B1, B3 - Complex roots of cubic and quartic equations | AQA A-Level Further Mathematics

How much can one known root tell you about an entire polynomial equation? B3 supplies a second root when the coefficients are real, and B1 uses the resulting factors to solve cubics and quartics. You will recover all the roots from the information supplied, including zero and repeated roots.

Conjugate roots and real coefficients

If a cubic with real coefficients has root $1+2\mathrm{i}$ , what other root must it have? The partner $1-2\mathrm{i}$ is forced by conjugation. This gives more information than an arbitrary single root would provide.

Root and factor

A root $r$ of $f(z)=0$ satisfies $f(r)=0$ . By the factor theorem, $z-r$ is then a factor of $f(z)$ . The theorem works for real and complex roots.

Suppose $f(z)=z^3-3z^2+7z-5$ and $f(1+2\mathrm{i})=0$ . Conjugating that equality changes every occurrence of $1+2\mathrm{i}$ to $1-2\mathrm{i}$ , while the real coefficients stay fixed. The result is $f(1-2\mathrm{i})=0$ .

More generally, write a polynomial with real coefficients as $f(z)=a_nz^n+\cdots+a_1z+a_0$ . Conjugation preserves sums and products, so

\begin{aligned} \bigl(f(r)\bigr)^* &=a_n(r^*)^n+\cdots+a_1r^*+a_0\\ &=f(r^*) \end{aligned}

If $f(r)=0$ , then $f(r^*)=0^*=0$ . When $r$ is non-real, $r^*\ne r$ , so these are two distinct roots.

Conjugate-root theorem

Non-real roots of a polynomial equation with real coefficients occur in conjugate pairs. If $p+\mathrm{i}q$ is a root, with real $p,q$ and $q\ne0$ , then $p-\mathrm{i}q$ is also a root. Together they give the real quadratic factor

[z-(p+\mathrm{i}q)][z-(p-\mathrm{i}q)]=(z-p)^2+q^2

Not in the formula booklet: learn it.

The real-coefficient condition is essential. The equation $z-\mathrm{i}=0$ has root $\mathrm{i}$ , but $-\mathrm{i}$ is not a root: substituting it gives $-2\mathrm{i}$ . Its constant coefficient is non-real, so the theorem does not apply. A real root is its own conjugate; the theorem does not turn it into a second distinct root.

For the opening pair, the quadratic factor is $(z-1)^2+4=z^2-2z+5$ . Two linear factors containing $\mathrm{i}$ have combined into a factor with real coefficients. This is the useful connection: a complex root can unlock ordinary real polynomial division.

Watch out

Wrong: “complex roots always come in pairs”. Correct: non-real conjugate roots pair up when the polynomial has real coefficients. A general non-real root need not be purely imaginary.

Apply the conditions as well as the sign change.

Factoring a cubic from a known root

A cubic becomes a quadratic after division by a known linear factor. If the given root is non-real and the coefficients are real, use its conjugate too: the quadratic factor leaves only a linear quotient.

You already know

Polynomial division finds $Q(z)$ and a remainder $R(z)$ such that $f(z)=D(z)Q(z)+R(z)$ , with the remainder's degree smaller than the divisor's. If $D$ is a factor, the remainder is zero. Expanding the proposed product checks the division.

Start with a real root so that the quadratic is obtained by ordinary division.

Worked example 1: a given real root

Question. It is given that $1$ is a root of $z^3-3z^2+7z-5=0$ . Find all the roots.

Plan. Divide by $z-1$ , then solve the quadratic quotient.

\begin{aligned} z^3-3z^2+7z-5 &=(z-1)(z^2-2z+5) && \text{polynomial division}\\ z^2-2z+5&=0 && \text{set the remaining factor to zero}\\ (z-1)^2&=-4 && \text{complete the square}\\ z&=1\pm2\mathrm{i} && \text{solve the quadratic} \end{aligned}

For the division, subtracting $z^2(z-1)$ first leaves $-2z^2+7z-5$ . Subtracting $-2z(z-1)$ then leaves $5z-5$ , so the last quotient term is $5$ and the remainder is zero.

Answer. $z=1,\ 1+2\mathrm{i},\ 1-2\mathrm{i}$ .

Check. Expanding $(z-1)(z^2-2z+5)$ gives $z^3-3z^2+7z-5$ . The given root makes the first factor zero; the two new roots make $(z-1)^2+4$ zero.

Keep the polynomial the same, but change the information supplied: this time you know a non-real root. That changes which factor is available first.

Worked example 2: a given non-real root

Question. It is given that $1+2\mathrm{i}$ is a root of $z^3-3z^2+7z-5=0$ . Find the other roots.

Plan. Use real coefficients to obtain the conjugate root, form their quadratic factor and divide it out.

\begin{aligned} \text{Conjugate root}&=1-2\mathrm{i} && \text{real coefficients}\\ [z-(1+2\mathrm{i})][z-(1-2\mathrm{i})] &=(z-1)^2+4=z^2-2z+5 && \text{combine the pair}\\ z^3-3z^2+7z-5 &=(z^2-2z+5)(z-1) && \text{divide out the quadratic}\\ z-1&=0\implies z=1 && \text{solve the linear factor} \end{aligned}

In this division, subtracting $z(z^2-2z+5)$ leaves $-z^2+2z-5$ , which is $-1$ times the divisor. That gives the quotient $z-1$ with zero remainder.

Answer. The other roots are $1-2\mathrm{i}$ and $1$ .

Check. Substitution of the real root gives $1-3+7-5=0$ . For each non-real root, the quadratic factor becomes $(\pm2\mathrm{i})^2+4=0$ .

Both routes reach the same factorisation. A cubic may also have three real roots; do not assume a non-real pair exists unless the information or the remaining quadratic establishes it.

Watch out

Wrong: a root $1+2\mathrm{i}$ gives the factor $z+1+2\mathrm{i}$ . Correct: subtract the whole root, $z-(1+2\mathrm{i})=z-1-2\mathrm{i}$ .

Use the non-real-root route for a new cubic.

Quartics from a complex root or quadratic factor

A quartic has a quadratic quotient after division by a quadratic factor. A given non-real root supplies such a factor through its conjugate; a given quadratic factor lets you start the division immediately. In either case, solve both quadratic factors to obtain the full root set.

Worked example 3: a quartic with a given complex root

Question. It is given that $1+2\mathrm{i}$ is a root of $z^4+3z^2+6z+10=0$ . Find all the roots.

Plan. Form the real quadratic factor from the conjugate pair, divide, then solve the other quadratic.

\begin{aligned} \text{Conjugate root}&=1-2\mathrm{i} && \text{real coefficients}\\ \text{Pair's factor}&=(z-1)^2+4=z^2-2z+5 && \text{multiply linear factors}\\ z^4+3z^2+6z+10 &=(z^2-2z+5)(z^2+2z+2) && \text{polynomial division}\\ z^2+2z+2&=0\implies (z+1)^2=-1 && \text{complete the square}\\ z&=-1\pm\mathrm{i} && \text{solve the quotient} \end{aligned}

Include the missing cubic term as $0z^3$ when dividing. The successive quotient terms are $z^2$ , $2z$ and $2$ : the successive remainders are $2z^3-2z^2+6z+10$ , then $2z^2-4z+10$ , then zero.

Answer. $z=1+2\mathrm{i},\ 1-2\mathrm{i},\ -1+\mathrm{i},\ -1-\mathrm{i}$ .

Check. In the product of quadratics, the cubic terms cancel, the quadratic coefficient is $2-4+5=3$ , the linear coefficient is $-4+10=6$ , and the constant is $10$ . Each listed pair makes its corresponding quadratic factor zero.

Now keep the same first quadratic factor but change the second factor's constant. The question supplies the first factor directly, so no conjugate-root deduction is needed to start.

Worked example 4: a quartic with a given quadratic factor

Question. The polynomial $z^4+6z^2+25$ has factor $z^2-2z+5$ . Solve $z^4+6z^2+25=0$ .

Plan. Divide by the supplied factor and solve each quadratic by completing the square.

\begin{aligned} z^4+6z^2+25 &=(z^2-2z+5)(z^2+2z+5) && \text{divide by the given factor}\\ z^2-2z+5=0&\implies (z-1)^2=-4 && \text{first quadratic}\\ z^2+2z+5=0&\implies (z+1)^2=-4 && \text{second quadratic} \end{aligned}

The division successively leaves $2z^3+z^2+25$ , then $5z^2-10z+25$ , then zero, giving quotient $z^2+2z+5$ .

Answer. $z=1\pm2\mathrm{i}$ or $z=-1\pm2\mathrm{i}$ .

Check. The product's cubic and linear coefficients are zero, its quadratic coefficient is $5-4+5=6$ , and its constant is $25$ . Substituting each root into its completed-square factor gives zero.

Changing the quotient changed only its own root pair. The known factor still supplied $1\pm2\mathrm{i}$ . A quadratic quotient can instead give real or repeated roots; inspect its discriminant rather than forcing every answer into a non-real pair.

Watch out

Wrong: after solving the quotient, report only its two roots. Correct: include the roots of the factor that was divided out as well. Division reduces the work, not the original equation's root set.

Recover both pairs for a new quartic.

Repeated roots and complete root sets

A factor can occur more than once, so the number of distinct root values can be smaller than the degree. The factorisation tells you both which values solve the equation and how many times each occurs.

Multiplicity

A root $r$ has multiplicity $m$ when $(z-r)^m$ is a factor but $(z-r)^{m+1}$ is not. A repeated root has multiplicity at least $2$ . Repeating a root in a list records its multiplicity; it does not create a new value.

For a product of complex numbers to be zero, at least one factor must be zero. Indeed, if a factor is nonzero, you can divide by it. But if you divide an equation by an expression involving $z$ , first keep the values that make that expression zero, or you may discard roots.

Worked example 5: a repeated conjugate pair

Question. It is given that $z^2+1$ is a factor of $z^4+2z^2+1$ . Solve $z^4+2z^2+1=0$ , stating multiplicities.

Plan. Divide out the given factor and check whether it appears again.

\begin{aligned} z^4+2z^2+1&=(z^2+1)^2 && \text{divide and recognise a square}\\ &=(z-\mathrm{i})^2(z+\mathrm{i})^2 && \text{factor each }z^2+1\\ z^2+1=0&\implies z=\pm\mathrm{i} && \text{solve the factor} \end{aligned}

Answer. $z=\mathrm{i}$ and $z=-\mathrm{i}$ , each of multiplicity $2$ . There are two distinct roots and four roots counted with multiplicity.

Check. Squaring $z^2+1$ recovers the polynomial. At either root $z^2=-1$ , so the original expression is $1-2+1=0$ . The displayed linear factor powers confirm the multiplicities.

Keep the supplied factor $z^2+1$ , but change the quotient from $z^2+1$ to $z^2$ . This replaces the repeated non-real pair with a repeated zero root.

Worked example 6: retaining zero roots

Question. Solve $z^4+z^2=0$ , stating multiplicities.

Plan. Factor out $z^2$ , retaining the values that make it zero.

\begin{aligned} z^4+z^2&=z^2(z^2+1) && \text{take out the common factor}\\ &=z^2(z-\mathrm{i})(z+\mathrm{i}) && \text{factor the quadratic}\\ z^2=0\quad\text{or}\quad z^2+1&=0 && \text{zero-product rule} \end{aligned}

Answer. $z=0$ , of multiplicity $2$ , and $z=\mathrm{i},-\mathrm{i}$ , each of multiplicity $1$ .

Check. Zero gives $0+0=0$ , while either non-real root gives $1-1=0$ . The multiplicities add to the quartic's degree: $2+1+1=4$ .

For the factored cubics and quartics here, the linear factor degrees account for the whole polynomial. This is a useful completeness check, but counting alone cannot prove that an incorrectly calculated value is a root: also check the factorisation or substitute back. Non-real pairs in a real-coefficient polynomial have matching multiplicities, because conjugating each repeated factor preserves its exponent.

Watch out

Wrong: divide $z^4+z^2=0$ by $z^2$ and report only $z=\pm\mathrm{i}$ . That division assumed $z\ne0$ . Correct: factor first, so the repeated root $z=0$ remains visible.

Find both the values and their multiplicities in this fresh case.

Quick Recap

A root $r$ gives a factor $z-r$ .
With real coefficients, a non-real root supplies its conjugate and a real quadratic factor.
Divide a cubic or quartic by the known factor, then solve every remaining factor.
Retain zero roots, repeated factors and the roots of factors already divided out.

One supplied root can reveal its partner and reduce the degree enough to solve the whole equation.

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B1, B3 - Complex roots of cubic and quartic equations

Conjugate roots and real coefficients

Factoring a cubic from a known root

Quartics from a complex root or quadratic factor

Repeated roots and complete root sets

Quick Recap

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