Pj3 - Solving cubics and quartics

Pj3 - Solving cubics and quartics

Use the conjugate-pair rule to find complex roots of polynomials with real coefficients. You will reduce a cubic to a quadratic using a real root, and turn a given non-real root of a quartic into a real quadratic factor.

Why non-real roots come in pairs

A quartic has real coefficients and one root is 1+2i1 + 2i. How much of the equation have you already solved? More than a quarter: 1−2i1 - 2i must be another root, and those two roots determine a whole quadratic factor.

Start with f(z)=z2−2z+5f(z) = z^2 - 2z + 5. Substituting 1+2i1 + 2i gives

f(1+2i)=(−3+4i)−(2+4i)+5=0f(1 + 2i) = (-3 + 4i) - (2 + 4i) + 5 = 0

Replacing every ii by −i-i gives the conjugate calculation, whose answer is still 00. That is the reason for the pairing.

You already know

For z=a+biz = a + bi, z∗=a−biz^* = a - bi. Products of conjugates are real: (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2. A root α\alpha of a polynomial corresponds to a factor z−αz - \alpha.

Conjugation respects addition and multiplication: expanding (x+yi)(u+vi)(x + yi)(u + vi) shows that conjugating the answer is the same as conjugating both factors first. Also, every real coefficient is its own conjugate. Consequently, for a polynomial ff with real coefficients,

f(z∗)=(f(z))∗f(z^*) = \bigl(f(z)\bigr)^*

If f(z)=0f(z) = 0, the right-hand side is 0∗=00^* = 0, so z∗z^* is a root too.

Conjugate-pair rule

If a polynomial has real coefficients and a+bia + bi is a root, where aa, bb are real and b≠0b \ne 0, then a−bia - bi is a distinct root. Together they give the real quadratic factor

(z−(a+bi))(z−(a−bi))=((z−a)−bi)((z−a)+bi)=(z−a)2+b2=z2−2az+a2+b2\begin{aligned} \bigl(z-(a+bi)\bigr)\bigl(z-(a-bi)\bigr) &= \bigl((z-a)-bi\bigr)\bigl((z-a)+bi\bigr) \\ &= (z-a)^2 + b^2 \\ &= z^2 - 2az + a^2 + b^2 \end{aligned}

Not in the formula booklet: learn it.

For instance, 1+2i1 + 2i and 1−2i1 - 2i give (z−1)2+4=z2−2z+5(z-1)^2 + 4 = z^2 - 2z + 5. Keep the factors bracketed: the imaginary terms cancel, and the final constant is positive.

Here is a useful surprise. A real cubic cannot have three non-real roots: non-real roots use up places in pairs, leaving at least one real root. A real quartic can instead have no real roots at all. For example, (z2+1)(z2+4)=0(z^2+1)(z^2+4)=0 has roots ii, −i-i, 2i2i, −2i-2i.

Key idea

The real-coefficient condition makes the pairing compulsory. For z−i=0z-i=0, the coefficient −i-i is not real, so ii is a root but −i-i is not.

A real root equals its own conjugate. That does not make it a repeated root: (z−2)(z2+1)=0(z-2)(z^2+1)=0 has the real root 22 only once.

Reduce a cubic using a real root

Once you have a real root, the factor theorem turns a cubic into a linear factor times a quadratic. Polynomial division or coefficient comparison then finds the quadratic, and the quadratic formula finds its roots, real or complex.

Factor theorem

For a polynomial ff,

f(r)=0⟺z−r is a factor of f(z)f(r)=0 \quad\Longleftrightarrow\quad z-r \text{ is a factor of } f(z)

Division by z−rz-r reduces the degree by one. Do not divide the equation by z−rz-r and then forget the root rr.

Not in the formula booklet: learn it.

For a monic cubic, write the remaining factor as z2+Az+Bz^2 + Az + B. Comparing coefficients gives a short, visible alternative to long division. If the leading coefficient is not 11, you can first divide the whole equation by that non-zero coefficient; this leaves its roots unchanged.

Worked example 1: a given real root

Question. Given that 22 is a root, solve z3−2z2+5z−10=0z^3 - 2z^2 + 5z - 10 = 0.

Plan. Factor out z−2z-2, find the remaining quadratic by coefficient comparison, then solve both factors.

f(2)=8−8+10−10=0check the supplied rootf(z)=(z−2)(z2+Az+B)factor theorem=z3+(A−2)z2+(B−2A)z−2Bexpand\begin{aligned} f(2) &= 8 - 8 + 10 - 10 = 0 && \text{check the supplied root} \\ f(z) &= (z-2)(z^2+Az+B) && \text{factor theorem} \\ &= z^3+(A-2)z^2+(B-2A)z-2B && \text{expand} \end{aligned}

Comparing coefficients gives A−2=−2A-2=-2, so A=0A=0, and −2B=−10-2B=-10, so B=5B=5. The coefficient of zz is then B−2A=5B-2A=5, as required.

(z−2)(z2+5)=0complete factorisationz=2orz2=−5set each factor to zeroz=2, 5 i, −5 itake both square roots\begin{aligned} (z-2)(z^2+5) &= 0 && \text{complete factorisation} \\ z &= 2 \quad\text{or}\quad z^2=-5 && \text{set each factor to zero} \\ z &= 2,\ \sqrt5\,i,\ -\sqrt5\,i && \text{take both square roots} \end{aligned}

Why this step? A cubic's three roots include the one already used to obtain the quadratic; factoring does not remove it from the answer.

Answer. z=2z=2 or z=±5 iz=\pm\sqrt5\,i.

Check. Expanding (z−2)(z2+5)(z-2)(z^2+5) reproduces the original cubic, and its non-real roots are a conjugate pair.

The next example changes how the real root is supplied: you find it by a simple substitution. Once it is found, the same factor-theorem method applies.

Worked example 2: finding a real root first

Question. In this question you must show detailed reasoning. Solve z3+z2+4z+4=0z^3 + z^2 + 4z + 4 = 0.

Plan. Try −1-1, factor out z+1z+1, and solve the resulting quadratic.

f(−1)=−1+1−4+4=0test a simple real valuef(z)=z2(z+1)+4(z+1)group terms=(z+1)(z2+4)factor out z+1z+1=0orz2=−4zero-product rulez=−1, 2i, −2isolve both factors\begin{aligned} f(-1) &= -1+1-4+4=0 && \text{test a simple real value} \\ f(z) &= z^2(z+1)+4(z+1) && \text{group terms} \\ &= (z+1)(z^2+4) && \text{factor out } z+1 \\ z+1 &= 0 \quad\text{or}\quad z^2=-4 && \text{zero-product rule} \\ z &= -1,\ 2i,\ -2i && \text{solve both factors} \end{aligned}

Why this step? Showing f(−1)=0f(-1)=0 justifies the linear factor; a calculator's list of roots would not show that reasoning.

Answer. z=−1z=-1, 2i2i or −2i-2i.

Check. (z+1)(z2+4)=z3+z2+4z+4(z+1)(z^2+4)=z^3+z^2+4z+4, including the correct constant term.

Always show the equation you solve after factorisation. This is especially useful when the question says “Determine” or asks for detailed reasoning.

Watch out

Wrong: from z2=−4z^2=-4, write only z=2iz=2i. Right: z=±2iz=\pm2i, since both (2i)2(2i)^2 and (−2i)2(-2i)^2 equal −4-4. Keep the real root as well.

Try another positive real root. Your answer should contain one real root and a non-real conjugate pair.

Turn a complex root into a quadratic factor

For a quartic, one non-real root supplies its conjugate and therefore a real quadratic factor. Divide by that quadratic, then solve the remaining quadratic. You will be given enough information to find at least one complex root or quadratic factor; a general quartic formula is unnecessary.

Use a monic quadratic z2+Az+Bz^2+Az+B for the remaining factor when the original quartic is monic. If the cubic term is missing, its coefficient is 00: retain that equation when comparing coefficients.

Worked example 3: two non-real pairs

Question. The equation z4−2z3+7z2−4z+10=0z^4-2z^3+7z^2-4z+10=0 has the root 1+2i1+2i. Determine all its roots.

Plan. Use the conjugate root to form z2−2z+5z^2-2z+5, find the other quadratic, then solve it.

The coefficients are real, so 1−2i1-2i is a root too. Their factor is

(z−1−2i)(z−1+2i)=(z−1)2+4=z2−2z+5(z-1-2i)(z-1+2i)=(z-1)^2+4=z^2-2z+5

Write the quartic as (z2−2z+5)(z2+Az+B)(z^2-2z+5)(z^2+Az+B). Comparing coefficients gives

A−2=−2coefficient of z3A=05B=10constant termB=2\begin{aligned} A-2 &= -2 && \text{coefficient of } z^3 \\ A &= 0 \\ 5B &= 10 && \text{constant term} \\ B &= 2 \end{aligned}

Why this step? The conjugate pair gives a real factor, so the remaining division stays in real algebra. The other coefficients check whether the factorisation is correct.

f(z)=(z2−2z+5)(z2+2)factorisez2+2=0⟹z=±2 isolve the remaining factor\begin{aligned} f(z) &= (z^2-2z+5)(z^2+2) && \text{factorise} \\ z^2+2 &= 0 \quad\Longrightarrow\quad z=\pm\sqrt2\,i && \text{solve the remaining factor} \end{aligned}

Answer. z=1±2iz=1\pm2i or z=±2 iz=\pm\sqrt2\,i.

Check. Expansion gives the middle terms (2+5)z2−4z(2+5)z^2-4z, matching 7z2−4z7z^2-4z. All four roots occur in conjugate pairs.

Change the remaining quadratic from z2+2z^2+2 to z2−3z^2-3. The supplied complex root is unchanged, but the other two roots become real.

Worked example 4: one non-real pair and two real roots

Question. The equation z4−2z3+2z2+6z−15=0z^4-2z^3+2z^2+6z-15=0 has the root 1+2i1+2i. Determine all its roots.

Plan. Reuse the conjugate-pair factor and find the new remaining quadratic.

The real coefficients give the additional root 1−2i1-2i, so z2−2z+5z^2-2z+5 is a factor. Put

f(z)=(z2−2z+5)(z2+Az+B)f(z)=(z^2-2z+5)(z^2+Az+B)

Then

A−2=−2⟹A=0coefficient of z35B=−15⟹B=−3constant termf(z)=(z2−2z+5)(z2−3)check the other coefficientsz2−3=0⟹z=±3real square roots\begin{aligned} A-2 &= -2 \quad\Longrightarrow\quad A=0 && \text{coefficient of } z^3 \\ 5B &= -15 \quad\Longrightarrow\quad B=-3 && \text{constant term} \\ f(z) &= (z^2-2z+5)(z^2-3) && \text{check the other coefficients} \\ z^2-3 &= 0 \quad\Longrightarrow\quad z=\pm\sqrt3 && \text{real square roots} \end{aligned}

Why this step? A real quartic need not have four non-real roots. The sign in the remaining quadratic determines whether its roots are real here.

Answer. z=1±2iz=1\pm2i or z=±3z=\pm\sqrt3.

Check. The product has z2z^2 coefficient 5−3=25-3=2, zz coefficient (−2)(−3)=6(-2)(-3)=6, and constant −15-15, as required.

The opening question is answered: one non-real root of a real quartic supplies two roots and a real quadratic factor. The equation that remains is only quadratic.

Exam tip

When using a conjugate root, say “because the coefficients are real”. For a “Determine” question, show the factorisation and a method of solving the remaining quadratic; a final list alone does not justify the roots.

In the last check, the supplied pair moves from 1±2i1\pm2i to 2±i2\pm i. Build its quadratic factor afresh.

Watch out

Wrong: (z−1−2i)(z−1+2i)=(z−1)2−4(z-1-2i)(z-1+2i)=(z-1)^2-4. The imaginary product is −(2i)2=+4-(2i)^2=+4, so the correct factor is (z−1)2+4=z2−2z+5(z-1)^2+4=z^2-2z+5.

Keep the conjugate factors bracketed when you form the next quadratic.

Quick Recap

  • Non-real roots of a polynomial with real coefficients occur in conjugate pairs; a real root need not be repeated.
  • The pair a±bia\pm bi gives the real factor (z−a)2+b2(z-a)^2+b^2.
  • For a cubic, use a real root and the factor theorem, then solve the remaining quadratic.
  • For a quartic, use the supplied root or quadratic factor, then solve the remaining quadratic and retain every root.

Knowing 1+2i1+2i is a root of a real quartic immediately gives 1−2i1-2i and the factor z2−2z+5z^2-2z+5: half the roots are already known.