How many roots does a polynomial of degree n have in the complex numbers?

Exactly n, when n is 1 or more. The book states this as the fundamental theorem of algebra, so the polynomial splits into n linear factors over ℂ.

What is a linear factor?

A factor of the form z − z1, where z1 is a root of the polynomial.

Can z² + 4 be factorised?

Not over the real numbers, but over ℂ it can: z² + 4 = (z + 2i)(z − 2i).

What are the roots of 2z² − 12z + 50 = 0 in book Example 10?

z = 3 + 4i and z = 3 − 4i, found by completing the square.

Which board sets my paper if I study in Gujrat?

Gujrat district is listed under the Board of Intermediate and Secondary Education, Gujranwala. PECTAA writes the books. Your board sets the paper and the dates, so follow its notices.

Polynomial Factors over Complex Numbers: PECTAA Class 11 Maths, Punjab Board

Over the real numbers, z² + 4 has no factors. Once i is allowed, every polynomial breaks into simple pieces. This note shows how, using the book's own examples.

Unit 1 of the PECTAA Mathematics 11 book (Complex Numbers) ends its algebra part with complex polynomials and their linear factors. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Complex polynomial. A polynomial whose coefficients may be complex numbers. A short example is (1 − i)z + 3i, which has degree 1. A polynomial like z² + (i − 3)z − 3i has degree 2.

Root (zero). A value z1 is a root of P(z) when P(z1) = 0. The words root and zero mean the same thing here.

Linear factor. An expression of the form z − z1. If z1 is a root of P, then z − z1 divides P exactly. This is the factor theorem, used here with complex numbers.

Fundamental theorem of algebra (as stated in the book). A polynomial of degree n ≥ 1 has exactly n roots in ℂ, counted with repetition. So it can always be written as

P(z) = a(z − z1)(z − z2)…(z − zn)

where a is the leading coefficient and z1, …, zn are the roots.

Why this matters. Over the real numbers, some quadratics have no roots, so they stay unfactorised. Over ℂ, no polynomial is "stuck". Each degree gives exactly that many linear factors.

Real coefficients give conjugate pairs. When every coefficient is real, the complex roots come in pairs a + bi and a − bi. You can see this in Examples 9 and 10 below: 2 ± i and 3 ± 4i.

Key points

Degree tells you the count. Degree 2 gives two roots, degree 3 gives three, and so on.
Find one root, then divide. For a cubic, test small integers such as ±1, ±2 first. Once a root is found, divide to get a quadratic, then solve the quadratic.
Completing the square works with complex answers. If (z − h)² equals a negative number, take the square root using i.

Worked examples

Example 1 (book Example 6). Factorise z² + (i − 3)z − 3i.

Look for two numbers whose product is −3i and whose sum is i − 3.
The pair i and −3 works: i × (−3) = −3i and i + (−3) = i − 3.
So z² + (i − 3)z − 3i = (z + i)(z − 3).
Check by expanding: z² − 3z + iz − 3i. This matches.

Example 2 (book Example 9). Factorise z³ − 3z² + z + 5 into linear factors.

Try z = −1: (−1)³ − 3(1) + (−1) + 5 = −1 − 3 − 1 + 5 = 0. So z + 1 is a factor.
Divide: z³ − 3z² + z + 5 = (z + 1)(z² − 4z + 5).
Solve z² − 4z + 5 = 0: z = (4 ± √(16 − 20))/2 = (4 ± 2i)/2 = 2 ± i.
So z³ − 3z² + z + 5 = (z + 1)(z − 2 − i)(z − 2 + i).

Example 3 (book Example 10). Solve 2z² − 12z + 50 = 0 by completing the square.

Divide by 2: z² − 6z + 25 = 0.
Move the constant: z² − 6z = −25.
Add 9 to both sides: (z − 3)² = −16.
Take square roots: z − 3 = ±4i, so z = 3 ± 4i.

Example 4 (in the style of the exercise). Factorise z² + 9 over ℂ.

Set z² + 9 = 0, so z² = −9 and z = ±3i.
So z² + 9 = (z + 3i)(z − 3i).
Check: z² − 9i² = z² + 9.

Common mistakes in the exam

Saying "no factors". Over ℂ every polynomial of degree 1 or more has linear factors. Only over the real numbers can a quadratic fail to split.
Sign slips in linear factors. A root 2 + i gives the factor z − (2 + i) = z − 2 − i, not z + 2 + i.
Losing the leading coefficient. In 2z² − 12z + 50 the full factor form is 2(z − 3 − 4i)(z − 3 + 4i). If you divide by 2 to solve, put the 2 back when you write factors.
Stopping after one root of a cubic. A cubic has three roots. Find one, divide, then finish the quadratic.
Writing √(−16) = −4. The square roots of −16 are ±4i.
Forgetting the conjugate. With real coefficients, if 2 + i is a root, 2 − i is a root too.

Exam technique

State the root you tested and show the substitution giving 0. The marker wants to see why z + 1 is a factor.
Show the division or the comparison of coefficients in one or two lines.
Always expand your final factors once to check. It takes half a minute and catches sign slips.
For completing the square, write each line separately as in Example 3.
Check your board's notice for the sections in your paper.

Practice questions

Factorise z² + 16 over ℂ. Answer: z = ±4i, so z² + 16 = (z + 4i)(z − 4i).
Solve z² − 2z + 5 = 0. Answer: (z − 1)² = −4, so z = 1 ± 2i.
Show that z − 1 is a factor of z³ − z² + 4z − 4 and find the other factors. Answer: At z = 1 the value is 1 − 1 + 4 − 4 = 0. Dividing gives z² + 4 = (z + 2i)(z − 2i). So the factors are (z − 1)(z + 2i)(z − 2i).
How many roots in ℂ does a polynomial of degree 6 have? Answer: Exactly 6, counted with repetition.
One root of a quadratic with real coefficients is 5 − 2i. Write the other root and the quadratic with leading coefficient 1. Answer: The other root is 5 + 2i. Sum = 10 and product = 25 + 4 = 29, so z² − 10z + 29.

Ready to practise this? Continue in 1st Year Mathematics.