1.1 - Root relations for quartics

1.1 - Root relations for quartics

A quartic stores four combinations of its roots in its coefficients. You will derive those combinations, read them when coefficients are missing and use constraints on the roots to recover unknown coefficients.

Expanding four factors

Can a quartic have a zero sum of roots but a nonzero product? Yes: the roots −2,−1,1,2-2,-1,1,2 add to zero and multiply to 44. Expanding their factors will show where both facts appear.

You already know how three factors give the three cubic relations. With four factors, each term still comes from selecting one term from every bracket. Selecting one negative root term gives a negative contribution; selecting two gives a positive one. The signs alternate all the way to the positive product of four roots.

Symmetric root combinations

For four roots α,β,γ,δ\alpha,\beta,\gamma,\delta, counted with multiplicity, write

S1=α+β+γ+δS2=αβ+αγ+αδ+βγ+βδ+γδS3=αβγ+αβδ+αγδ+βγδS4=αβγδ.\begin{aligned} S_1&=\alpha+\beta+\gamma+\delta \\ 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 \\ S_4&=\alpha\beta\gamma\delta. \end{aligned}

Each distinct selection of root positions occurs once, even when some root values are equal.

These are symmetric combinations: swapping the labels of roots leaves each complete sum or product unchanged.

There are four single selections, six pairs, four triples and one selection of all four roots. These are the coefficients in

(x−α)(x−β)(x−γ)(x−δ)=x4−S1x3+S2x2−S3x+S4.(x-\alpha)(x-\beta)(x-\gamma)(x-\delta) =x^4-S_1x^3+S_2x^2-S_3x+S_4.

The subscripts count how many root terms are multiplied, rather than the power of xx.

Worked example 1: collecting four combinations

Question. Expand (x−1)(x−2)(x−4)(x−5)(x-1)(x-2)(x-4)(x-5) using the root combinations.

Plan. Find the single, pair, triple and four-root products, then use alternating signs.

S1=1+2+4+5=12S2=2+4+5+8+10+20=49S3=8+10+20+40=78S4=40.\begin{aligned} S_1&=1+2+4+5=12 \\ S_2&=2+4+5+8+10+20=49 \\ S_3&=8+10+20+40=78 \\ S_4&=40. \end{aligned}

Answer. The expansion is x4−12x3+49x2−78x+40x^4-12x^3+49x^2-78x+40.

Check. Pairing the factors gives (x2−6x+5)(x2−6x+8)(x^2-6x+5)(x^2-6x+8). Multiplication gives x4−12x3+(36+13)x2−78x+40x^4-12x^3+(36+13)x^2-78x+40, confirming every coefficient by a different grouping.

Now change only the last root from 55 to −5-5. The alternating pattern stays the same; the numerical root combinations change.

Worked example 2: a negative root

Question. Expand (x−1)(x−2)(x−4)(x+5)(x-1)(x-2)(x-4)(x+5).

Plan. Include the negative root −5-5 inside each combination before applying the coefficient signs.

S1=1+2+4−5=2S2=2+4−5+8−10−20=−21S3=8−10−20−40=−62S4=−40.\begin{aligned} S_1&=1+2+4-5=2 \\ S_2&=2+4-5+8-10-20=-21 \\ S_3&=8-10-20-40=-62 \\ S_4&=-40. \end{aligned}

Answer. The expansion is x4−2x3−21x2+62x−40x^4-2x^3-21x^2+62x-40.

Check. Directly multiplying (x2−3x+2)(x2+x−20)(x^2-3x+2)(x^2+x-20) gives the same coefficients. In particular, the coefficient of xx is 60+2=6260+2=62.

Keep the sign of a root separate from the sign supplied by the expansion.

Watch out

The coefficient of xx is −S3-S_3. If S3=−62S_3=-62, that coefficient is +62+62, not −62-62.

Use the combinations or a correct factor expansion in this check.

Reading all four relations

Comparing the expansion with a quartic equation gives all four relations at once. As with cubics, divide by the leading coefficient before reading the root combinations.

Quartic root relations

For ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0, where a≠0a\ne0, with four roots counted with multiplicity:

S1=−ba,S2=ca,S3=−da,S4=ea.S_1=-\frac ba,\qquad S_2=\frac ca,\qquad S_3=-\frac da,\qquad S_4=\frac ea.

Not in the formula booklet: learn it.

A missing power has coefficient zero. Keep its place: the constant never moves into an empty coefficient slot.

Worked example 3: two missing coefficients

Question. State the four root combinations for 2x4−10x2+8=02x^4-10x^2+8=0.

Plan. Write the coefficients in descending powers, including both zeros.

(a,b,c,d,e)=(2,0,−10,0,8)S1=0,S2=−5,S3=0,S4=4.\begin{aligned} (a,b,c,d,e)&=(2,0,-10,0,8) \\ S_1&=0,&S_2&=-5,&S_3&=0,&S_4&=4. \end{aligned}

Answer. The four values are 0,−5,0,40,-5,0,4, respectively.

Check. The polynomial is 2(x2−1)(x2−4)2(x^2-1)(x^2-4), so its roots are −2,−1,1,2-2,-1,1,2. Their six pair products add to −5-5, their four triple products cancel and their product is 44. This answers the opening question.

The next equation still has a missing coefficient, but only the triple-product sum must now be zero. A zero coefficient does not make every root combination zero.

Worked example 4: only the linear term is absent

Question. State the four root combinations for 3x4−30x3+75x2−108=03x^4-30x^3+75x^2-108=0.

Plan. Insert 0x0x, then divide each signed coefficient by 33.

S1=−−303=10S2=753=25S3=−03=0S4=−1083=−36.\begin{aligned} S_1&=-\frac{-30}{3}=10 \\ S_2&=\frac{75}{3}=25 \\ S_3&=-\frac03=0 \\ S_4&=\frac{-108}{3}=-36. \end{aligned}

Answer. The values are 10,25,0,−3610,25,0,-36.

Check. The factorisation is 3(x+1)(x−2)(x−3)(x−6)3(x+1)(x-2)(x-3)(x-6). Its triple products are −6,−12,−18,36-6,-12,-18,36, which add to zero, although none of the four roots is zero.

The constant, rather than the linear coefficient, determines whether zero is a root.

Key idea

For a genuine quartic, a zero root occurs exactly when e=0e=0, because S4=e/aS_4=e/a is the product of all four roots. A zero sum of triple products does not imply a zero root.

Read the slots carefully in a different non-monic equation.

Constraints on four roots

When some roots are unknown, their sum and product may be enough to find coefficients without finding the roots individually. Group the known roots first, then express the six pairs and four triples in terms of the unknown pair.

For a root list r,s,u,vr,s,u,v, put U=u+vU=u+v and V=uvV=uv. Listing products gives

S1=r+s+US2=rs+(r+s)U+VS3=rsU+(r+s)VS4=rsV.\begin{aligned} S_1&=r+s+U \\ S_2&=rs+(r+s)U+V \\ S_3&=rsU+(r+s)V \\ S_4&=rsV. \end{aligned}

For example, ru+rv+su+sv=(r+s)(u+v)ru+rv+su+sv=(r+s)(u+v). This grouping reduces a long list to two small calculations.

Worked example 5: unknown roots need not be solved

Question. A quartic 2x4+px3+qx2+tx−48=02x^4+px^3+qx^2+tx-48=0 has roots 1,−2,u,v1,-2,u,v with root sum 66. Find p,q,tp,q,t.

Plan. Use the sum and product to find U=u+vU=u+v and V=uvV=uv, then obtain the coefficients.

U=6−(1−2)=7−2V=−48/2=−24product relationV=12S2=−2+(−1)(7)+12=3S3=(−2)(7)+(−1)(12)=−26.\begin{aligned} U&=6-(1-2)=7 \\ -2V&=-48/2=-24 &&\text{product relation} \\ V&=12 \\ S_2&=-2+(-1)(7)+12=3 \\ S_3&=(-2)(7)+(-1)(12)=-26. \end{aligned}

Answer. p=−2(6)=−12p=-2(6)=-12, q=2(3)=6q=2(3)=6 and t=−2(−26)=52t=-2(-26)=52.

Check. As u+v=7u+v=7 and uv=12uv=12, the unknown pair has factor x2−7x+12x^2-7x+12. Multiplying 2(x2+x−2)(x2−7x+12)2(x^2+x-2)(x^2-7x+12) gives 2x4−12x3+6x2+52x−482x^4-12x^3+6x^2+52x-48.

Now the known roots coincide. The grouping still works, but the two copies must both appear.

Worked example 6: a repeated known root

Question. The quartic 3x4+px3+qx2+tx−144=03x^4+px^3+qx^2+tx-144=0 has roots 2,2,u,v2,2,u,v and root sum 88. Find p,q,tp,q,t.

Plan. Count both copies of 22, then use the same grouped relations.

U=8−4=44V=−144/3=−48⇒V=−12S2=4+4(4)−12=8S3=4(4)+4(−12)=−32.\begin{aligned} U&=8-4=4 \\ 4V&=-144/3=-48\quad\Rightarrow\quad V=-12 \\ S_2&=4+4(4)-12=8 \\ S_3&=4(4)+4(-12)=-32. \end{aligned}

Answer. p=−24p=-24, q=24q=24 and t=96t=96.

Check. The unknown pair has factor x2−4x−12x^2-4x-12. Expanding 3(x−2)2(x2−4x−12)3(x-2)^2(x^2-4x-12) gives 3x4−24x3+24x2+96x−1443x^4-24x^3+24x^2+96x-144. Both the constant and repeated factor agree with the question.

This method uses the information requested by the coefficients; solving two additional roots would add work without adding information.

Problem-solving

Choose the relation with the fewest unknowns first. A known constant and two known nonzero roots often reveal the product of the other pair immediately.

Try the same strategy with a new repeated root.

Quick Recap

  • Four factors give alternating signs: x4−S1x3+S2x2−S3x+S4x^4-S_1x^3+S_2x^2-S_3x+S_4.
  • Divide coefficients by the nonzero leading coefficient.
  • Keep a zero in every missing coefficient slot.
  • Group an unknown pair by its sum and product when finding coefficients.

The roots −2,−1,1,2-2,-1,1,2 show that a zero sum and a nonzero product can occur together: their monic equation is x4−5x2+4=0x^4-5x^2+4=0.