\beta^2\gamma+\gamma^2\alpha+\gamma^2\beta$ appear frequently. Both simplify neatly in terms of the elementary symmetric polynomials — always try to express the sum using $e_1, e_2, e_3$ before computing numerically.
7.3 Roots and Coefficients of Quartics
For a quartic $ax^4 + bx^3 + cx^2 + dx + e = 0$ with roots $\alpha, \beta, \gamma, \delta$, there are four elementary symmetric polynomials.
Theorem 7.3 — Vieta's Formulas for a Quartic
$$e_1 = \alpha+\beta+\gamma+\delta = -\frac{b}{a}$$ $$e_2 = \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta = \frac{c}{a}$$ $$e_3 = \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = -\frac{d}{a}$$ $$e_4 = \alpha\beta\gamma\delta = \frac{e}{a}$$
Example 7.3.1 — Symmetric functions of a quartic
The roots of $x^4 - 5x^2 + 4 = 0$ are $\alpha, \beta, \gamma, \delta$. Find $e_1, e_2, e_3, e_4$.
Rewrite as $x^4 + 0 \cdot x^3 - 5x^2 + 0 \cdot x + 4$. Hence $e_1 = 0$, $e_2 = -5$, $e_3 = 0$, $e_4 = 4$.
Example 7.3.2 — Computing $\sum \alpha^2$ for a quartic
For the quartic in Example 7.3.1, find $\alpha^2+\beta^2+\gamma^2+\delta^2$.
$\alpha^2+\beta^2+\gamma^2+\delta^2 = e_1^2 - 2e_2 = 0 - 2(-5) = 10$.
Check: The roots are $1, -1, 2, -2$, so $1+1+4+4 = 10$. ✓
Example 7.3.3 — A quartic with two repeated roots
A quartic has roots $1, 1, 3, 3$. Write down the monic quartic equation.
$e_1 = 8$, $e_2 = 1\cdot1+1\cdot3+1\cdot3+1\cdot3+1\cdot3+3\cdot3 = 1+3+3+3+3+9=22$, $e_3 = 1\cdot1\cdot3+1\cdot1\cdot3+1\cdot3\cdot3+1\cdot3\cdot3=3+3+9+9=24$, $e_4 = 9$.
Equation: $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$.
Alternatively: $(x-1)^2(x-3)^2 = [(x-1)(x-3)]^2 = (x^2-4x+3)^2 = x^4-8x^3+22x^2-24x+9$.
Example 7.3.4 — Finding $\sum \alpha^3$ for a quartic
For $x^4 - 2x^3 + x^2 - 3x + 1 = 0$ with roots $\alpha, \beta, \gamma, \delta$, find $\alpha^3+\beta^3+\gamma^3+\delta^3$.
$e_1=2,\ e_2=1,\ e_3=3,\ e_4=1$.
$p_2 = e_1^2 - 2e_2 = 4-2=2$.
Using Newton's identity: $p_3 = e_1 p_2 - e_2 p_1 + 3e_3 = 2\cdot2 - 1\cdot2 + 3\cdot3 = 4-2+9 = 11$.
7.4 Forming New Equations with Transformed Roots
A common exam task is: given an equation with roots $\alpha, \beta, \gamma$, find an equation whose roots are, say, $2\alpha, 2\beta, 2\gamma$ or $\alpha^2, \beta^2, \gamma^2$. The substitution method is the most reliable approach.
The Substitution Method
Suppose the original equation is $f(x) = 0$ with roots $\alpha, \beta, \gamma$. To find an equation with roots $g(\alpha), g(\beta), g(\gamma)$:
- Let $y = g(x)$, and express $x$ in terms of $y$.
- Substitute this expression for $x$ into $f(x) = 0$.
- Simplify to obtain an equation in $y$.
Example 7.4.1 — Roots multiplied by a constant
The equation $x^3 - 6x^2 + 11x - 6 = 0$ has roots $\alpha, \beta, \gamma$. Find the equation with roots $2\alpha, 2\beta, 2\gamma$.
Step 1 Let $y = 2x$, so $x = y/2$.
Step 2 Substitute: $\bigl(\tfrac{y}{2}\bigr)^3 - 6\bigl(\tfrac{y}{2}\bigr)^2 + 11\bigl(\tfrac{y}{2}\bigr) - 6 = 0$.
Step 3 Multiply through by $8$: $y^3 - 12y^2 + 44y - 48 = 0$.
Example 7.4.2 — Reciprocal roots
The equation $x^3 + 2x^2 - 5x + 3 = 0$ has roots $\alpha, \beta, \gamma$. Find the equation with roots $\tfrac{1}{\alpha}, \tfrac{1}{\beta}, \tfrac{1}{\gamma}$.
Step 1 Let $y = 1/x$, so $x = 1/y$.
Step 2 Substitute: $\tfrac{1}{y^3} + \tfrac{2}{y^2} - \tfrac{5}{y} + 3 = 0$.
Step 3 Multiply by $y^3$: $3y^3 - 5y^2 + 2y + 1 = 0$.
Example 7.4.3 — Squared roots
The roots of $x^3 - 7x + 6 = 0$ are $\alpha, \beta, \gamma$. Find an equation with roots $\alpha^2, \beta^2, \gamma^2$.
Let $y = x^2$, so $x = \sqrt{y}$. Substitute: $y\sqrt{y} - 7\sqrt{y} + 6 = 0 \Rightarrow \sqrt{y}(y - 7) = -6 \Rightarrow y(y-7)^2 = 36$.
Expand: $y^3 - 14y^2 + 49y - 36 = 0$.
Example 7.4.4 — Shifted roots
The equation $x^3 - 3x + 1 = 0$ has roots $\alpha, \beta, \gamma$. Find an equation with roots $\alpha - 1, \beta - 1, \gamma - 1$.
Let $y = x - 1$, so $x = y + 1$. Substitute:
$(y+1)^3 - 3(y+1) + 1 = y^3+3y^2+3y+1-3y-3+1 = y^3+3y^2-1 = 0$.
Example 7.4.5 — Transformation $y = \alpha + 1/\alpha$
The equation $x^3 - 6x^2 + 11x - 6 = 0$ has roots $1, 2, 3$. Find the monic cubic whose roots are $1+1, 2+\tfrac{1}{2}, 3+\tfrac{1}{3}$, i.e. $2, \tfrac{5}{2}, \tfrac{10}{3}$.
Let $y = x + 1/x$, so $xy = x^2 + 1$, giving $x^2 - xy + 1 = 0$. This approach yields a degree-6 equation which factors; for this particular equation, we can multiply the roots directly. Sum $= 2 + \tfrac{5}{2} + \tfrac{10}{3} = \tfrac{12+15+20}{6} = \tfrac{47}{6}$. Product of pairs and product can be computed similarly, giving $y^3 - \tfrac{47}{6}y^2 + \tfrac{79}{9}y - \tfrac{100}{9} = 0$, or equivalently the monic cubic $9y^3 - \tfrac{141}{2}y^2 + \tfrac{79}{1}y - 100 = 0$ (clearing fractions step by step).
Example 7.4.6 — Using Vieta's formulas directly
For the cubic $x^3+px+q=0$ with roots $\alpha,\beta,\gamma$, find the cubic with roots $\alpha+\beta, \alpha+\gamma, \beta+\gamma$.
Since $\alpha+\beta+\gamma=0$ (no $x^2$ term), we have $\alpha+\beta = -\gamma$, $\alpha+\gamma=-\beta$, $\beta+\gamma=-\alpha$. So the new roots are $-\gamma,-\beta,-\alpha$, i.e. the negatives of the original roots. Replacing $x$ with $-x$: $(-x)^3+p(-x)+q = -x^3-px+q=0$, or $x^3+px-q=0$.
7.5 Complex Roots of Polynomials
Theorem 7.4 — Conjugate Root Theorem
If $p(x)$ is a polynomial with real coefficients and $z = a + bi$ (with $b \ne 0$) is a root, then the complex conjugate $\bar{z} = a - bi$ is also a root.
Consequence: A polynomial of degree $n$ with real coefficients has either 0, 2, 4, … complex (non-real) roots. Complex roots always occur in conjugate pairs.
Building the Polynomial from a Complex Root
If $z = a + bi$ is a root of a real polynomial, then both $z$ and $\bar{z}$ are roots, contributing the real quadratic factor:
$$(x - z)(x - \bar{z}) = x^2 - 2ax + (a^2 + b^2)$$
This quadratic has real coefficients and an always-positive constant term $a^2+b^2 > 0$.
Example 7.5.1 — Cubic with one complex root
Given that $2 + i$ is a root of $x^3 - 7x^2 + 17x - 15 = 0$, find all roots.
Step 1 By the conjugate root theorem, $2 - i$ is also a root.
Step 2 The corresponding quadratic factor is $(x-(2+i))(x-(2-i)) = x^2 - 4x + 5$.
Step 3 Divide: $x^3-7x^2+17x-15 = (x^2-4x+5)(x-3)$.
The three roots are $2+i$, $2-i$, and $3$.
Example 7.5.2 — Quartic with two complex conjugate pairs
A real quartic has roots $1+2i$ and $3i$. Write down all four roots and hence write the quartic as a product of two real quadratic factors.
By the conjugate root theorem the four roots are $1+2i,\ 1-2i,\ 3i,\ -3i$.
Factors: $(x^2-2x+5)(x^2+9)$.
The quartic is $x^4-2x^3+14x^2-18x+45 = 0$.
Example 7.5.3 — Finding a cubic given one complex root
A monic cubic with real coefficients has a root $1 - 3i$ and a real root $\alpha$. Given that the product of all roots is $-20$, find the cubic.
The three roots are $1-3i$, $1+3i$, and $\alpha$. Product: $(1-3i)(1+3i)\alpha = 10\alpha = -20$, so $\alpha = -2$.
Quadratic factor from complex pair: $x^2-2x+10$. Cubic: $(x+2)(x^2-2x+10) = x^3-3x^2+6x+20 = 0$. Wait — checking sign: $(x+2)(x^2-2x+10)= x^3-2x^2+10x+2x^2-4x+20 = x^3+6x+20$. Let me recompute: constant = $20$, product of roots should be $-d/a$, so product $= -20$ means $d=20$; the factorisation gives $x^3+6x+20=0$ with product of roots $= -20$. ✓
Example 7.5.4 — Using Vieta's formulas with complex roots
A quartic $x^4 + ax^3 + bx^2 + cx + 10 = 0$ with real coefficients has roots $1+i$, $1-i$, $\alpha$, $\beta$ where $\alpha, \beta$ are real. Find $a, b, c$ given that $\alpha + \beta = 0$.
$e_1 = (1+i)+(1-i)+\alpha+\beta = 2+0 = 2$, so $a = -2$.
$e_4 = (1+i)(1-i)\alpha\beta = 2\alpha\beta = 10$, so $\alpha\beta = 5$.
Since $\alpha+\beta=0$ and $\alpha\beta=5$, then $\alpha,\beta$ satisfy $t^2-0\cdot t+5=0$, i.e. $t^2+5=0$. But these must be real — contradiction! So if $\alpha+\beta=0$ these are in fact the pair $\pm\sqrt{5}\,i$, meaning the quartic is $(x^2+2x+2)(x^2+5)$; expanding gives $a=-2$ is wrong… Checking the sum properly: four roots $1\pm i, \pm\sqrt{5}i$ sum to $2$, so $a = -2$. Then $b = e_2 = (1+i)(1-i) + (1+i)(\sqrt{5}i) + (1+i)(-\sqrt{5}i) + (1-i)(\sqrt{5}i) + (1-i)(-\sqrt{5}i) + (\sqrt{5}i)(-\sqrt{5}i) = 2 + \sqrt{5}i(2i) + (-\sqrt{5}i)(2i) + 5 = 2+5+$ (imaginary terms cancel) $= 7$. So $b=7$. And $c=0$ by symmetry ($e_3 = 0$ since real part sum of $e_3$ terms cancels).
Exam Tip
In a polynomial with real coefficients, never state just one complex root — always state the conjugate pair together. Forgetting this is one of the most common errors in A-Level Further Maths. If you find a quadratic factor $x^2 - 2ax + (a^2+b^2)$ with negative discriminant, that is the correct real quadratic factor encoding the complex pair $a \pm bi$.
Practice Problems
Problem 1
The roots of $3x^2 + 7x - 6 = 0$ are $\alpha$ and $\beta$. Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
Show solution
$\alpha^2+\beta^2 = \tfrac{49}{9} - 2(-2) = \tfrac{49}{9}+4 = \tfrac{85}{9}$.
$\alpha^3+\beta^3 = (-\tfrac{7}{3})^3 - 3(-2)(-\tfrac{7}{3}) = -\tfrac{343}{27} - 14 = -\tfrac{343}{27} - \tfrac{378}{27} = -\tfrac{721}{27}$.
Problem 2
The roots of $x^3 - 4x^2 + x + 6 = 0$ are $\alpha, \beta, \gamma$. Find $\alpha^2+\beta^2+\gamma^2$ and $\alpha\beta\gamma(\alpha+\beta+\gamma)$.
Show solution
$\alpha^2+\beta^2+\gamma^2 = e_1^2 - 2e_2 = 16-2 = 14$.
$\alpha\beta\gamma(\alpha+\beta+\gamma) = e_3 \cdot e_1 = (-6)(4) = -24$.
Problem 3
The equation $2x^3 - x^2 - 13x - 6 = 0$ has roots $\alpha, \beta, \gamma$. Without solving, find $\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$.
Show solution
$\sum\tfrac{1}{\alpha^2} = \frac{(\alpha\beta)^2+(\alpha\gamma)^2+(\beta\gamma)^2}{(\alpha\beta\gamma)^2} = \frac{e_2^2-2e_1 e_3}{e_3^2} = \frac{\frac{169}{4}-3}{9} = \frac{\frac{157}{4}}{9} = \frac{157}{36}$.
Problem 4
The roots of $x^4 + 3x^3 - 2x^2 + x - 5 = 0$ are $\alpha,\beta,\gamma,\delta$. Find $\alpha^2+\beta^2+\gamma^2+\delta^2$.
Show solution
$\sum\alpha^2 = e_1^2 - 2e_2 = 9-2(-2) = 9+4 = 13$.
Problem 5
The equation $x^3 + px + q = 0$ has roots $\alpha, \beta, \gamma$. Show that $\alpha^2+\beta^2+\gamma^2 = -2p$ and find $\alpha^4+\beta^4+\gamma^4$ in terms of $p$ and $q$.
Show solution
$\alpha^4+\beta^4+\gamma^4 = (\alpha^2+\beta^2+\gamma^2)^2 - 2(\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2) = 4p^2 - 2e_2^2 = 4p^2 - 2p^2 = 2p^2$.
(Here $\alpha^2\beta^2+\cdots = (\alpha\beta+\alpha\gamma+\beta\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma) = p^2-0=p^2$.)
Problem 6
The equation $x^3 - 2x^2 - 5x + 6 = 0$ has roots $\alpha, \beta, \gamma$. Find the cubic equation with roots $\alpha+1, \beta+1, \gamma+1$.
Show solution
$(y-1)^3 - 2(y-1)^2 - 5(y-1) + 6 = 0$
$= y^3-3y^2+3y-1-2y^2+4y-2-5y+5+6$
$= y^3-5y^2+2y+8 = 0$.
Problem 7
The equation $x^3 - 5x^2 + 8x - 4 = 0$ has roots $\alpha, \beta, \gamma$. Find the cubic with roots $\alpha^2, \beta^2, \gamma^2$.
Show solution
$\sqrt{y}(y-5\sqrt{y}+8) = 4 \Rightarrow y\sqrt{y}-5y+8\sqrt{y}=4$
$\sqrt{y}(y+8) = 5y+4$
Square both sides: $y(y+8)^2 = (5y+4)^2$
$y^3+16y^2+64y = 25y^2+40y+16$
$y^3-9y^2+24y-16 = 0$.
Problem 8
Given that $3-i$ is a root of $x^3 - 8x^2 + 22x - 20 = 0$, find all roots.
Show solution
Quadratic factor: $(x-(3-i))(x-(3+i)) = x^2-6x+10$.
Divide: $x^3-8x^2+22x-20 = (x^2-6x+10)(x-2)$.
The three roots are $3-i,\ 3+i,\ 2$.
Problem 9
A quartic with real coefficients has roots $2+3i$ and $1-i$. Write the quartic as a product of two real quadratic factors and expand to find the quartic equation.
Show solution
Factors: $(x^2-4x+13)(x^2-2x+2)$.
Expand: $x^4-2x^3+2x^2-4x^3+8x^2-8x+13x^2-26x+26$
$= x^4-6x^3+23x^2-34x+26 = 0$.
Problem 10
The roots of $x^4 - 2x^3 - 3x^2 + 4x + 4 = 0$ include a repeated root. By writing the quartic as $(x-a)^2(x^2+bx+c)$, find all roots and verify Vieta's formulas $e_1=2$ and $e_4=4$.
Show solution
So the quartic is $(x+1)^2(x-2)^2$, with roots $-1,-1,2,2$.
$e_1 = -1-1+2+2 = 2$ ✓. $e_4 = (-1)(-1)(2)(2) = 4$ ✓.