### Sums and products of roots

Suppose we have a monic cubic polynomial $$f(x)$$ with roots 2, 5, 7:

$f(x) = (x-2)(x-5)(x-7).$

If we expand out these brackets, we see something interesting:

$f(x) = x^3 - (2+5+7) x^2 + (2 \cdot 5 + 2 \cdot 7 + 5 \cdot 7) x - 2 \cdot 5 \cdot 7.$

Each coefficient is written in terms of the roots. In particular, the coefficient of $$x^2$$ is the (negative) sum of the roots, and the constant term is the (negative) product of the roots.

More generally, suppose we have a cubic polynomial $$f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$$ with roots $$\alpha, \beta, \gamma$$. Then

$f(x) = c (x-\alpha) (x-\beta) (x-\gamma)$

for some real number constant $$c$$. Expanding this out gives

$f(x) = c \big[ x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) x - \alpha \beta \gamma \big].$

Comparing the two expressions for $$f(x)$$, we obtain

$a_3 = c, \qquad a_2 = -c(\alpha + \beta + \gamma), \qquad a_1 = c(\alpha \beta + \beta \gamma + \gamma \alpha), \qquad a_0 = - c \alpha \beta \gamma.$

From these equalities we obtain the following theorem, relating the sums and products of the roots to the coefficients of $$f(x)$$.

###### Theorem (Vieta's formulas for cubics)

Let $$f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$$ be a cubic polynomial with roots $$\alpha, \beta, \gamma$$. Then

$\alpha + \beta + \gamma = - \frac{a_2}{a_3}, \qquad \alpha \beta + \beta \gamma + \gamma \alpha = \frac{a_1}{a_3}, \qquad \alpha \beta \gamma = - \frac{a_0}{a_3}.$
###### Technical notes.
1. The roots of $$f(x)$$ referred to in this theorem are all the complex number roots of $$f(x)$$. If we restrict attention to real roots, the result is not true.
2. In the theorem we must take all the roots of $$f(x)$$ with multiplicities.

#### Example

Let $$f(x) = 2x^3 + 3x^2 - 4x + 7$$.

1. What is the sum of the roots of $$f(x)$$?
2. What is the product of the roots of $$f(x)$$?

#### Solution

By Vieta's formulas, the sum of the roots is $$-\frac{3}{2}$$ and the product of the roots is $$-\frac{7}{2}$$.

(We can also deduce that, if the three roots of $$f(x)$$ are $$\alpha, \beta, \gamma$$, then $$\alpha \beta + \beta \gamma + \gamma \alpha = -2$$.)

Vieta's formulas apply not just to cubics but to polynomials of any degree. For instance, consider the quartic polynomial

$f(x) = 2(x-1)(x-3)(x-6)(x-7).$

Expanding out this expression gives

\begin{align*} f(x) = 2 \Big( x^4 &- \big(1+3+6+7\big) x^3 + \big(1 \cdot 3 + 1 \cdot 6 + 1 \cdot 7 + 3 \cdot 6 + 3 \cdot 7 + 6 \cdot 7 \big) x^2 \\ &- \big(1 \cdot 3 \cdot 6 + 1 \cdot 3 \cdot 7 + 1 \cdot 6 \cdot 7 + 3 \cdot 6 \cdot 7\big) x + 1 \cdot 3 \cdot 6 \cdot 7 \Big). \end{align*}

The coefficient of $$x^3$$ is $$-2(1+3+6+7)$$, which is two (the leading coefficient) times the (negative) sum of the roots. And the constant term is two times the product of the roots. So again the coefficients of $$f(x)$$ can be described in terms of the sums and products of the roots.

For a general polynomial we have the following theorem. The proof is left to you.

###### Theorem (Vieta's formulas)

Let $$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$. Then

1. the sum of the roots of $$f(x)$$, counted with multiplicities, is $$-\dfrac{a_{n-1}}{a_n}$$, and
2. the product of the roots, again counted with multiplicities, is $$(-1)^n \dfrac{a_0}{a_n}$$.

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