CP 4.1 - Roots and coefficients of polynomials

CP 4.1 - Roots and coefficients of polynomials

A polynomial's coefficients contain information about all its roots together, even when the individual roots are awkward. You will extract sums and products for equations up to degree four, then use them to evaluate expressions involving powers, reciprocals and shifted roots.

Reading symmetric sums from coefficients

The roots of x3−3x2+2x−4=0x^3-3x^2+2x-4=0 are α,β,γ\alpha,\beta,\gamma. Can you find α3+β3+γ3\alpha^3+\beta^3+\gamma^3 without solving the cubic? The coefficients will let you do exactly that.

First look at a smaller case:

(x−2)(x−5)=x2−7x+10(x-2)(x-5)=x^2-7x+10

The coefficient of xx is the negative of the root sum, and the constant is their product. Expanding one more factor reveals the same alternating pattern.

You already know

The factor theorem gives a factor x−αx-\alpha for a root α\alpha. A degree-nn polynomial has nn complex roots counting multiplicity; repeated roots occupy repeated places in the list. Real coefficients make non-real roots occur in conjugate pairs.

For a cubic,

a(x−α)(x−β)(x−γ)=a[x3−(α+β+γ)x2+(αβ+αγ+βγ)x−αβγ]\begin{aligned} a(x-\alpha)(x-\beta)(x-\gamma) &=a\bigl[x^3-(\alpha+\beta+\gamma)x^2\\ &\quad +(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma\bigr] \end{aligned}

Every pair appears once. A term in xx comes from choosing a root from two brackets; a constant comes from choosing all three negative roots. This explains the signs, rather than leaving them as a memory test.

Symmetric sums

A symmetric expression is unchanged when the roots are relabelled. Write S1S_1 for the sum of roots, S2S_2 for the sum of all distinct pair products, S3S_3 for the sum of distinct triple products and, for a quartic, S4S_4 for the product of all four roots. For three roots, S2=αβ+αγ+βγS_2=\alpha\beta+\alpha\gamma+\beta\gamma; it does not include α2\alpha^2 or count αβ\alpha\beta twice.

Comparing the expansion with the supplied coefficients gives the rules below. Dividing by the leading coefficient is essential; multiplying an entire polynomial by a non-zero constant does not change its roots.

Roots and coefficients

For a quadratic ax2+bx+c=0ax^2+bx+c=0, with a≠0a\ne0:

S1=−ba,S2=caS_1=-\frac ba,\qquad S_2=\frac ca

For a cubic ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0:

S1=−ba,S2=ca,S3=−daS_1=-\frac ba,\qquad S_2=\frac ca,\qquad S_3=-\frac da

For a quartic ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0:

S1=−ba,S2=ca,S3=−da,S4=eaS_1=-\frac ba,\quad S_2=\frac ca,\quad S_3=-\frac da,\quad S_4=\frac ea

Roots are counted with multiplicity. The coefficients and roots may be complex.

Not in the formula booklet: learn it.

For four roots α,β,γ,δ\alpha,\beta,\gamma,\delta, there are six pair products and four triple products:

S2=αβ+αγ+αδ+βγ+βδ+γδS3=αβγ+αβδ+αγδ+βγδ\begin{aligned} S_2&=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta\\ S_3&=\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta \end{aligned}

The product of four negative roots is positive, explaining the quartic constant's positive sign. A missing power has coefficient zero, so its corresponding symmetric sum is zero too.

Key idea

Read the coefficients in descending powers, alternate the signs and divide every one by the leading coefficient. You can know these collective properties without knowing a single individual root.

Here the leading coefficient and the quartic product sign both matter.

Powers, reciprocals and shifted products

To evaluate a symmetric expression, rewrite it in terms of the sums you can read from the coefficients; calculating the roots first usually makes the task harder.

Squaring α+β+γ\alpha+\beta+\gamma produces the individual squares plus twice the pair products. For cubes, expand S13S_1^3 and subtract 3S1S23S_1S_2: the mixed terms disappear, leaving the individual cubes minus 3S33S_3. Reciprocals are handled by one common denominator.

Useful identities for three roots

For roots α,β,γ\alpha,\beta,\gamma:

α2+β2+γ2=S12−2S2\alpha^2+\beta^2+\gamma^2=S_1^2-2S_2 α3+β3+γ3=S13−3S1S2+3S3\alpha^3+\beta^3+\gamma^3=S_1^3-3S_1S_2+3S_3

If no root is zero,

1α+1β+1γ=αβ+αγ+βγαβγ=S2S3\frac1\alpha+\frac1\beta+\frac1\gamma =\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma} =\frac{S_2}{S_3}

For any number tt, expanding the three brackets gives

(t+α)(t+β)(t+γ)=t3+t2S1+tS2+S3(t+\alpha)(t+\beta)(t+\gamma)=t^3+t^2S_1+tS_2+S_3

Not in the formula booklet: learn it.

The reciprocal condition can be checked without solving: a polynomial with non-zero constant has no zero root. State the coefficient sums first so that sign errors have fewer places to hide.

Worked example 1: four expressions from one cubic

Question. The roots of x3−3x2+2x−4=0x^3-3x^2+2x-4=0 are α,β,γ\alpha,\beta,\gamma. Find the sum of their squares, the sum of their cubes, the sum of their reciprocals and (3+α)(3+β)(3+γ)(3+\alpha)(3+\beta)(3+\gamma).

Plan. Read S1,S2,S3S_1,S_2,S_3, then substitute into the four identities.

S1=3,S2=2,S3=4S_1=3,\qquad S_2=2,\qquad S_3=4

Thus

α2+β2+γ2=32−2(2)=5square-sum identityα3+β3+γ3=33−3(3)(2)+3(4)=21cube-sum identity1α+1β+1γ=24=12common denominator(3+α)(3+β)(3+γ)=27+9(3)+3(2)+4=64expand in symmetric sums\begin{aligned} \alpha^2+\beta^2+\gamma^2&=3^2-2(2)=5 &&\text{square-sum identity}\\ \alpha^3+\beta^3+\gamma^3&=3^3-3(3)(2)+3(4)=21 &&\text{cube-sum identity}\\ \frac1\alpha+\frac1\beta+\frac1\gamma&=\frac24=\frac12 &&\text{common denominator}\\ (3+\alpha)(3+\beta)(3+\gamma)&=27+9(3)+3(2)+4=64 &&\text{expand in symmetric sums} \end{aligned}

Why this step? Each requested expression is unchanged by relabelling the roots. Rewriting it using S1,S2,S3S_1,S_2,S_3 exposes exactly the information the polynomial provides.

Answer. The four values are 55, 2121, 1/21/2 and 6464, respectively.

Check. For the monic cubic ff, f(−3)=(−3−α)(−3−β)(−3−γ)=−64f(-3)=(-3-\alpha)(-3-\beta)(-3-\gamma)=-64. Direct substitution gives −27−27−6−4=−64-27-27-6-4=-64, confirming the shifted product.

Worked example 2 changes only the leading coefficient of the polynomial, from 11 to 22. The roots change, and every coefficient ratio must now include that 22.

Worked example 2: the leading coefficient changes

Question. The roots of 2x3−3x2+2x−4=02x^3-3x^2+2x-4=0 are α,β,γ\alpha,\beta,\gamma. Find the same four expressions as in Worked example 1.

Plan. Divide each relevant coefficient by 22 before using the identities.

S1=32,S2=1,S3=2S_1=\frac32,\qquad S_2=1,\qquad S_3=2

Therefore

α2+β2+γ2=94−2=14use S12−2S2α3+β3+γ3=278−92+6=398use the cube identity1α+1β+1γ=12use S2/S3(3+α)(3+β)(3+γ)=27+272+3+2=912use t=3\begin{aligned} \alpha^2+\beta^2+\gamma^2&=\frac94-2=\frac14 &&\text{use }S_1^2-2S_2\\ \alpha^3+\beta^3+\gamma^3&=\frac{27}{8}-\frac92+6=\frac{39}{8} &&\text{use the cube identity}\\ \frac1\alpha+\frac1\beta+\frac1\gamma&=\frac12 &&\text{use }S_2/S_3\\ (3+\alpha)(3+\beta)(3+\gamma)&=27+\frac{27}{2}+3+2=\frac{91}{2} &&\text{use }t=3 \end{aligned}

Why this step? The factorisation begins 2(x−α)(x−β)(x−γ)2(x-\alpha)(x-\beta)(x-\gamma). Ignoring the 22 would double the coefficient sums, not merely change the appearance of the equation.

Answer. The values are 1/41/4, 39/839/8, 1/21/2 and 91/291/2, respectively.

Check. Direct substitution gives f(−3)=−54−27−6−4=−91f(-3)=-54-27-6-4=-91. Since f(−3)=−2(3+α)(3+β)(3+γ)f(-3)=-2(3+\alpha)(3+\beta)(3+\gamma), the product is 91/291/2.

Here is a useful surprise: a sum of squares of complex roots can be negative. For x2+1=0x^2+1=0, the roots are i\mathrm{i} and −i-\mathrm{i}, and their squares sum to −2-2. The rule “a sum of squares is non-negative” requires real numbers; polynomial roots need not be real.

Watch out

Wrong: 1/α+1/β+1/γ=3/(α+β+γ)1/\alpha+1/\beta+1/\gamma=3/(\alpha+\beta+\gamma). Fractions do not combine that way. Right: use the common denominator αβγ\alpha\beta\gamma, giving S2/S3S_2/S_3.

A “Hence” instruction may require these coefficient relationships, even when a calculator could approximate the roots. Keep the identity and the substitution visible.

Using all four quartic relationships

A quartic gives one extra layer of products: the triple sum and the product of all four roots are different quantities, and both are useful.

The square-sum identity still holds, because every distinct pair occurs twice when the root sum is squared. For reciprocals, putting four fractions over their common denominator leaves four triple products in the numerator.

Quartic square and reciprocal sums

For four roots α,β,γ,δ\alpha,\beta,\gamma,\delta:

α2+β2+γ2+δ2=S12−2S2\alpha^2+\beta^2+\gamma^2+\delta^2=S_1^2-2S_2

If no root is zero,

1α+1β+1γ+1δ=S3S4\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta=\frac{S_3}{S_4}

Not in the formula booklet: learn it.

Use the cubic coefficient for S1S_1, the quadratic coefficient for S2S_2, the linear coefficient for S3S_3 and the constant for S4S_4. Writing the four values on one line prevents accidental mixing.

Worked example 3: a non-monic quartic

Question. The roots of 2x4−6x3+4x2+10x−8=02x^4-6x^3+4x^2+10x-8=0 are α,β,γ,δ\alpha,\beta,\gamma,\delta. Find their square sum and reciprocal sum.

Plan. Extract all four relationships, then use S12−2S2S_1^2-2S_2 and S3/S4S_3/S_4.

S1=3,S2=2,S3=−5,S4=−4S_1=3,\quad S_2=2,\quad S_3=-5,\quad S_4=-4

Hence

α2+β2+γ2+δ2=32−2(2)=5use the pair sum1α+1β+1γ+1δ=−5−4=54use triple sum over product\begin{aligned} \alpha^2+\beta^2+\gamma^2+\delta^2&=3^2-2(2)=5 &&\text{use the pair sum}\\ \frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta&=\frac{-5}{-4}=\frac54 &&\text{use triple sum over product} \end{aligned}

Why this step? For four reciprocals, each numerator is a product of the other three roots. The numerator is S3S_3, not the pair sum S2S_2 used for a cubic.

Answer. The square sum is 55, and the reciprocal sum is 5/45/4.

Check. The coefficient ratio for the reciprocal sum simplifies directly to (−10/2)/(−8/2)=10/8=5/4(-10/2)/(-8/2)=10/8=5/4. The leading coefficient cancels, but the signs still matter.

Worked example 4 changes the information sought: a condition on the square sum now determines an unknown coefficient. The same identity becomes an equation rather than a final calculation.

Worked example 4: finding an unknown coefficient

Question. The roots of x4+px3+px2−2x+3=0x^4+px^3+px^2-2x+3=0 have square sum 88. Find the possible real values of pp.

Plan. Express the root sum and pair sum in terms of pp, then solve the square-sum condition.

S1=−p,S2=pread the coefficients8=(−p)2−2puse the given square sump2−2p−8=0collect terms(p−4)(p+2)=0factorise\begin{aligned} S_1&=-p,\qquad S_2=p &&\text{read the coefficients}\\ 8&=(-p)^2-2p &&\text{use the given square sum}\\ p^2-2p-8&=0 &&\text{collect terms}\\ (p-4)(p+2)&=0 &&\text{factorise} \end{aligned}

Why this step? The square is (−p)2=p2(-p)^2=p^2, while the pair-sum term keeps its sign. Writing both substitutions explicitly avoids losing the linear term's minus sign.

Answer. p=4p=4 or p=−2p=-2.

Check. 42−2(4)=84^2-2(4)=8 and (−2)2−2(−2)=8(-2)^2-2(-2)=8. No claim that all four roots are real was needed.

The opening cubic's cube sum was 2121, found from three coefficient ratios. Quartics work in the same spirit: a little information about every root together often answers the question more cleanly than solving for each root separately.

Watch out

Wrong: a quartic has four roots, so the cross-products in S12S_1^2 need coefficient 44. Right: each pair occurs twice, regardless of the number of roots, so the identity is always S12=∑α2+2S2S_1^2=\sum \alpha^2+2S_2.

The following parameter sits only in the cubic coefficient. That makes the final condition especially short.

Quick Recap

  • Divide by the leading coefficient and use alternating signs for successive symmetric sums.
  • A cubic supplies a root sum, pair sum and product; a quartic also distinguishes triple sum from four-root product.
  • Square and cube sums come from expanding symmetric identities, while reciprocal sums need a common denominator.
  • Expand shifted products in symmetric sums, and keep repeated roots counted with multiplicity.

The opening roots need not be found: their cube sum is 33−3(3)(2)+3(4)=213^3-3(3)(2)+3(4)=21.