Question

@ParthKohli Vieta's Formula

Answer 1

.

Answer 2

Hello.

Answer 3

Let's first think about a quadratic-polynomial.\[ax^2 + bx + c = a\left(x - \alpha\right)\left( x - \beta \right) \]\(\alpha, \beta\) are the roots of the polynomial. Now if we expand the right-hand-side,\[ax^2 + bx + c = ax^2 - a(\alpha + \beta )x + a \alpha \beta \]Comparing the coefficients,\[\alpha + \beta = -\frac{b}{a}\]\[\alpha \beta = \frac{c}{a}\]

Answer 4

not \(\alpha\beta = \frac{1}{a}\) ?

Answer 5

no

Answer 6

Now let's think about a cubic.\[ax^3 +bx^2 + cx + d = a(x- \alpha) (x - \beta ) (x - \gamma)\]\[= ax^3 - a(\alpha + \beta + \gamma) x^2+ a( \alpha \beta + \beta \gamma + \gamma \alpha) x - a(\alpha \beta \gamma ) \]

Answer 7

And this would work the same way