Roots of Polynomials

This lesson covers the relationship between the roots of a polynomial and its coefficients. These relationships, known as Vieta's formulae (or Newton's identities), allow us to find symmetric functions of roots — such as their sum, product, and sum of squares — without solving the polynomial.

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Vieta's Formulae for Quadratics

For a quadratic $ax^2 + bx + c = 0$ax2+bx+c=0 with roots $\alpha$α and $\beta$β:

Symmetric function	Formula
$\alpha + \beta$α+β	$-b/a$−b/a
$\alpha\beta$αβ	$c/a$c/a

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Vieta's Formulae for Cubics

For $ax^3 + bx^2 + cx + d = 0$ax3+bx2+cx+d=0 with roots $\alpha, \beta, \gamma$α,β,γ:

Symmetric function	Formula
$\alpha + \beta + \gamma$α+β+γ	$-b/a$−b/a
$\alpha\beta + \beta\gamma + \gamma\alpha$αβ+βγ+γα	$c/a$c/a
$\alpha\beta\gamma$αβγ	$-d/a$−d/a

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Vieta's Formulae for Quartics

For $ax^4 + bx^3 + cx^2 + dx + e = 0$ax4+bx3+cx2+dx+e=0 with roots $\alpha, \beta, \gamma, \delta$α,β,γ,δ:

Symmetric function	Formula
$\sum \alpha$∑α	$-b/a$−b/a
$\sum \alpha\beta$∑αβ	$c/a$c/a
$\sum \alpha\beta\gamma$∑αβγ	$-d/a$−d/a
$\alpha\beta\gamma\delta$αβγδ	$e/a$e/a

Pattern: The signs alternate: $-, +, -, +, \ldots$−,+,−,+,… when dividing by the leading coefficient.

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Finding Other Symmetric Functions

Given the basic symmetric functions (sums and products of roots), we can derive others:

Sum of Squares

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$α2+β2=(α+β)2−2αβ

For three roots:

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)

Sum of Cubes

$\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)^3 - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma$α3+β3+γ3=(α+β+γ)3−3(α+β+γ)(αβ+βγ+γα)+3αβγ

Sum of Reciprocals

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$α1​+β1​=αβα+β​

$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}$α1​+β1​+γ1​=αβγαβ+βγ+γα​

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Worked Examples

Example 1: The equation $x^3 - 6x^2 + 11x - 6 = 0$x3−6x2+11x−6=0 has roots $\alpha, \beta, \gamma$α,β,γ. Find $\alpha^2 + \beta^2 + \gamma^2$α2+β2+γ2 and $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$α1​+β1​+γ1​.

From Vieta's: $\alpha + \beta + \gamma = 6$α+β+γ=6, $\alpha\beta + \beta\gamma + \gamma\alpha = 11$αβ+βγ+γα=11, $\alpha\beta\gamma = 6$αβγ=6.

$\alpha^2 + \beta^2 + \gamma^2 = 6^2 - 2(11) = 36 - 22 = 14$α2+β2+γ2=62−2(11)=36−22=14

$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{11}{6}$α1​+β1​+γ1​=611​

Example 2: The equation $2x^2 - 5x + 1 = 0$2x2−5x+1=0 has roots $\alpha, \beta$α,β. Find the equation whose roots are $\alpha^2$α2 and $\beta^2$β2.

$\alpha + \beta = 5/2$α+β=5/2, $\alpha\beta = 1/2$αβ=1/2.

New roots sum: $\alpha^2 + \beta^2 = (5/2)^2 - 2(1/2) = 25/4 - 1 = 21/4$α2+β2=(5/2)2−2(1/2)=25/4−1=21/4.

New roots product: $(\alpha\beta)^2 = 1/4$(αβ)2=1/4.

New equation: $x^2 - \frac{21}{4}x + \frac{1}{4} = 0$x2−421​x+41​=0, or $4x^2 - 21x + 1 = 0$4x2−21x+1=0.

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Forming New Equations from Transformations of Roots

If a polynomial has roots $\alpha, \beta, \gamma, \ldots$α,β,γ,… and we want the polynomial with roots $f(\alpha), f(\beta), f(\gamma), \ldots$f(α),f(β),f(γ),…:

1. Compute the symmetric functions of the new roots using the known symmetric functions of the old roots.
2. Alternatively, use the substitution method: if the new roots are $y = f(x)$y=f(x), invert to get $x = f^{-1}(y)$x=f−1(y) and substitute into the original equation.

Example 3: $x^3 - 3x + 1 = 0$x3−3x+1=0 has roots $\alpha, \beta, \gamma$α,β,γ. Find the equation with roots $2\alpha, 2\beta, 2\gamma$2α,2β,2γ.

Substitute $x = y/2$x=y/2: $(y/2)^3 - 3(y/2) + 1 = 0$(y/2)3−3(y/2)+1=0, so $y^3/8 - 3y/2 + 1 = 0$y3/8−3y/2+1=0, i.e., $y^3 - 12y + 8 = 0$y3−12y+8=0.

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Practice Problems

1. The equation $x^3 + 2x^2 - 5x - 6 = 0$x3+2x2−5x−6=0 has roots $\alpha, \beta, \gamma$α,β,γ. Find $\alpha + \beta + \gamma$α+β+γ, $\alpha\beta\gamma$αβγ, and $\alpha^2 + \beta^2 + \gamma^2$α2+β2+γ2.
2. The roots of $x^2 - 3x + 1 = 0$x2−3x+1=0 are $\alpha, \beta$α,β. Find $\alpha^3 + \beta^3$α3+β3.
3. Find the equation whose roots are $\alpha + 1, \beta + 1, \gamma + 1$α+1,β+1,γ+1 if $\alpha, \beta, \gamma$α,β,γ are roots of $x^3 - 4x^2 + x + 6 = 0$x3−4x2+x+6=0.

Solutions:

1. $\alpha + \beta + \gamma = -2$α+β+γ=−2, $\alpha\beta + \beta\gamma + \gamma\alpha = -5$αβ+βγ+γα=−5, $\alpha\beta\gamma = 6$αβγ=6. $\alpha^2 + \beta^2 + \gamma^2 = (-2)^2 - 2(-5) = 4 + 10 = 14$α2+β2+γ2=(−2)2−2(−5)=4+10=14.
2. $\alpha + \beta = 3$α+β=3, $\alpha\beta = 1$αβ=1. $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 27 - 9 = 18$α3+β3=(α+β)3−3αβ(α+β)=27−9=18.
3. Substitute $x = y - 1$x=y−1: $(y-1)^3 - 4(y-1)^2 + (y-1) + 6 = 0$(y−1)3−4(y−1)2+(y−1)+6=0. Expanding: $y^3 - 7y^2 + 14y - 8 = 0$y3−7y2+14y−8=0, i.e., $y^3 - 7y^2 + 14y - 8 = 0$y3−7y2+14y−8=0.

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Summary

• Vieta's formulae relate roots to coefficients without solving.
• Use identities like $\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$∑α2=(∑α)2−2∑αβ to find higher symmetric functions.
• To form new equations: either compute new symmetric functions or use substitution.

Exam Tip: Write down Vieta's formulae for the given polynomial at the start. Then use standard identities to build the required symmetric function step by step.