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3. Roots of Polynomial Equations0%

Complex Numbers · Topic 3 of 5

3. Roots of Polynomial Equations

Video coming soon3 worked examples

Theory

The Fundamental Theorem of Algebra guarantees that every polynomial equation of degree nn has exactly nn roots in the complex numbers (counting repeats). Some may be real, some may not — but the count is always nn.

When the coefficients are real, one extra fact does most of the work: complex roots always occur in conjugate pairs. If x+yix + yi is a root, so is x−yix - yi. Multiplying their two factors together produces a quadratic with real coefficients:

(z−(x+yi))(z−(x−yi))=z2−2xz+(x2+y2)\bigl(z - (x+yi)\bigr)\bigl(z - (x-yi)\bigr) = z^2 - 2xz + (x^2 + y^2)

That gives a reliable method: find one root, pair it with its conjugate, form the real quadratic factor, divide, and solve what's left.

A useful consequence is that you can predict the shape of the answer. A cubic with real coefficients has either three real roots, or one real root and one conjugate pair — it can never have exactly two non-real roots and one non-real left over, because they must pair up.

The Golden Rule: the conjugate-pair result holds only when every coefficient is real. Check that first; if the equation contains an ii in its coefficients, the pairing does not apply and you must solve directly.

⚠️ Common Examiner Traps

  • Forgetting the real root: a cubic has three roots. Finding the conjugate pair is only two thirds of the answer.
  • Sign slip forming the quadratic: the middle coefficient is −2x-2x — twice the real part, negated — and the constant is x2+y2x^2 + y^2, a sum, not a difference.
  • Verifying a root: if asked to verify, substitute and show the result is 00; compute the powers of (x+yi)(x+yi) step by step, and show the real and imaginary parts cancelling separately.
  • Dividing carelessly: after dividing by the quadratic factor the remainder must be exactly zero. If it isn't, the root or the factor is wrong — go back rather than pressing on.

Worked examples

Example 1

Find all the roots of z3−4z2+14z−20=0z^3 - 4z^2 + 14z - 20 = 0.

Step 1: Look for a real root among the factors of the constant term 2020. Trying z=2z = 2:

8−16+28−20=0 ✓8 - 16 + 28 - 20 = 0 \ \checkmark

So z=2z = 2 is a root and (z−2)(z - 2) is a factor.

Step 2: Divide to find the remaining quadratic factor:

z3−4z2+14z−20=(z−2)(z2−2z+10)z^3 - 4z^2 + 14z - 20 = (z-2)(z^2 - 2z + 10)

Step 3: Solve the quadratic. The discriminant is negative, so the remaining roots are non-real:

z=2±4−402=2±−362=2±6i2=1±3iz = \frac{2 \pm \sqrt{4 - 40}}{2} = \frac{2 \pm \sqrt{-36}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i

Step 4: The three roots are z=2z = 2, z=1+3iz = 1 + 3i and z=1−3iz = 1 - 3i — one real root and a conjugate pair, exactly as expected for a real cubic.

Example 2

Verify that z=2+iz = 2 + i is a root of z4−2z3−6z2+22z−15=0z^4 - 2z^3 - 6z^2 + 22z - 15 = 0, and hence find all the roots.

Step 1: Build up the powers of 2+i2+i one at a time:

(2+i)2=4+4i+i2=3+4i(2+i)3=(3+4i)(2+i)=6+3i+8i+4i2=2+11i(2+i)4=(3+4i)2=9+24i+16i2=−7+24i\begin{aligned} (2+i)^2 &= 4 + 4i + i^2 = 3 + 4i \\ (2+i)^3 &= (3+4i)(2+i) = 6 + 3i + 8i + 4i^2 = 2 + 11i \\ (2+i)^4 &= (3+4i)^2 = 9 + 24i + 16i^2 = -7 + 24i \end{aligned}

Step 2: Substitute into the equation and collect real and imaginary parts separately:

(−7+24i)−2(2+11i)−6(3+4i)+22(2+i)−15(-7+24i) - 2(2+11i) - 6(3+4i) + 22(2+i) - 15
real:−7−4−18+44−15=0imaginary:24−22−24+22=0\begin{aligned} \text{real}: &\quad -7 - 4 - 18 + 44 - 15 = 0 \\ \text{imaginary}: &\quad 24 - 22 - 24 + 22 = 0 \end{aligned}

Both parts are zero, so z=2+iz = 2+i is a root.

Step 3: The coefficients are real, so z=2−iz = 2 - i is also a root. Their combined factor is:

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

Step 4: Divide the quartic by this quadratic:

z4−2z3−6z2+22z−15=(z2−4z+5)(z2+2z−3)z^4 - 2z^3 - 6z^2 + 22z - 15 = (z^2 - 4z + 5)(z^2 + 2z - 3)

Step 5: Factorise the remaining quadratic:

z2+2z−3=(z+3)(z−1)  ⟹  z=−3  or  z=1z^2 + 2z - 3 = (z+3)(z-1) \implies z = -3 \ \text{ or } \ z = 1

Step 6: The four roots are 2+i2 + i, 2−i2 - i, 11 and −3-3.

Example 3

A polynomial equation of degree 4 has real coefficients. Two of its roots are 3i3i and 1−i1 - i. Write down the other two roots and hence find the equation.

Step 1: Because the coefficients are real, each given root brings its conjugate with it:

the other roots are  −3i  and  1+i\text{the other roots are } \ -3i \ \text{ and } \ 1 + i

Step 2: Pair each root with its conjugate to form two real quadratic factors. For ±3i\pm 3i we have x=0x = 0, y=3y = 3:

z2−2(0)z+(0+9)=z2+9z^2 - 2(0)z + (0 + 9) = z^2 + 9

Step 3: For 1±i1 \pm i we have x=1x = 1, y=1y = 1:

z2−2(1)z+(1+1)=z2−2z+2z^2 - 2(1)z + (1 + 1) = z^2 - 2z + 2

Step 4: Multiply the two factors together:

(z2+9)(z2−2z+2)=z4−2z3+2z2+9z2−18z+18(z^2+9)(z^2-2z+2) = z^4 - 2z^3 + 2z^2 + 9z^2 - 18z + 18

Step 5: Collect like terms:

z4−2z3+11z2−18z+18=0z^4 - 2z^3 + 11z^2 - 18z + 18 = 0