2.7.4 Roots of Polynomials

Suppose $P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ is a polynomial of degree n with n roots $α_{1}, α_{2}, . . ., α_{n}$
- The polynomial is written as $\sum_{r = 0}^{n} a_{r} x^{r} = 0, a_{n} \neq 0$ in the formula booklet
- a_n is the coefficient of the leading term
- a_n_-1 is the coefficient of the xⁿ^-1 term
  - Be careful: this could be equal to zero
- a₀is the constant term
  - Be careful: this could be equal to zero
In factorised form: $P (x) = a_{n} (x - α_{1}) (x - α_{2}) . . . (x - α_{n})$
- Comparing coefficients of the xⁿ^-1 term and the constant term gives
  - $a_{n - 1} = a_{n} (- α_{1} - α_{2} - . . . - α_{n})$
  - $a_{0} = a_{n} (- α_{1}) \times (- α_{2}) \times . . . \times (- α_{n})$
The sum of the roots is given by:
- $α_{1} + α_{2} + \dots + α_{n} = - \frac{a_{n - 1}}{a_{n}}$
The product of the roots is given by:
- $α_{1} \times α_{2} \times \dots \times α_{n} = \frac{{(- 1)}^{n} a_{0}}{a_{n}}$
  - both of these formulae are in your formula booklet

If you know a complex root of a real polynomial then its complex conjugate is another root
Form two equations using the roots

Examiners might trick you by not having an x^n-1 term or a constant term
To make sure you do not get tricked you can write out the full polynomial using 0 as a coefficient where needed
- For example: Write $x^{4} + 2 x^{2} - 5 x$ as $x^{4} + 0 x^{3} + 2 x^{2} - 5 x + 0$

$2 - 3 i$ , $\frac{5}{3} i$ and $α$ are three roots of the equation $18 x^{5} - 9 x^{4} + 32 x^{3} + 794 x^{2} - 50 x + k = 0$ .

Use the sum of all the roots to find the value of

α

2-7-4-ib-aa-hl-sum-product-roots-a-we-solution

Use the product of all the roots to find the value of

k

2-7-4-ib-aa-hl-sum-product-roots-b-we-solution

DP IB Maths: AA HL

2.7.4 Roots of Polynomials