Question

If alpha andd beta are zeroes of a polynomial x2+6x+9 then form a polynomial whose zeroes are - alpha and -beta

Answer 1

Answer:

Polynomial = x² - 6x + 9.

Step-by-step explanation:

ATQ,

p(x) = x² + 6x + 9.

And the zeroes of another polynomial are -α and -β.

By factoring p(x) = x² + 6x + 9, we can derive values for α and β.

$\Longrightarrow \ \sf x^2 + 6x + 9\\ \\ \\ \boxed{\begin{minipage}{4 cm}\sf \ \ \ \ \ \ \ \ \ \ Sum \longrightarrow 6\\ \\\sf {\ \ \ \ \ \ \ \ Product \longrightarrow 9}\\ \\\sf {\ \ \ \ \ \ \ \ Split \longrightarrow 3 \times 3}\end{minipage}}\\ \\ \\ \Longrightarrow \ \sf x^2 + 3x + 3x + 9\\ \\\Longrightarrow \ \sf x(x + 3) + 3(x + 3)\\ \\\Longrightarrow \ \sf (x + 3) \ (x + 3)$

∴ x + 3 = 0

  x = -3

∴ x + 3 = 0

  x = -3

Considering 'α' and 'β' to be the zeroes we get,

α = -3, β = -3

General form of a quadratic equation:

$\boxed{\sf {k(x^{2} - (\alpha + \beta)x + \alpha \beta)}}$

$\sf Polynomial \ = \ k(x^2 - (-\alpha + -\beta)x \ + \ - \alpha \times -\beta)\\ \\\sf Polynomial \ = \ k(x^2 - (-(-3) + -(-3))x \ + \ -(-3) \times -(-3))\\ \\\sf Polynomial \ = \ k(x^2 - (3+3)x \ + 3 \times 3)\\ \\\sf Polynomial \ = \ k(x^2 - 6x + 9)\\ \\\sf If \ k = 1,\\ \\\sf \underline{\underline{Polynomial \ = \ x^2 - 6x + 9\\ \\}}$