Factorise polynomials over the complex numbers

What complete factorisation means

The factor theorem works over C: P(α)=0 exactly when z−α divides P(z). A root has multiplicity m when (z−α)^m divides P but (z−α)^(m+1) does not. Complete factorisation over C uses linear factors, their multiplicities and the original nonzero leading coefficient. A list of roots without the leading scalar does not specify the polynomial

Use the fundamental theorem with its hypotheses

Every nonconstant polynomial with complex coefficients has a complex root. Repeated division by root factors gives exactly n roots counting multiplicity for a nonzero degree-n polynomial. This is an informal use of the fundamental theorem of algebra, not a proof of it. The n roots need not be distinct. A nonzero constant has no roots; the zero polynomial vanishes everywhere and is excluded from this finite count

A degree-five polynomial has one root of multiplicity three and one root of multiplicity two. How many distinct roots does it have?

Apply the conjugate theorem only when justified

If P has real coefficients, P(conjugate z)=conjugate(P(z)). Therefore a nonreal root a+bi has the conjugate a−bi with the same multiplicity. A supplied root without an exact multiplicity does not establish that either root is simple. A real root is its own conjugate and does not force another root

Do not infer the coefficients from an unnormalised root set

The polynomial i(z²+1) has the conjugate roots ±i but nonreal coefficients. A monic polynomial whose roots are closed under conjugation with matching multiplicities does have real coefficients. Multiplying it by a nonreal scalar preserves its roots and can destroy real coefficients. The real-coefficient theorem does not apply automatically to an arbitrary complex-coefficient polynomial

Use a conjugate pair as a quadratic factor

For real P and a known nonreal root a+bi, b≠0, the monic real quadratic Q(z)=(z−a)²+b² divides P. Divide by Q or compare all coefficients to find the residual. Its degree drops by two; solve that residual before claiming the root list is complete. The earlier Cartesian lesson provides the exact quadratic and conjugate-factor practice

Work a quartic with two nonreal pairs

Let P(z)=z⁴−2z³+4z²−6z+35 and suppose −1+2i is a root. Its conjugate is −1−2i, so Q(z)=z²+2z+5. Division gives z²−4z+7=(z−2)²+3. Hence the remaining roots are 2±√3 i. All four values are distinct and simple

Polynomial

\(\displaystyle \begin{aligned}P(z)&=z^4-2z^3+4z^2\\&\quad-6z+35\end{aligned}\)

Known root

\(\displaystyle P(-1+2i)=0\)

Give the linear coefficient of the residual monic quadratic after removing the conjugate-pair factor

Write the full factorisation over C

The real quadratic factors are useful intermediate expressions, but neither is linear over C. Split both using the four root values. The leading coefficient here is 1. If the original polynomial were −3P(z), retain −3 in front of the same four factors

A residual quadratic can have real roots

For P(z)=z⁴+2z³−z²+16z+10, the known root 1+2i gives Q=z²−2z+5. The quotient is z²+4z+2=(z+2)²−2, so the other roots are −2±√2. The full root set has two nonreal and two real values

Polynomial

\(\displaystyle \begin{aligned}P(z)&=z^4+2z^3-z^2\\&\quad+16z+10\end{aligned}\)

Known root

\(\displaystyle P(1+2i)=0\)

Give the centre of the two residual real roots after removing the conjugate-pair factor

When the residual discriminant is zero

The polynomial z⁴+2z³−2z²−6z+5 factors as (z²+4z+5)(z−1)². Thus −2±i are simple roots and 1 is a double root. There are three distinct points and four roots counting multiplicity. Zero discriminant in the residual does not repeat the nonreal pair

How should this quartic be represented in the Argand plane?

Read repeated quadratic factors

For P(z)=(z²−2z+5)³, first factor z²−2z+5=(z−1−2i)(z−1+2i). Cubing the product gives each linear factor exponent three. There are two distinct roots, each of multiplicity three, and total multiplicity six

How many distinct root values are shown here?

Complete the square with complex coefficients

Complete the square after dividing by the nonzero leading coefficient. The two square roots are negatives of each other and need not be conjugates

1. 1Normalise The coefficients may be complex \(\displaystyle \begin{gathered}Az^2+Bz+C=0,\quad A\ne0\\z^2+\frac BA z=-\frac CA\end{gathered}\)
2. 2Complete the square Use half the linear coefficient \(\displaystyle \begin{gathered}H=-\frac B{2A}\\z^2-2Hz+H^2=H^2-\frac CA\\(z-H)^2=W,\quad W=H^2-\frac CA\end{gathered}\)
3. 3Solve For a nonzero right side use both square-root signs; a zero right side gives a double root \(\displaystyle \begin{gathered}w^2=W\ne0\quad\Longrightarrow\quad z=H\pm w\\W=0\quad\Longrightarrow\quad z=H\ \text{(multiplicity 2)}\end{gathered}\)

Find a complex square root systematically

Equate the real and imaginary components, then use the modulus to determine a square-root branch

1. 1Component equations The unknown components are real \(\displaystyle \begin{gathered}(u+vi)^2=X+Yi\\u^2-v^2=X,\quad2uv=Y\end{gathered}\)
2. 2Modulus Square and add the component equations, then take a nonnegative root \(\displaystyle \begin{gathered}(u^2+v^2)^2=X^2+Y^2\\u^2+v^2=\sqrt{X^2+Y^2}\end{gathered}\)
3. 3Nonreal target Choose the branch with positive real part, then find its imaginary part \(\displaystyle \begin{gathered}Y\ne0\\u=\sqrt{\frac{\sqrt{X^2+Y^2}+X}{2}}>0\\v=\frac{Y}{2u}\end{gathered}\)
4. 4Real target Avoid division by a zero real component; use both signs for a nonzero target \(\displaystyle \begin{gathered}Y=0\\X>0:\ \pm\sqrt X\\X<0:\ \pm i\sqrt{-X}\\X=0:\ 0\end{gathered}\)

Work a complex-coefficient quadratic

For z²−(2+2i)z−3+6i=0, completing the square gives (z−1−i)²=3−4i. Since (2−i)²=3−4i, the roots are (1+i)±(2−i), namely 3 and −1+2i. These are not a conjugate pair

Give the imaginary part of the root with smaller real part

Use a substitution without discarding branches

For z⁴+z²−6=0, put y=z². Then y²+y−6=(y−2)(y+3), giving y=2 or −3. Thus z=±√2 or ±√3 i. Both signs are required for each nonzero y-value. Factorisation over C has four linear factors

Multiplicity changes at y=0

For (z²−2)², each of ±√2 is a double root. For z²(z²+3), zero is a double root and ±√3 i are simple. For z⁴=(z²)², zero is the only root and has multiplicity four. A simple zero of the y-polynomial becomes a double zero in z because y=z²

Polynomial

\(\displaystyle z^2(z^2+3)\)

What is the multiplicity of zero?

Factor a polynomial with complex coefficients, ungraded

Given P(z)=z³−(2+3i)z²+(−5+8i)z+6+3i and the root i, factor P completely over C. Show the division or coefficient comparison, then solve the residual quadratic. Explain why conjugating i is not a valid way to assert another root

Model: use the stated root, then solve the residual

Use the supplied root to extract a linear factor, compare every coefficient and solve the remaining quadratic over the complex numbers

1. 1Given polynomial The known root supplies the first factor \(\displaystyle \begin{aligned}P(z)&=z^3-(2+3i)z^2\\&\quad+(-5+8i)z+6+3i\\P(i)&=0\end{aligned}\)
2. 2Set the residual Expand the product before comparing coefficients \(\displaystyle \begin{aligned}P(z)&=(z-i)(z^2+Bz+C)\\&=z^3+(B-i)z^2\\&\quad+(C-iB)z-iC\end{aligned}\)
3. 3Find the coefficients Match the quadratic and linear coefficients \(\displaystyle \begin{gathered}B-i=-2-3i\\B=-2-2i\\C-iB=-5+8i\\C=(-5+8i)+(2-2i)=-3+6i\end{gathered}\)
4. 4Check the constant The remaining coefficient also agrees \(\displaystyle -iC=-i(-3+6i)=6+3i\)
5. 5Complete the square Use the residual coefficients \(\displaystyle \begin{gathered}z^2-(2+2i)z-3+6i=0\\(z-1-i)^2=(1+i)^2+3-6i\\=3-4i\end{gathered}\)
6. 6Find a square root Use positive real part for one branch \(\displaystyle \begin{gathered}u^2-v^2=3,\quad2uv=-4\\u^2+v^2=\sqrt{3^2+(-4)^2}=5\\2u^2=8,\quad u=2,\quad v=-1\end{gathered}\)
7. 7Complete factorisation Both square-root signs give distinct residual roots \(\displaystyle \begin{gathered}z=(1+i)\pm(2-i)\\z=3\quad\text{or}\quad z=-1+2i\\P(z)=(z-i)(z-3)(z+1-2i)\end{gathered}\)
8. 8Conjugation does not apply The coefficients are not all real, and the conjugate of the supplied root is not a root \(\displaystyle \begin{aligned}P(-i)&=(-2i)(-3-i)(1-3i)\\&=16+12i\ne0\end{aligned}\)

Prove the conjugate theorem and multiplicity, ungraded

For a nonzero real-coefficient polynomial P, prove P(conjugate z)=conjugate(P(z)). Then prove that a root α and its conjugate have the same exact multiplicity. State what changes when α is real and why a known root alone does not specify its multiplicity

Model: conjugate the factorisation as well as the value

Conjugation preserves sums and products and fixes every real coefficient, giving the value identity term by term. If P(z)=(z−α)^m Q(z) with Q(α)≠0, define Q*(z) by conjugating each coefficient of Q. Conjugating the coefficient identity gives P(z)=(z−conjugate α)^m Q*(z). Moreover Q*(conjugate α)=conjugate(Q(α))≠0, so the multiplicity is exactly m. For real α the two roots coincide; no extra distinct value follows

Explain real and complex factorisation, ungraded

Explain why a real-coefficient polynomial can be factored over R into linear factors and irreducible quadratic factors, and why a real polynomial of odd degree has a real root. Give the leading-scalar condition for the converse from conjugate-paired roots to real coefficients

Model: organise the complete complex factorisation

Use the complete complex factorisation and pair each nonreal root with its conjugate of equal multiplicity

1. 1Conjugate factor A real root supplies a real linear factor; a nonreal pair supplies this real quadratic \(\displaystyle \begin{gathered}a,b\in\mathbb R,\quad b\ne0\\(z-a-bi)(z-a+bi)\\=(z-a)^2+b^2\end{gathered}\)
2. 2Irreducibility over the reals The discriminant is strictly negative \(\displaystyle \Delta=(-2a)^2-4(a^2+b^2)=-4b^2<0\)
3. 3Multiplicity and odd degree A nonreal pair contributes an even total multiplicity, so an odd degree leaves at least one real root \(\displaystyle \bigl((z-a)^2+b^2\bigr)^m\quad\text{contributes }2m\)
4. 4Converse A conjugation-invariant root multiset gives a monic real polynomial. The leading scalar must also be real and nonzero to preserve real coefficients \(\displaystyle P(z)=A\prod_j(z-\alpha_j)^{m_j},\quad A\in\mathbb R\setminus\{0\}\)

Factor and sketch a higher-degree example, ungraded

Factor z⁸−16 over R and then completely over C. List every distinct root with its multiplicity and sketch them with equal Argand scales. Explain how the total multiplicity checks the degree. Use the established polar-root method where helpful

Model: combine algebra and the root geometry

Repeated difference of squares gives (z²−2)(z²+2)(z⁴+4). The last factor is (z²−2z+2)(z²+2z+2). The roots are ±√2, ±√2 i, and all four values ±1±i. Each is simple; they are eight equally spaced points of radius √2, with spacing π/4. Over C use the eight corresponding (z−root) factors. Over R keep z²+2 and the last two quadratics irreducible. The entire learner sketch remains ungraded

Check a repeated-root polynomial with technology, ungraded

For (z²−2z+5)³, compute approximate roots and residuals, compare the total multiplicity with the degree, and explain why a numerical root list alone may not reveal exact repetition. Check the exact quadratic factor separately and retain your own technology output

Model: exact factors establish repetition

Exact factorisation gives 1±2i with exponent three each. Substitution makes P and its first two derivatives vanish at both roots, while the third derivative is nonzero there. A numerical solver may split a repeated value into nearby approximations; small residuals do not establish exact multiplicity or a complete root list. Use the exact factors and degree count to interpret the output. This model does not establish that the learner ran the calculation

Factor values and complete explanations

Find factor coefficients, root components, multiplicities, theorem conditions and biquadratic groups. Use quadratic, conjugate and polar-root methods where needed. General proofs, arbitrary complex-polynomial division, unrestricted root-set entry, complete sketches and technology execution remain ungraded