# Let P(z) = z⁴ + az³ + bz² + cz + d be a quartic polynomial with real coefficients. Let two of the roots of P(z) = 0 be 2 – i and -1 + 2i. Find a, b, c and d.

This is a question from a FP1 paper. Here i denotes √-1.

Fact that should be burned into your soul: **‘Complex roots of a polynomial equation with real coefficients form conjugate pairs’**. This tells you if z = x + yi is a root then so is its conjugate z* = x – yi. Using this fact, we can deduce that 2 + i and -1 – 2i are also roots of P(z) = 0.

So we now have 4 roots of our **quartic **equation P(z) = 0, so that’s all of them! We can now employ the **factor theorem** that you (probably) met in C1. Remember, this states that if z = α is a root of a polynomial then (z – α) is a factor of that polynomial.

So, since we’re told the leading coefficient of P(z) is 1 we can apply the factor theorem to deduce that

P(z) = (z – (2 - i))(z – (2 + i))(z – (-1 + 2i))(z – (-1 - 2i))

So now all it comes down to is some tedious expansion of brackets and a bit of simplification. It’s a good exercise to build your confidence with complex numbers.

⇒ P(z) = (z – 2 + i)(z – 2 – i)(z + 1 – 2i)(z + 1 + 2i)

⇒ P(z) = (z² - 2z – zi – 2z + 4 + 2i + zi - 2i - i²)(z² + z + 2zi + z + 1 + 2i – 2zi – 2i - 4i²)

⇒ P(z) = (z² - 4z + 5)(z² + 2z + 5) [Remember: i² = - 1 by definition]

⇒ P(z) = z⁴ + 2z³ + 5z² - 4z³ - 8z² - 20z + 5z² + 10z + 25

⇒ P(z) = z⁴ - 2z³ + 2z² - 10z + 25

Thus a = -2, b = 2, c = -10, d = 25 and we’re done.